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, 654 geometric series, 669 graph hole in, 305 horizontal scaling, 132 horizontal shift, 123 of a function, 93 of a relation, 20 of an equation, 23 rational function, 321 reﬂection about an axis, 126 transformations, 135 vertical scaling, 130 vertical shift, 121 greatest integer function, 67 growth model limited, 475 logistic, 475 uninhibited, 472 guide rectangle for a hyperbola, 532 for an ellipse, 519 Half-Angle Formulas, 779 harmonic motion, 885 Henderson-Hasselbalch Equation, 446 Heron’s Formula, 914 hole in a graph, 305 location of, 306 Hooke’s Law, 350 horizontal asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 308 horizontal line, 23 Horizontal Line Test (HLT), 381 hyperbola asymptotes, 531 branch, 531 center, 531 conjugate axis, 532 deﬁnition of, 531 foci, 531 from slicing a cone, 496 guide rectangle, 532 standard equation horizontal, 534 vertical, 534 transverse axis, 531 vertices, 531 hyperbolic cosine, 1062 hyperbolic sine, 1062 hyperboloid, 542 identity function, 367 matrix, additive, 579 Index 1075 matrix, multiplicative, 585 statement which is always true, 549 imaginary axis, 991 imaginary part of a complex number, 991 imaginary unit, i, 287 implied domain of a function, 58 inconsistent system, 553 increasing function interval deﬁnition of, 3 notation for, 3 notation, extended, 756 inverse matrix, additive, 579, 581 matrix, multiplicative, 602 of a function formal deﬁnition of, 101 intuitive deﬁnition of, 100 independent system, 554 independent variable, 55 index of a root, 397 induction base step, 673 induction hypothesis, 673 inductive step, 673 inequality absolute value, 211 graphical interpretation, 209 non-linear, 643 quadratic, 215 sign diagram, 214 inﬂection point, 477 information entropy, 477 initial side of an angle, 698 instantaneous rate of change, 161, 472, 707 integer deﬁnition of, 2 greatest
integer function, 67 set of, 2 intercept deﬁnition of, 25 location of, 25 interest compound, 470 compounded continuously, 472 simple, 469 Intermediate Value Theorem polynomial zero version, 241 interrobang, 321 intersection of two sets, 4 deﬁnition of, 379 properties of, 379 solving for, 384 uniqueness of, 380 inverse variation, 350 invertibility function, 382 invertible function, 379 matrix, 602 irrational number deﬁnition of, 2 set of, 2 irreducible quadratic, 291 joint variation, 350 Kepler’s Third Law of Planetary Motion, 355 Kirchhoﬀ’s Voltage Law, 605 latus rectum of a parabola, 507 Law of Cosines, 910 Law of Sines, 897 leading coeﬃcient of a polynomial, 236 leading term of a polynomial, 236 Learning Curve Equation, 315 least squares regression line, 225 lemniscate, 950 lima¸con, 950 line horizontal, 23 least squares regression, 225 linear function, 156 of best ﬁt, 225 parallel, 166 1076 Index perpendicular, 167 point-slope form, 155 slope of, 151 slope-intercept form, 155 vertical, 23 linear equation n variables, 554 two variables, 549 linear function, 156 local maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 local minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 logarithm algebraic properties of, 438 change of base formula, 442 common, 422 general, “base b”, 422 graphical properties of, 423 inverse properties of, 437 natural, 422 one-to-one properties of, 437 solving equations with, 459 logarithmic scales, 431 logistic growth, 475 LORAN, 538 lower triangular matrix, 593 main diagonal, 585 major axis of an ellipse, 516 Markov Chain, 592 mathematical model, 60 matrix addition associative property, 579 commutative property, 579 deﬁnition of, 578 properties of, 579 additive identity, 579 additive inverse, 579 adjoint, 622 augmented, 568 characteristic polynomial, 626 cofactor, 616 de�
��nition, 567 determinant deﬁnition of, 614 properties of, 616 dimension, 567 entry, 567 equality, 578 invertible, 602 leading entry, 569 lower triangular, 593 main diagonal, 585 matrix multiplication associative property of, 585 deﬁnition of, 584 distributive property, 585 identity for, 585 properties of, 585 minor, 616 multiplicative inverse, 602 product of row and column, 584 reduced row echelon form, 570 rotation, 986 row echelon form, 569 row operations, 568 scalar multiplication associative property of, 581 deﬁnition of, 580 distributive properties, 581 identity for, 581 properties of, 581 zero product property, 581 size, 567 square matrix, 586 sum, 578 upper triangular, 593 maximum formal deﬁnition of, 102 intuitive deﬁnition of, 101 measure of an angle, 693 1077 Index midpoint deﬁnition of, 12 midpoint formula, 13 minimum formal deﬁnition of, 102 intuitive deﬁnition of, 101 minor, 616 minor axis of an ellipse, 516 model mathematical, 60 modulus of a complex number deﬁnition of, 991 properties of, 993 multiplicity eﬀect on the graph of a polynomial, 245, 249 of a zero, 244 natural base, 420 natural logarithm, 422 natural number deﬁnition of, 2 set of, 2 negative angle, 698 Newton’s Law of Cooling, 421, 474 Newton’s Law of Universal Gravitation, 351 oblique asymptote, 311 obtuse angle, 694 odd function, 95 Ohm’s Law, 350, 605 one-to-one function, 381 ordered pair, 6 ordinary frequency, 708 ordinate, 6 orientation, 1048 oriented angle, 697 oriented arc, 704 origin, 7 orthogonal projection, 1038 orthogonal vectors, 1037 overdetermined system, 554 parabola axis of symmetry, 191 deﬁnition of, 505 directrix, 505 focal diameter, 507 focal length, 506 focus, 505 from slicing a cone, 496 graph of
a quadratic function, 188 latus rectum, 507 reﬂective property, 510 standard equation horizontal, 508 vertical, 506 vertex, 188, 505 vertex formulas, 194 paraboloid, 510 parallel vectors, 1030 parameter, 1048 parametric equations, 1048 parametric solution, 552 parametrization, 1048 partial fractions, 628 Pascal’s Triangle, 688 password strength, 477 period circular motion, 708 of a function, 790 of a sinusoid, 881 periodic function, 790 pH, 432 phase, 795, 881 phase shift, 795, 881 pi, π, 700 piecewise-deﬁned function, 62 point of diminishing returns, 477 point-slope form of a line, 155 polar coordinates conversion into rectangular, 924 deﬁnition of, 919 equivalent representations of, 923 polar axis, 919 pole, 919 1078 Index polar form of a complex number, 995 polar rose, 950 polynomial division dividend, 258 divisor, 258 factor, 258 quotient, 258 remainder, 258 synthetic division, 260 polynomial function completely factored over the complex numbers, 291 over the real numbers, 291 constant term, 236 deﬁnition of, 235 degree, 236 end behavior, 239 leading coeﬃcient, 236 leading term, 236 variations in sign, 273 zero lower bound, 274 multiplicity, 244 upper bound, 274 positive angle, 698 Power Reduction Formulas, 778 power rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 price-demand function, 82 principal, 469 principal nth root, 397 principal argument of a complex number, 991 principal unit vectors, ˆı, ˆ, 1024 Principle of Mathematical Induction, 673 product rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 Product to Sum Formulas, 780 proﬁt function, 82 projection x−axis, 45 y−axis, 46 orthogonal, 1038 Pythagorean Conjugates, 751 Pythagorean Identities, 749 quad
rantal angle, 698 quadrants, 8 quadratic formula, 194 quadratic function deﬁnition of, 188 general form, 190 inequality, 215 irreducible quadratic, 291 standard form, 190 quadratic regression, 228 Quotient Identities, 745 quotient rule for absolute value, 173 for complex numbers, 997 for exponential functions, 437 for logarithms, 438 for radicals, 398 for the modulus of a complex number, 993 radian measure, 701 radical properties of, 398 radicand, 397 radioactive decay, 473 radius of a circle, 498 range deﬁnition of, 45 rate of change average, 160 Index 1079 instantaneous, 161, 472 slope of a line, 154 rational exponent, 398 rational functions, 301 rational number deﬁnition of, 2 set of, 2 Rational Zeros Theorem, 269 ray deﬁnition of, 693 initial point, 693 real axis, 991 Real Factorization Theorem, 292 real number deﬁnition of, 2 set of, 2 real part of a complex number, 991 Reciprocal Identities, 745 rectangular coordinates also known as Cartesian coordinates, 919 conversion into polar, 924 rectangular form of a complex number, 991 recursion equation, 654 reduced row echelon form, 570 reference angle, 721 Reference Angle Theorem for cosine and sine, 722 for the circular functions, 747 reﬂection of a function graph, 126 of a point, 10 regression coeﬃcient of determination, 226 correlation coeﬃcient, 226 least squares line, 225 quadratic, 228 total squared error, 225 relation algebraic description, 23 deﬁnition, 20 Fundamental Graphing Principle, 23 Remainder Theorem, 258 revenue function, 82 Richter Scale, 431 right angle, 694 root index, 397 radicand, 397 Roots of Unity, 1006 rotation matrix, 986 rotation of axes, 974 row echelon form, 569 row operations for a matrix, 568 scalar multiplication matrix associative property of, 581 deﬁnition of, 580 distributive properties of, 581 properties of, 581 vector associative property of, 1018 deﬁnition of, 1017 distributive properties of, 101
8 properties of, 1018 scalar projection, 1039 secant graph of, 800 of an angle, 744, 752 properties of, 802 secant line, 160 sequence nth term, 652 alternating, 652 arithmetic common diﬀerence, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 deﬁnition of, 652 geometric common ratio, 654 deﬁnition of, 654 formula for nth term, 656 sum of ﬁrst n terms, 666 1080 recursive, 654 series, 668 set deﬁnition of, 1 empty, 2 intersection, 4 roster method, 1 set-builder notation, 1 sets of numbers, 2 union, 4 verbal description, 1 set-builder notation, 1 Side-Angle-Side triangle, 910 Side-Side-Side triangle, 910 sign diagram algebraic function, 399 for quadratic inequality, 214 polynomial function, 242 rational function, 321 simple interest, 469 sine graph of, 792 of an angle, 717, 730, 744 properties of, 791 sinusoid amplitude, 794, 881 baseline, 881 frequency angular, 881 ordinary, 881 graph of, 795, 882 period, 881 phase, 881 phase shift, 795, 881 properties of, 881 vertical shift, 881 slant asymptote, 311 slant asymptote determination of, 312 formal deﬁnition of, 311 slope deﬁnition, 151 Index of a line, 151 rate of change, 154 slope-intercept form of a line, 155 smooth, 241 sound intensity level decibel, 431 square matrix, 586 standard position of a vector, 1019 standard position of an angle, 698 start-up cost, 82 steady state, 592 stochastic process, 592 straight angle, 693 Sum Identity for cosine, 771, 775 for sine, 773, 775 for tangent, 775 Sum to Product Formulas, 781 summation notation deﬁnition of, 661 index of summation, 661 lower limit of summation, 661 properties of, 664 upper limit of summation, 661 supplementary angles, 696 symmetry about the x-axis
, 9 about the y-axis, 9 about the origin, 9 testing a function graph for, 95 testing an equation for, 26 synthetic division tableau, 260 system of equations back-substitution, 560 coeﬃcient matrix, 590 consistent, 553 constant matrix, 590 deﬁnition, 549 dependent, 554 free variable, 552 Gauss-Jordan Elimination, 571 Gaussian Elimination, 557 Index 1081 inconsistent, 553 independent, 554 leading variable, 556 linear n variables, 554 two variables, 550 linear in form, 646 non-linear, 637 overdetermined, 554 parametric solution, 552 triangular form, 556 underdetermined, 554 unknowns matrix, 590 tangent graph of, 804 of an angle, 744, 752 properties of, 806 terminal side of an angle, 698 Thurstone, Louis Leon, 315 total squared error, 225 transformation non-rigid, 129 rigid, 129 transformations of function graphs, 120, 135 transverse axis of a hyperbola, 531 Triangle Inequality, 183 triangular form, 556 underdetermined system, 554 uninhibited growth, 472 union of two sets, 4 Unit Circle deﬁnition of, 501 important points, 724 unit vector, 1023 Upper and Lower Bounds Theorem, 274 upper triangular matrix, 593 variable dependent, 55 independent, 55 variable cost, 159 variation constant of proportionality, 350 direct, 350 inverse, 350 joint, 350 variations in sign, 273 vector x-component, 1012 y-component, 1012 addition associative property, 1015 commutative property, 1015 deﬁnition of, 1014 properties of, 1015 additive identity, 1015 additive inverse, 1015, 1018 angle between two, 1035, 1036 component form, 1012 Decomposition Theorem Generalized, 1040 Principal, 1024 deﬁnition of, 1012 direction deﬁnition of, 1020 properties of, 1020 dot product commutative property of, 1034 deﬁnition of, 1034 distributive property of, 1034 geometric interpretation, 1035 properties of, 1034 relation to magnitude, 1034 relation to orthogonality, 1037 work, 1042 head, 1012 initial point, 1012 magnitude deﬁnition of, 10
20 properties of, 1020 relation to dot product, 1034 normalization, 1024 orthogonal projection, 1038 1082 Index multiplicity of, 244 of a function, 95 upper and lower bounds, 274 orthogonal vectors, 1037 parallel, 1030 principal unit vectors, ˆı, ˆ, 1024 resultant, 1013 scalar multiplication associative property of, 1018 deﬁnition of, 1017 distributive properties, 1018 identity for, 1018 properties of, 1018 zero product property, 1018 scalar product deﬁnition of, 1034 properties of, 1034 scalar projection, 1039 standard position, 1019 tail, 1012 terminal point, 1012 triangle inequality, 1044 unit vector, 1023 velocity average angular, 707 instantaneous, 707 instantaneous angular, 707 vertex of a hyperbola, 531 of a parabola, 188, 505 of an angle, 693 of an ellipse, 516 vertical asymptote formal deﬁnition of, 304 intuitive deﬁnition of, 304 location of, 306 vertical line, 23 Vertical Line Test (VLT), 43 whole number deﬁnition of, 2 set of, 2 work, 1041 wrapping function, 704 zero1 2 3 4 x −1 −2 −3 −4 The horizontal number line is usually called the x-axis while the vertical number line is usually called the y-axis.6 As with the usual number line, we imagine these axes extending oﬀ indeﬁnitely in both directions.7 Having two number lines allows us to locate the positions of points oﬀ of the number lines as well as points on the lines themselves. For example, consider the point P on the next page. To use the numbers on the axes to label this point, we imagine dropping a vertical line from the x-axis to P and extending a horizontal line from the y-axis to P. This process is sometimes called ‘projecting’ the point P to the x- (respectively y-) axis. We then describe the point P using the ordered pair (2, −4). The ﬁrst number in the ordered pair is called the abscissa or x-coordinate and the second is called the ordinate or y-coordinate.8 Taken together, the ordered pair (2, −
4) comprise the Cartesian coordinates9 of the point P. In practice, the distinction between a point and its coordinates is blurred; for example, we often speak of ‘the point (2, −4).’ We can think of (2, −4) as instructions on how to 5So named in honor of Ren´e Descartes. 6The labels can vary depending on the context of application. 7Usually extending oﬀ towards inﬁnity is indicated by arrows, but here, the arrows are used to indicate the direction of increasing values of x and y. 8Again, the names of the coordinates can vary depending on the context of the application. If, for example, the horizontal axis represented time we might choose to call it the t-axis. The ﬁrst number in the ordered pair would then be the t-coordinate. 9Also called the ‘rectangular coordinates’ of P – see Section 11.4 for more details. 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 7 reach P from the origin (0, 0) by moving 2 units to the right and 4 units downwards. Notice that the order in the ordered pair is important − if we wish to plot the point (−4, 2), we would move to the left 4 units from the origin and then move upwards 2 units, as below on the right. y 4 3 2 1 (−4, 2) y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 P −1 −2 −3 −4 P (2, −4) When we speak of the Cartesian Coordinate Plane, we mean the set of all possible ordered pairs (x, y) as x and y take values from the real numbers. Below is a summary of important facts about Cartesian coordinates. Important Facts about the Cartesian Coordinate Plane (a, b) and (c, d) represent the same point in the plane if and only if a = c and b = d. (x, y) lies on the x-axis if and only if y = 0. (x, y) lies on the y-axis if and only if x = 0. The origin is the point (0, 0). It is the only point common to both axes. Example 1.1.2. Plot the following points
: A(5, 8), B − 5 F (0, 5), G(−7, 0), H(0, −9), O(0, 0).10 2, 3, C(−5.8, −3), D(4.5, −1), E(5, 0), Solution. To plot these points, we start at the origin and move to the right if the x-coordinate is positive; to the left if it is negative. Next, we move up if the y-coordinate is positive or down if it is negative. If the x-coordinate is 0, we start at the origin and move along the y-axis only. If the y-coordinate is 0 we move along the x-axis only. 10The letter O is almost always reserved for the origin. 8 Relations and Functions y F (0, 5) A(5, 8) O(0, 0) E(5, 0(−7, 0) −9 −8 −7 −6 −5 −4 −3 −2 −(−5.8, −3) D(4.5, −1) −1 −2 −3 −4 −5 −6 −7 −8 −9 H(0, −9) The axes divide the plane into four regions called quadrants. They are labeled with Roman numerals and proceed counterclockwise around the plane: Quadrant II x < 0, y > 0 y 4 3 2 1 Quadrant I x > 0, y > 0 −4 −3 −2 −1 1 2 3 4 x Quadrant III x < 0, y < 0 −1 −2 −3 −4 Quadrant IV x > 0, y < 0 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 9 For example, (1, 2) lies in Quadrant I, (−1, 2) in Quadrant II, (−1, −2) in Quadrant III and (1, −2) in Quadrant IV. If a point other than the origin happens to lie on the axes, we typically refer to that point as lying on the positive or negative x-axis (if y = 0) or on the positive or negative y-axis (if x = 0). For example, (0, 4) lies on the positive y-axis whereas (−117, 0) lies on the negative x-axis. Such points do not belong to any of the four
quadrants. One of the most important concepts in all of Mathematics is symmetry.11 There are many types of symmetry in Mathematics, but three of them can be discussed easily using Cartesian Coordinates. Deﬁnition 1.3. Two points (a, b) and (c, d) in the plane are said to be symmetric about the x-axis if a = c and b = −d symmetric about the y-axis if a = −c and b = d symmetric about the origin if a = −c and b = −d Schematically, Q(−x, y) y 0 P (x, y) x R(−x, −y) S(x, −y) In the above ﬁgure, P and S are symmetric about the x-axis, as are Q and R; P and Q are symmetric about the y-axis, as are R and S; and P and R are symmetric about the origin, as are Q and S. Example 1.1.3. Let P be the point (−2, 3). Find the points which are symmetric to P about the: 1. x-axis 2. y-axis 3. origin Check your answer by plotting the points. Solution. The ﬁgure after Deﬁnition 1.3 gives us a good way to think about ﬁnding symmetric points in terms of taking the opposites of the x- and/or y-coordinates of P (−2, 3). 11According to Carl. Jeﬀ thinks symmetry is overrated. 10 Relations and Functions 1. To ﬁnd the point symmetric about the x-axis, we replace the y-coordinate with its opposite to get (−2, −3). 2. To ﬁnd the point symmetric about the y-axis, we replace the x-coordinate with its opposite to get (2, 3). 3. To ﬁnd the point symmetric about the origin, we replace the x- and y-coordinates with their opposites to get (2, −3). y 3 2 1 (2, 3) P (−2, 3) −3 −2 −1 1 2 3 x −1 −2 −3 (−2, −3) (2, −3) One way to visualize the processes in the previous example is with the concept of a reﬂection
. If we start with our point (−2, 3) and pretend that the x-axis is a mirror, then the reﬂection of (−2, 3) across the x-axis would lie at (−2, −3). If we pretend that the y-axis is a mirror, the reﬂection of (−2, 3) across that axis would be (2, 3). If we reﬂect across the x-axis and then the y-axis, we would go from (−2, 3) to (−2, −3) then to (2, −3), and so we would end up at the point symmetric to (−2, 3) about the origin. We summarize and generalize this process below. Reﬂections To reﬂect a point (x, y) about the: x-axis, replace y with −y. y-axis, replace x with −x. origin, replace x with −x and y with −y. 1.1.3 Distance in the Plane Another important concept in Geometry is the notion of length. If we are going to unite Algebra and Geometry using the Cartesian Plane, then we need to develop an algebraic understanding of what distance in the plane means. Suppose we have two points, P (x0, y0) and Q (x1, y1), in the plane. By the distance d between P and Q, we mean the length of the line segment joining P with Q. (Remember, given any two distinct points in the plane, there is a unique line containing both 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 11 points.) Our goal now is to create an algebraic formula to compute the distance between these two points. Consider the generic situation below on the left. Q (x1, y1) Q (x1, y1) d d P (x0, y0) P (x0, y0) (x1, y0) With a little more imagination, we can envision a right triangle whose hypotenuse has length d as drawn above on the right. From the latter ﬁgure, we see that the lengths of the legs of the triangle are \|x1 − x0\| and \|y1 − y0\| so the Pythagorean Theorem gives us \|x1 − x0\|2 + \|y1 − y0\|2 =
d2 (x1 − x0)2 + (y1 − y0)2 = d2 (Do you remember why we can replace the absolute value notation with parentheses?) By extracting the square root of both sides of the second equation and using the fact that distance is never negative, we get Equation 1.1. The Distance Formula: The distance d between the points P (x0, y0) and Q (x1, y1) is: d = (x1 − x0)2 + (y1 − y0)2 It is not always the case that the points P and Q lend themselves to constructing such a triangle. If the points P and Q are arranged vertically or horizontally, or describe the exact same point, we cannot use the above geometric argument to derive the distance formula. It is left to the reader in Exercise 35 to verify Equation 1.1 for these cases. Example 1.1.4. Find and simplify the distance between P (−2, 3) and Q(1, −3). Solution. d = = = So the distance is 3 √ 5. = 3 √ (x1 − x0)2 + (y1 − y0)2 (1 − (−2))2 + (−3 − 3)2 9 + 36 √ 5 12 Relations and Functions Example 1.1.5. Find all of the points with x-coordinate 1 which are 4 units from the point (3, 2). Solution. We shall soon see that the points we wish to ﬁnd are on the line x = 1, but for now we’ll just view them as points of the form (1, y). Visually, y 3 2 1 −1 −2 −3 (3, 2) distance is 4 units 2 3 x (1, y) We require that the distance from (3, 2) to (1, y) be 4. The Distance Formula, Equation 1.1, yields d = (x1 − x0)2 + (y1 − y0)2 (1 − 3)2 + (y − 2)2 4 + (y − 2)2 4 + (y − 2)22 4 = 4 = 42 = 16 = 4 + (y − 2)2 12 = (y − 2)2 √ (y − 2)2 = 12 2 12 √ 3 √ y = 2 ± 2 3 squaring both sides extracting the square
root We obtain two answers: (1, 2 + 2 why there are two answers. √ 3) and (1, 2 − 2 √ 3). The reader is encouraged to think about Related to ﬁnding the distance between two points is the problem of ﬁnding the midpoint of the line segment connecting two points. Given two points, P (x0, y0) and Q (x1, y1), the midpoint M of P and Q is deﬁned to be the point on the line segment connecting P and Q whose distance from P is equal to its distance from Q. 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 13 Q (x1, y1) M P (x0, y0) If we think of reaching M by going ‘halfway over’ and ‘halfway up’ we get the following formula. Equation 1.2. The Midpoint Formula: The midpoint M of the line segment connecting P (x0, y0) and Q (x1, y1) is: M = x0 + x1 2, y0 + y1 2 If we let d denote the distance between P and Q, we leave it as Exercise 36 to show that the distance between P and M is d/2 which is the same as the distance between M and Q. This suﬃces to show that Equation 1.2 gives the coordinates of the midpoint. Example 1.1.6. Find the midpoint of the line segment connecting P (−2, 3) and Q(1, −3). Solution. M = = = x0 + x1, 2 (−2) + 1 2 y0 + y1 2, 3 + (−3 The midpoint is − 1 2, 0. We close with a more abstract application of the Midpoint Formula. We will revisit the following example in Exercise 72 in Section 2.1. Example 1.1.7. If a = b, prove that the line y = x equally divides the line segment with endpoints (a, b) and (b, a). Solution. To prove the claim, we use Equation 1.2 to ﬁnd the midpoint Since the x and y coordinates of this point are the same, we ﬁnd that the midpoint lies on the line y = x, as required. 14 Relations and Functions 1.1.
4 Exercises 1. Fill in the chart below: Set of Real Numbers Interval Notation Region on the Real Number Line {x \| − 1 ≤ x < 5} {x \| − 5 < x ≤ 0} {x \| x ≤ 3} {x \| x ≥ −3} [0, 3) (−3, 3) (−∞, 9) 7 7 2 5 4 In Exercises 2 - 7, ﬁnd the indicated intersection or union and simplify if possible. Express your answers in interval notation. 2. (−1, 5] ∩ [0, 8) 3. (−1, 1) ∪ [0, 6] 4. (−∞, 4] ∩ (0, ∞) 5. (−∞, 0) ∩ [1, 5] 6. (−∞, 0) ∪ [1, 5] 7. (−∞, 5] ∩ [5, 8) In Exercises 8 - 19, write the set using interval notation. 8. {x \| x = 5} 9. {x \| x = −1} 10. {x \| x = −3, 4} 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 15 11. {x \| x = 0, 2} 12. {x \| x = 2, −2} 13. {x \| x = 0, ±4} 14. {x \| x ≤ −1 or x ≥ 1} 15. {x \| x < 3 or x ≥ 2} 16. {x \| x ≤ −3 or x > 0} 17. {x \| x ≤ 5 or x = 6} 18. {x \| x > 2 or x = ±1} 19. {x \| − 3 < x < 3 or x = 4} 20. Plot and label the points A(−3, −7), B(1.3, −2), C(π, √ 10), D(0, 8), E(−5.5, 0), F (−8, 4), G(9.2, −7.8) and H(7, 5) in the Cartesian Coordinate Plane given below9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −3 −4 −5 −6 −7 −8 −9 21. For each point given in Exercise 20 above Identify the quadrant or axis in
/on which the point lies. Find the point symmetric to the given point about the x-axis. Find the point symmetric to the given point about the y-axis. Find the point symmetric to the given point about the origin. 16 Relations and Functions In Exercises 22 - 29, ﬁnd the distance d between the points and the midpoint M of the line segment which connects them. 22. (1, 2), (−3, 5) 23. (3, −10), (−1, 2) 24. 26. 1 2 24 5 √, − 45, 11 −, 5 12, √ √ 28. 2. 19 5 √ 25. − 2 3 27. √ 2, − √ 8, − 12 20, 27. 29. (0, 0), (x, y) 30. Find all of the points of the form (x, −1) which are 4 units from the point (3, 2). 31. Find all of the points on the y-axis which are 5 units from the point (−5, 3). 32. Find all of the points on the x-axis which are 2 units from the point (−1, 1). 33. Find all of the points of the form (x, −x) which are 1 unit from the origin. 34. Let’s assume for a moment that we are standing at the origin and the positive y-axis points due North while the positive x-axis points due East. Our Sasquatch-o-meter tells us that Sasquatch is 3 miles West and 4 miles South of our current position. What are the coordinates of his position? How far away is he from us? If he runs 7 miles due East what would his new position be? 35. Verify the Distance Formula 1.1 for the cases when: (a) The points are arranged vertically. (Hint: Use P (a, y0) and Q(a, y1).) (b) The points are arranged horizontally. (Hint: Use P (x0, b) and Q(x1, b).) (c) The points are actually the same point. (You shouldn’t need a hint for this one.) 36. Verify the Midpoint Formula by showing the distance between P (x1, y1) and M and the distance between M and Q(x2, y2) are both half of the distance between P
and Q. 37. Show that the points A, B and C below are the vertices of a right triangle. (a) A(−3, 2), B(−6, 4), and C(1, 8) (b) A(−3, 1), B(4, 0) and C(0, −3) 38. Find a point D(x, y) such that the points A(−3, 1), B(4, 0), C(0, −3) and D are the corners of a square. Justify your answer. 39. Discuss with your classmates how many numbers are in the interval (0, 1). 40. The world is not ﬂat.12 Thus the Cartesian Plane cannot possibly be the end of the story. Discuss with your classmates how you would extend Cartesian Coordinates to represent the three dimensional world. What would the Distance and Midpoint formulas look like, assuming those concepts make sense at all? 12There are those who disagree with this statement. Look them up on the Internet some time when you’re bored. 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 17 1.1.5 Answers 1. Set of Real Numbers Interval Notation Region on the Real Number Line {x \| − 1 ≤ x < 5} [−1, 5) {x \| 0 ≤ x < 3} {x \| 2 < x ≤ 7} [0, 3) (2, 7] {x \| − 5 < x ≤ 0} (−5, 0] {x \| − 3 < x < 3} (−3, 3) {x \| 5 ≤ x ≤ 7} [5, 7] {x \| x ≤ 3} (−∞, 3] {x \| x < 9} (−∞, 9) {x \| x > 4} (4, ∞) {x \| x ≥ −3} [−3, ∞) 5 3 7 0 3 7 3 9 −1 0 2 −5 −3 5 4 −3 2. (−1, 5] ∩ [0, 8) = [0, 5] 3. (−1, 1) ∪ [0, 6] = (−1, 6] 4. (−∞, 4] ∩ (0, ∞) = (0, 4] 5. (−∞, 0) ∩ [1, 5] = ∅ 6. (−�
�, 0) ∪ [1, 5] = (−∞, 0) ∪ [1, 5] 7. (−∞, 5] ∩ [5, 8) = {5} 8. (−∞, 5) ∪ (5, ∞) 9. (−∞, −1) ∪ (−1, ∞) 10. (−∞, −3) ∪ (−3, 4) ∪ (4, ∞) 11. (−∞, 0) ∪ (0, 2) ∪ (2, ∞) 12. (−∞, −2) ∪ (−2, 2) ∪ (2, ∞) 13. (−∞, −4) ∪ (−4, 0) ∪ (0, 4) ∪ (4, ∞) 18 Relations and Functions 14. (−∞, −1] ∪ [1, ∞) 15. (−∞, ∞) 16. (−∞, −3] ∪ (0, ∞) 17. (−∞, 5] ∪ {6} 18. {−1} ∪ {1} ∪ (2, ∞) 19. (−3, 3) ∪ {4} 20. The required points A(−3, −7), B(1.3, −2), C(π, √ 10), D(0, 8), E(−5.5, 0), F (−8, 4), G(9.2, −7.8), and H(7, 5) are plotted in the Cartesian Coordinate Plane below. y D(0, 8) H(7, 5) √ 10) C(π (−8, 4) E(−5.5, 0) −9 −8 −7 −6 −5 −4 −3 −2 −(1.3, −2) −1 −2 −3 −4 −5 −6 −7 −8 −9 A(−3, −7) G(9.2, −7.8) 1.1 Sets of Real Numbers and the Cartesian Coordinate Plane 19 21. (a) The point A(−3, −7) is (b) The point B(1.3, −2) is in Quadrant III symmetric about x-axis with (−3, 7) symmetric about
y-axis with (3, −7) symmetric about origin with (3, 7) in Quadrant IV symmetric about x-axis with (1.3, 2) symmetric about y-axis with (−1.3, −2) symmetric about origin with (−1.3, 2) (c) The point C(π, √ 10) is (d) The point D(0, 8) is in Quadrant I symmetric about x-axis with (π, − symmetric about y-axis with (−π, symmetric about origin with (−π, − √ √ √ 10) 10) 10) on the positive y-axis symmetric about x-axis with (0, −8) symmetric about y-axis with (0, 8) symmetric about origin with (0, −8) (e) The point E(−5.5, 0) is (f) The point F (−8, 4) is on the negative x-axis symmetric about x-axis with (−5.5, 0) symmetric about y-axis with (5.5, 0) symmetric about origin with (5.5, 0) in Quadrant II symmetric about x-axis with (−8, −4) symmetric about y-axis with (8, 4) symmetric about origin with (8, −4) (g) The point G(9.2, −7.8) is in Quadrant IV symmetric about x-axis with (9.2, 7.8) symmetric about y-axis with (−9.2, −7.8) symmetric about origin with (−9.2, 7.8) (h) The point H(7, 5) is in Quadrant I symmetric about x-axis with (7, −5) symmetric about y-axis with (−7, 5) symmetric about origin with (−7, −5) 22. d = 5, M = −1, 7 2 26, M = 1, 3 2 24. d = √ √ √ 26. d = 28. d = 30. (3 + 74, M = 13 10, − 13 √ 10 83, M = 4 √ 2 5, 5 √ √ 7, −1), (3 − √ 32. (−1 + 3, 0), (−1 − 7, −1) �
� 3, 0) 34. (−3, −4), 5 miles, (4, −4) 3 29. d = √ 23. d = 4 √ 25. d = 27. d = 3 37 10, M = (1, −4, − x2 + y2, M = x 5 31. (0, 3) √ 33 37. (a) The distance from A to B is \|AB\| = 52, and the distance from B to C is \|BC\| =, we are guaranteed by the converse of the Pythagorean Theorem that the triangle is a right triangle. 13, the distance from A to C is \|AC\| = 652 132 65. Since √ √ = √ + √ 522 √ √ (b) Show that \|AC\|2 + \|BC\|2 = \|AB\|2 20 1.2 Relations Relations and Functions From one point of view,1 all of Precalculus can be thought of as studying sets of points in the plane. With the Cartesian Plane now fresh in our memory we can discuss those sets in more detail and as usual, we begin with a deﬁnition. Deﬁnition 1.4. A relation is a set of points in the plane. Since relations are sets, we can describe them using the techniques presented in Section 1.1.1. That is, we can describe a relation verbally, using the roster method, or using set-builder notation. Since the elements in a relation are points in the plane, we often try to describe the relation graphically or algebraically as well. Depending on the situation, one method may be easier or more convenient to use than another. As an example, consider the relation R = {(−2, 1), (4, 3), (0, −3)}. As written, R is described using the roster method. Since R consists of points in the plane, we follow our instinct and plot the points. Doing so produces the graph of R. y 4 3 2 1 (−2, 1) (4, 3) −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 (0, −3) The graph of R. In the following example, we graph a variety of relations. Example 1.2.1. Graph the following relations. 1. A = {(0, 0), (−3, 1), (
4, 2), (−3, 2)} 2. HLS1 = {(x, 3) \| − 2 ≤ x ≤ 4} 3. HLS2 = {(x, 3) \| − 2 ≤ x < 4} 4. V = {(3, y) \| y is a real number} 5. H = {(x, y) \| y = −2} 6. R = {(x, y) \| 1 < y ≤ 3} 1Carl’s, of course. 1.2 Relations Solution. 21 1. To graph A, we simply plot all of the points which belong to A, as shown below on the left. 2. Don’t let the notation in this part fool you. The name of this relation is HLS1, just like the name of the relation in number 1 was A. The letters and numbers are just part of its name, just like the numbers and letters of the phrase ‘King George III’ were part of George’s name. In words, {(x, 3) \| − 2 ≤ x ≤ 4} reads ‘the set of points (x, 3) such that −2 ≤ x ≤ 4.’ All of these points have the same y-coordinate, 3, but the x-coordinate is allowed to vary between −2 and 4, inclusive. Some of the points which belong to HLS1 include some friendly points like: (−2, 3), (−1, 3), (0, 3), (1, 3), (2, 3), (3, 3), and (4, 3). However, HLS1 also contains the points (0.829, 3), − 5 It is impossible2 to list all of these points, which is why the variable x is used. Plotting several friendly representative points should convince you that HLS1 describes the horizontal line segment from the point (−2, 3) up to and including the point (4, 3). π, 3), and so on. 6, 34 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x The graph of A The graph of HLS1 3. HLS2 is hauntingly similar to HLS1. In fact, the only diﬀerence between the two is that instead of ‘−2 ≤ x ≤ 4’ we have ‘−2 ≤ x <
4’. This means that we still get a horizontal line segment which includes (−2, 3) and extends to (4, 3), but we do not include (4, 3) because of the strict inequality x < 4. How do we denote this on our graph? It is a common mistake to make the graph start at (−2, 3) end at (3, 3) as pictured below on the left. The problem with this graph is that we are forgetting about the points like (3.1, 3), (3.5, 3), (3.9, 3), (3.99, 3), and so forth. There is no real number that comes ‘immediately before’ 4, so to describe the set of points we want, we draw the horizontal line segment starting at (−2, 3) and draw an open circle at (4, 3) as depicted below on the right. 2Really impossible. The interested reader is encouraged to research countable versus uncountable sets. 22 Relations and Functions 4 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x This is NOT the correct graph of HLS2 The graph of HLS2 4. Next, we come to the relation V, described as the set of points (3, y) such that y is a real number. All of these points have an x-coordinate of 3, but the y-coordinate is free to be whatever it wants to be, without restriction.3 Plotting a few ‘friendly’ points of V should convince you that all the points of V lie on the vertical line4 x = 3. Since there is no restriction on the y-coordinate, we put arrows on the end of the portion of the line we draw to indicate it extends indeﬁnitely in both directions. The graph of V is below on the left. 5. Though written slightly diﬀerently, the relation H = {(x, y) \| y = −2} is similar to the relation V above in that only one of the coordinates, in this case the y-coordinate, is speciﬁed, leaving x to be ‘free’. Plotting some representative points gives us the horizontal line y = −2. y 4 3 2 1 −1 −2 −3 −4 1 2 3 4 x y −4 −3 −2 −1
1 2 3 4 x −1 −2 −3 −4 The graph of H The graph of V 6. For our last example, we turn to R = {(x, y) \| 1 < y ≤ 3}. As in the previous example, x is free to be whatever it likes. The value of y, on the other hand, while not completely free, is permitted to roam between 1 and 3 excluding 1, but including 3. After plotting some5 friendly elements of R, it should become clear that R consists of the region between the horizontal 3We’ll revisit the concept of a ‘free variable’ in Section 8.1. 4Don’t worry, we’ll be refreshing your memory about vertical and horizontal lines in just a moment! 5The word ‘some’ is a relative term. It may take 5, 10, or 50 points until you see the pattern. 1.2 Relations 23 lines y = 1 and y = 3. Since R requires that the y-coordinates be greater than 1, but not equal to 1, we dash the line y = 1 to indicate that those points do not belong to R. y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x The graph of R The relations V and H in the previous example lead us to our ﬁnal way to describe relations: algebraically. We can more succinctly describe the points in V as those points which satisfy the equation ‘x = 3’. Most likely, you have seen equations like this before. Depending on the context, ‘x = 3’ could mean we have solved an equation for x and arrived at the solution x = 3. In this case, however, ‘x = 3’ describes a set of points in the plane whose x-coordinate is 3. Similarly, the relation H above can be described by the equation ‘y = −2’. At some point in your mathematical upbringing, you probably learned the following. Equations of Vertical and Horizontal Lines The graph of the equation x = a is a vertical line through (a, 0). The graph of the equation y = b is a horizontal line through (0, b). Given that the very simple equations x = a and y = b produced lines, it’s natural to wonder what shapes other equations might yield. Thus our next objective is to study the graphs of equations in a more general setting as we
continue to unite Algebra and Geometry. 1.2.1 Graphs of Equations In this section, we delve more deeply into the connection between Algebra and Geometry by focusing on graphing relations described by equations. The main idea of this section is the following. The Fundamental Graphing Principle The graph of an equation is the set of points which satisfy the equation. That is, a point (x, y) is on the graph of an equation if and only if x and y satisfy the equation. Here, ‘x and y satisfy the equation’ means ‘x and y make the equation true’. It is at this point that we gain some insight into the word ‘relation’. If the equation to be graphed contains both x and y, then the equation itself is what is relating the two variables. More speciﬁcally, in the next two examples, we consider the graph of the equation x2 + y3 = 1. Even though it is not speciﬁcally 24 Relations and Functions spelled out, what we are doing is graphing the relation R = {(x, y) \| x2 + y3 = 1}. The points (x, y) we graph belong to the relation R and are necessarily related by the equation x2 + y3 = 1, since it is those pairs of x and y which make the equation true. Example 1.2.2. Determine whether or not (2, −1) is on the graph of x2 + y3 = 1. Solution. We substitute x = 2 and y = −1 into the equation to see if the equation is satisﬁed. Hence, (2, −1) is not on the graph of x2 + y3 = 1. (2)2 + (−1)3? = 1 3 = 1 We could spend hours randomly guessing and checking to see if points are on the graph of the equation. A more systematic approach is outlined in the following example. Example 1.2.3. Graph x2 + y3 = 1. Solution. To eﬃciently generate points on the graph of this equation, we ﬁrst solve for y x2 + y3 = 1 y3 = 1 − x2 y3 = 3√ y = 3√ We now substitute a value in for x, determine the corresponding value y, and plot the resulting point (x,
y). For example, substituting x = −3 into the equation yields 1 − x2 1 − x2 3 y = 3 1 − x2 = 3 1 − (−3)2 = 3√ −8 = −2, so the point (−3, −2) is on the graph. Continuing in this manner, we generate a table of points which are on the graph of the equation. These points are then plotted in the plane as shown below. y y x −2 −3 √ −2 − 3 −2 3 (x, y) (−3, −2) √ 3 (−2, − 3 3) (−1, 0) (0, 1) (1, 0) √ (2, − 3 3) (3, −2) 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 Remember, these points constitute only a small sampling of the points on the graph of this equation. To get a better idea of the shape of the graph, we could plot more points until we feel comfortable 1.2 Relations 25 ‘connecting the dots’. Doing so would result in a curve similar to the one pictured below on the far left. y 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 Don’t worry if you don’t get all of the little bends and curves just right − Calculus is where the art of precise graphing takes center stage. For now, we will settle with our naive ‘plug and plot’ approach to graphing. If you feel like all of this tedious computation and plotting is beneath you, then you can reach for a graphing calculator, input the formula as shown above, and graph. Of all of the points on the graph of an equation, the places where the graph crosses or touches the axes hold special signiﬁcance. These are called the intercepts of the graph. Intercepts come in two distinct varieties: x-intercepts and y-intercepts. They are deﬁned below. Deﬁnition 1.5. Suppose the graph of an equation is given. A point on a graph which is also on the x-axis is called an x-intercept of the graph. A point on a graph which is also on the y-axis is called an y-intercept of the graph. In our
previous example the graph had two x-intercepts, (−1, 0) and (1, 0), and one y-intercept, (0, 1). The graph of an equation can have any number of intercepts, including none at all! Since x-intercepts lie on the x-axis, we can ﬁnd them by setting y = 0 in the equation. Similarly, since y-intercepts lie on the y-axis, we can ﬁnd them by setting x = 0 in the equation. Keep in mind, intercepts are points and therefore must be written as ordered pairs. To summarize, Finding the Intercepts of the Graph of an Equation Given an equation involving x and y, we ﬁnd the intercepts of the graph as follows: x-intercepts have the form (x, 0); set y = 0 in the equation and solve for x. y-intercepts have the form (0, y); set x = 0 in the equation and solve for y. Another fact which you may have noticed about the graph in the previous example is that it seems to be symmetric about the y-axis. To actually prove this analytically, we assume (x, y) is a generic point on the graph of the equation. That is, we assume x2 + y3 = 1 is true. As we learned in Section 1.1, the point symmetric to (x, y) about the y-axis is (−x, y). To show that the graph is 26 Relations and Functions symmetric about the y-axis, we need to show that (−x, y) satisﬁes the equation x2 + y3 = 1, too. Substituting (−x, y) into the equation gives (−x)2 + (y)3? = 1 x2 + y3 = 1 Since we are assuming the original equation x2 + y3 = 1 is true, we have shown that (−x, y) satisﬁes the equation (since it leads to a true result) and hence is on the graph. In this way, we can check whether the graph of a given equation possesses any of the symmetries discussed in Section 1.1. We summarize the procedure in the following result. Testing the Graph of an Equation for Symmetry To test the graph of an equation for symmetry about the y-axis − substitute (−x, y)
into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the y-axis. about the x-axis – substitute (x, −y) into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the x-axis. about the origin - substitute (−x, −y) into the equation and simplify. If the result is equivalent to the original equation, the graph is symmetric about the origin. Intercepts and symmetry are two tools which can help us sketch the graph of an equation analytically, as demonstrated in the next example. Example 1.2.4. Find the x- and y-intercepts (if any) of the graph of (x − 2)2 + y2 = 1. Test for symmetry. Plot additional points as needed to complete the graph. Solution. To look for x-intercepts, we set y = 0 and solve (x − 2)2 + y2 = 1 (x − 2)2 + 02 = 1 (x − 2)2 = 1 √ (x − 2)2 = 1 x − 2 = ±1 extract square roots x = 2 ± 1 x = 3, 1 We get two answers for x which correspond to two x-intercepts: (1, 0) and (3, 0). Turning our attention to y-intercepts, we set x = 0 and solve 1.2 Relations 27 (x − 2)2 + y2 = 1 (0 − 2)2 + y2 = 1 4 + y2 = 1 y2 = −3 Since there is no real number which squares to a negative number (Do you remember why?), we are forced to conclude that the graph has no y-intercepts. Plotting the data we have so far, we get y 1 −1 (1, 0) (3, 0) 1 2 3 4 x Moving along to symmetry, we can immediately dismiss the possibility that the graph is symmetric about the y-axis or the origin. If the graph possessed either of these symmetries, then the fact that (1, 0) is on the graph would mean (−1, 0) would have to be on the graph. (Why?) Since (−1, 0) would be another x-intercept (and we’ve found all of these), the graph can’t have y-axis or origin symmetry.
The only symmetry left to test is symmetry about the x-axis. To that end, we substitute (x, −y) into the equation and simplify (x − 2)2 + y2 = 1? = 1 (x − 2)2 + (−y)2 (x − 2)2 + y2 = 1 Since we have obtained our original equation, we know the graph is symmetric about the x-axis. This means we can cut our ‘plug and plot’ time in half: whatever happens below the x-axis is reﬂected above the x-axis, and vice-versa. Proceeding as we did in the previous example, we obtain y 1 −1 1 2 3 4 x 28 Relations and Functions A couple of remarks are in order. First, it is entirely possible to choose a value for x which does not correspond to a point on the graph. For example, in the previous example, if we solve for y as is our custom, we get y = ± 1 − (x − 2)2. Upon substituting x = 0 into the equation, we would obtain y = ± 1 − (0 − 2)3, which is not a real number. This means there are no points on the graph with an x-coordinate of 0. When this happens, we move on and try another point. This is another drawback of the ‘plug-and-plot’ approach to graphing equations. Luckily, we will devote much of the remainder of this book to developing techniques which allow us to graph entire families of equations quickly.6 Second, it is instructive to show what would have happened had we tested the equation in the last example for symmetry about the y-axis. Substituting (−x, y) into the equation yields (x − 2)2 + y2 =. (−x − 2)2 + y2 ((−1)(x + 2))2 + y2 (x + 2)2 + y2 This last equation does not appear to be equivalent to our original equation. However, to actually prove that the graph is not symmetric about the y-axis, we need to ﬁnd a point (x, y) on the graph whose reﬂection (−x, y) is not. Our x-intercept (1, 0) ﬁts this bill nicely, since if we substitute (−1, 0) into the equation we get (x − 2)2
+ y2 (−1 − 2)2 + 02? = 1 = 1 9 = 1. This proves that (−1, 0) is not on the graph. 6Without the use of a calculator, if you can believe it! 1.2 Relations 1.2.2 Exercises In Exercises 1 - 20, graph the given relation. 1. {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} 2. {(−2, 0), (−1, 1), (−1, −1), (0, 2), (0, −2), (1, 3), (1, −3)} 29 3. {(m, 2m) \| m = 0, ±1, ±2} 4. 6 k, k \| k = ±1, ±2, ±3, ±4, ±5, ±6 5. n, 4 − n2 \| n = 0, ±1, ±2 6. √ j, j \| j = 0, 1, 4, 9 7. {(x, −2) \| x > −4} 8. {(x, 3) \| x ≤ 4} 9. {(−1, y) \| y > 1} 10. {(2, y) \| y ≤ 5} 11. {(−2, y) \| − 3 < y ≤ 4} 12. {(3, y) \| − 4 ≤ y < 3} 13. {(x, 2) \| − 2 ≤ x < 3} 14. {(x, −3) \| − 4 < x ≤ 4} 15. {(x, y) \| x > −2} 16. {(x, y) \| x ≤ 3} 17. {(x, y) \| y < 4} 18. {(x, y) \| x ≤ 3, y < 2} 19. {(x, y) \| x > 0, y < 4} 20. {(x, y } In Exercises 21 - 30, describe the given relation using either the roster or set-builder method. 21. y 4 3 2 1 −4 −3 −2 −1 −1 1 x Relation A 22. y 3 2 1 −4 −3 −2 −1 1 2 3 4 x Relation B 30 23. y 5 4
3 2 1 −1 −2 −3 1 2 3 x Relations and Functions 24. y 3 2 1 −3 −2 −1 −1 x −2 −3 −4 Relation C Relation D 25. 27. y 3 2 1 −4 −3 −2 −1 1 2 3 4 x Relation E y 3 2 1 26. 28. y 4 3 2 1 −3 −2 −1 1 2 3 x Relation F y 3 2 1 −3 −2 −1 −1 1 2 3 x −4 −3 −2 −1 −1 1 2 3 x −2 −3 Relation G −2 −3 Relation H 1.2 Relations 29. y 5 4 3 2 1 −1 −1 1 2 3 4 5 x Relation I In Exercises 31 - 36, graph the given line. 31. x = −2 33. y = 3 35. x = 0 31 30. y 2 1 −4 −3 −2 −1 −1 1 2 3 4 5 x −2 −3 Relation J 32. x = 3 34. y = −2 36. y = 0 Some relations are fairly easy to describe in words or with the roster method but are rather diﬃcult, if not impossible, to graph. Discuss with your classmates how you might graph the relations given in Exercises 37 - 40. Please note that in the notation below we are using the ellipsis,..., to denote that the list does not end, but rather, continues to follow the established pattern indeﬁnitely. For the relations in Exercises 37 and 38, give two examples of points which belong to the relation and two points which do not belong to the relation. 37. {(x, y) \| x is an odd integer, and y is an even integer.} 38. {(x, 1) \| x is an irrational number } 39. {(1, 0), (2, 1), (4, 2), (8, 3), (16, 4), (32, 5),...} 40. {..., (−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9),...} For each equation given in Exercises 41 - 52: Find the x- and y-intercept(s) of the graph, if
any exist. Follow the procedure in Example 1.2.3 to create a table of sample points on the graph of the equation. Plot the sample points and create a rough sketch of the graph of the equation. Test for symmetry. If the equation appears to fail any of the symmetry tests, ﬁnd a point on the graph of the equation whose reﬂection fails to be on the graph as was done at the end of Example 1.2.4 32 Relations and Functions 41. y = x2 + 1 43. y = x3 − x 45. y = √ x − 2 47. 3x − y = 7 49. (x + 2)2 + y2 = 16 51. 4y2 − 9x2 = 36 42. y = x2 − 2x − 8 44. y = x3 4 − 3x √ 46. y = 2 x + 4 − 2 48. 3x − 2y = 10 50. x2 − y2 = 1 52. x3y = −4 The procedures which we have outlined in the Examples of this section and used in Exercises 41 - 52 all rely on the fact that the equations were “well-behaved”. Not everything in Mathematics is quite so tame, as the following equations will show you. Discuss with your classmates how you might approach graphing the equations given in Exercises 53 - 56. What diﬃculties arise when trying to apply the various tests and procedures given in this section? For more information, including pictures of the curves, each curve name is a link to its page at www.wikipedia.org. For a much longer list of fascinating curves, click here. 53. x3 + y3 − 3xy = 0 Folium of Descartes 54. x4 = x2 + y2 Kampyle of Eudoxus 55. y2 = x3 + 3x2 Tschirnhausen cubic 56. (x2 + y2)2 = x3 + y3 Crooked egg 57. With the help of your classmates, ﬁnd examples of equations whose graphs possess symmetry about the x-axis only symmetry about the y-axis only symmetry about the origin only symmetry about the x-axis, y-axis, and origin Can you ﬁnd an example of an equation whose graph possesses exactly two of the symmetries listed above? Why or why not? 1.2 Relations 1.
2.3 Answers 1. 5. −3 −2 −2 −1 −1 1 2 x −2 −3 −4 y 4 3 2 1 −2 −1 1 2 x 33 2. 4. y 3 2 1 −2 −1 −1 1 2 x −2 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 6 34 7. Relations and Functions y 8. −4 −3 −2 −1 −3 −4 −3 −2 −1 1 2 3 4 x 9. y 101 1 2 x 11. y 4 3 2 1 −3 −2 −1 −1 x −2 −3 5 4 3 2 1 −1 −2 −3 1 2 3 x 12. y 3 2 1 −1 −2 −3 −4 1 2 3 x 1.2 Relations 35 14. 16. 18. 13. 15. 17. y 3 2 1 −4 −3 −2 −2 −1 −1 1 2 3 x −2 −3 y 4 3 2 1 −3 −2 −1 1 2 3 x y −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 y y 3 2 1 −1 −2 −3 3 2 1 −1 −2 −3 1 2 3 x 1 2 3 x 19. y 20. y 4 3 2 1 −2 −1 1 x 36 Relations and Functions 21. A = {(−4, −1), (−2, 1), (0, 3), (1, 4)} 22. B = {(x, 3) \| x ≥ −3} 23. C = {(2, y) \| y > −3} 24. D = {(−2, y) \| − 4 ≤ y < 3} 25. E = {(x, 2) \| − 4 ≤ x < 3} 26. F = {(x, y) \| y ≥ 0} 27. G = {(x, y) \| x > −2} 28. H = {(x, y) \| − 3 < x ≤ 2} 29. I = {(x, y) \| x ≥ 0,y ≥ 0} 30. J = {(x, y) \| − 4 < x < 5, −3 < y < 2} 31. 33. 35. y 3 2 1 −3 −2 −1 −1 x −2 −
3 The line x = −2 y 3 2 1 −3 −2 −1 1 2 3 x The line y = 3 y 3 2 1 32. y 3 2 1 −1 −2 −3 1 2 3 x The line x = 3 34. y −3 −2 −1 −1 1 2 3 x −2 −3 The line y = −2 36. y 3 2 1 −3 −2 −1 −1 1 2 3 x −3 −2 −1 −1 1 2 3 x −2 −3 −2 −3 The line x = 0 is the y-axis The line y = 0 is the x-axis 1.2 Relations 41. y = x2 + 1 37 42. y = x2 − 2x − 8 The graph has no x-intercepts x-intercepts: (4, 0), (−2, 0) y-intercept: (0, 1) x −2 −1 0 1 2 y (x, y) 5 (−2, 5) 2 (−1, 2) (0, 1) 1 (1, 2) 2 (2, 5) 5 y 5 4 3 2 1 −2−1 1 2 x The graph is not symmetric about the x-axis (e.g. (2, 5) is on the graph but (2, −5) is not) The graph is symmetric about the y-axis The graph is not symmetric about the origin (e.g. (2, 5) is on the graph but (−2, −5) is not) y-intercept: (0, −8) y 7 0 (x, y) x (−3, 7) −3 −2 (−2, 0) −1 −5 (−1, −5) (0, −8) (1, −9) (2, −8) (3, −5) (4, 0) (5, 7) 0 −8 1 −9 2 −8 3 −3−2−1 −2 −3 −4 −5 −6 −7 −8 −9 The graph is not symmetric about the x-axis (e.g. (−3, 7) is on the graph but (−3, −7) is not) The graph is not symmetric about the y-axis (e.g. (−3, 7) is on the graph but (3, 7) is not) The graph is not
symmetric about the origin (e.g. (−3, 7) is on the graph but (3, −7) is not) 38 43. y = x3 − x x-intercepts: (−1, 0), (0, 0), (1, 0) y-intercept: (0, 0) x y (x, y) −2 −6 (−2, −6) (−1, 0) −1 (0, 0) 0 (1, 0) 1 (2, 62−1 −1 1 2 x −2 −3 −4 −5 −6 The graph is not symmetric about the x-axis. (e.g. (2, 6) is on the graph but (2, −6) is not) The graph is not symmetric about the y-axis. (e.g. (2, 6) is on the graph but (−2, 6) is not) The graph is symmetric about the origin. Relations and Functions 44. y = x3 4 − 3x x-intercepts: ±2 √ 3, 0, (0, 0) y-intercept: (0, 0) x y −4 −4 9 −3 4 −2 4 11 −1 4 0 0 1 − 11 4 2 −4 3 − 9 4 4 4 (x, y) (−4, −4) −3, 9 4 (−2, 4) −1, 11 4 (0, 0) 1, − 11 4 (2, −4) 3, − 9 4 (4, 4) y 4 3 2 1 −4−3−2−1 −1 1 2 3 4 x −2 −3 −4 The graph is not symmetric about the x-axis (e.g. (−4, −4) is on the graph but (−4, 4) is not) The graph is not symmetric about the y-axis (e.g. (−4, −4) is on the graph but (4, −4) is not) The graph is symmetric about the origin 1.2 Relations 45. y = √ x − 2 x-intercept: (2, 0) The graph has no y-intercepts √ 46-intercept: (−3, 0) y-intercept: (0, 2) 39 y (x, y) 0 (2, 0) 1 (3, 1)
(6, 2) 2 3 (11, 3) x 2 3 6 11 10 11 x The graph is not symmetric about the x-axis (e.g. (3, 1) is on the graph but (3, −1) is not) The graph is not symmetric about the y-axis (e.g. (3, 1) is on the graph but (−3, 1) is not) The graph is not symmetric about the origin (e.g. (3, 1) is on the graph but (−3, −1) is not) x −4 −3 −2 −1 0 1 y −2 0 2 − 2 −2, 3 − 2 −2, 2 5 − 2 −2, y (x, y) (−4, −2) (−3, 00, 24−3−2−1 −1 1 2 x −2 −3 The graph is not symmetric about the x-axis (e.g. (−4, −2) is on the graph but (−4, 2) is not) The graph is not symmetric about the y-axis (e.g. (−4, −2) is on the graph but (4, −2) is not) The graph is not symmetric about the origin (e.g. (−4, −2) is on the graph but (4, 2) is not) 40 Relations and Functions 47. 3x − y = 7 Re-write as: y = 3x − 7. 48. 3x − 2y = 10 Re-write as: y = 3x−10 2. x-intercept: ( 7 3, 0) y-intercept: (0, −7) x y (x, y) −2 −13 (−2, −13) −1 −10 (−1, −10) (0, −7) (1, −4) (2, −1) (3, 2) 0 −7 1 −4 2 −1 2 3 y 3 2 1 −2−1 −1 1 2 3 x −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 x-intercepts: 10 3, 0 y-intercept: (0, −5) x y −2 −8 −1 − 13 2 0 −5 1 − 7 2 2 −2 (x, y) (−2, −
8) −1, − 13 2 (0, −5) 1, − 7 2 (2, −2) y 2 1 −3−2−1 −1 1 2 3 4 x −2 −3 −4 −5 −6 −7 −8 −9 The graph is not symmetric about the x-axis (e.g. (3, 2) is on the graph but (3, −2) is not) The graph is not symmetric about the y-axis (e.g. (3, 2) is on the graph but (−3, 2) is not) The graph is not symmetric about the origin (e.g. (3, 2) is on the graph but (−3, −2) is not) The graph is not symmetric about the x-axis (e.g. (2, −2) is on the graph but (2, 2) is not) The graph is not symmetric about the y-axis (e.g. (2, −2) is on the graph but (−2, −2) is not) The graph is not symmetric about the origin (e.g. (2, −2) is on the graph but (−2, 2) is not) 1.2 Relations 41 49. (x + 2)2 + y2 = 16 Re-write as y = ± 16 − (x + 2)2. x-intercepts: (−6, 0), (2, 0) y-intercepts: 0, ±2 3 √ y x −6 0 √ −4 ±2 −2 ±4 √ 0 ±2 0 2 (x, y) (−6, 0) √ 3 −4, ±2 3 3 √ (−2, ±4) 3 0, ±2 (2, 0) y 5 4 3 2 1 50. x2 − y2 = 1 √ Re-write as: y = ± x2 − 1. x-intercepts: (−1, 0), (1, 0) The graph has no y-intercepts x −3 ± −2 ± −x, y) √ √ 8 (−3, ± 3 (−2, ± 8) 3) (−1, 0) (1, 0) √ √ (2, ± (3, ± 3) 8) 3 8 −7−6−5−4−3−2
−1 −1 1 2 3 x −2 −3 −4 −5 The graph is symmetric about the x-axis The graph is not symmetric about the y-axis (e.g. (−6, 0) is on the graph but (6, 0) is not) The graph is not symmetric about the origin (e.g. (−6, 0) is on the graph but (6, 0) is not) y 3 2 1 −3−2−1 −1 1 2 3 x −2 −3 The graph is symmetric about the x-axis The graph is symmetric about the y-axis The graph is symmetric about the origin 42 51. 4y2 − 9x2 = 36 Re-write as: y = ± √ 9x2+36 2. The graph has no x-intercepts y-intercepts: (0, ±3) Relations and Functions 52. x3y = −4 Re-write as: y = − 4 x3. The graph has no x-intercepts The graph has no y-intercepts x y √ −4 ±3 √ −2 ±3 ±3 √ √ 0 2 ±3 4 ±3 √ √ (x, y) 5 −4, ±3 2 −2, ±3 (0, ±3) √ √ 2, ±3 4, ±3 2 5 2 5 5 2 x −2 −1 − 1 2 1 y 1 2 4 32 2 −32 ( 1 1 −4 2 − 1 2 (x, y) (−2, 1 2 ) (−1, 4) (− 1 2, 32) 2, −32) (1, −4) (24−3−2−1 −1 1 2 3 4 x −2 −3 −4 −5 −6 −7 y 32 4 −4 −2 −1 1 x 2 The graph is symmetric about the x-axis −32 The graph is symmetric about the y-axis The graph is symmetric about the origin The graph is not symmetric about the x-axis (e.g. (1, −4) is on the graph but (1, 4) is not) The graph is not symmetric about the y-axis (e.g. (1, −4) is on the graph but (−1, −4) is not) The graph is symmetric about the origin 1
.3 Introduction to Functions 43 1.3 Introduction to Functions One of the core concepts in College Algebra is the function. There are many ways to describe a function and we begin by deﬁning a function as a special kind of relation. Deﬁnition 1.6. A relation in which each x-coordinate is matched with only one y-coordinate is said to describe y as a function of x. Example 1.3.1. Which of the following relations describe y as a function of x? 1. R1 = {(−2, 1), (1, 3), (1, 4), (3, −1)} 2. R2 = {(−2, 1), (1, 3), (2, 3), (3, −1)} Solution. A quick scan of the points in R1 reveals that the x-coordinate 1 is matched with two diﬀerent y-coordinates: namely 3 and 4. Hence in R1, y is not a function of x. On the other hand, every x-coordinate in R2 occurs only once which means each x-coordinate has only one corresponding y-coordinate. So, R2 does represent y as a function of x. Note that in the previous example, the relation R2 contained two diﬀerent points with the same y-coordinates, namely (1, 3) and (2, 3). Remember, in order to say y is a function of x, we just need to ensure the same x-coordinate isn’t used in more than one point.1 To see what the function concept means geometrically, we graph R1 and R2 in the plane2 −1 −1 1 2 3 x −2 −1 −1 1 2 3 x The graph of R1 The graph of R2 The fact that the x-coordinate 1 is matched with two diﬀerent y-coordinates in R1 presents itself graphically as the points (1, 3) and (1, 4) lying on the same vertical line, x = 1. If we turn our attention to the graph of R2, we see that no two points of the relation lie on the same vertical line. We can generalize this idea as follows Theorem 1.1. The Vertical Line Test: A set of points in the plane represents y as a function of x if and only if no two points lie
on the same vertical line. 1We will have occasion later in the text to concern ourselves with the concept of x being a function of y. In this case, R1 represents x as a function of y; R2 does not. 44 Relations and Functions It is worth taking some time to meditate on the Vertical Line Test; it will check to see how well you understand the concept of ‘function’ as well as the concept of ‘graph’. Example 1.3.2. Use the Vertical Line Test to determine which of the following relations describes y as a function of x. y 4 3 2 1 −1 1 x −1 The graph of R The graph of S Solution. Looking at the graph of R, we can easily imagine a vertical line crossing the graph more than once. Hence, R does not represent y as a function of x. However, in the graph of S, every vertical line crosses the graph at most once, so S does represent y as a function of x. In the previous test, we say that the graph of the relation R fails the Vertical Line Test, whereas the graph of S passes the Vertical Line Test. Note that in the graph of R there are inﬁnitely many vertical lines which cross the graph more than once. However, to fail the Vertical Line Test, all you need is one vertical line that ﬁts the bill, as the next example illustrates. Example 1.3.3. Use the Vertical Line Test to determine which of the following relations describes y as a function of x1 1 x −1 −1 1 x −1 The graph of S1 The graph of S2 1.3 Introduction to Functions 45 Solution. Both S1 and S2 are slight modiﬁcations to the relation S in the previous example whose graph we determined passed the Vertical Line Test. In both S1 and S2, it is the addition of the point (1, 2) which threatens to cause trouble. In S1, there is a point on the curve with x-coordinate 1 just below (1, 2), which means that both (1, 2) and this point on the curve lie on the vertical line x = 1. (See the picture below and the left.) Hence, the graph of S1 fails the Vertical Line Test, so y is not a function of x here. However, in S2 notice that the point with x-coordinate 1 on the curve has been omitted,
leaving an ‘open circle’ there. Hence, the vertical line x = 1 crosses the graph of S2 only at the point (1, 2). Indeed, any vertical line will cross the graph at most once, so we have that the graph of S2 passes the Vertical Line Test. Thus it describes y as a function of x1 −1 x −1 1 x −1 S1 and the line x = 1 The graph of G for Ex. 1.3.4 Suppose a relation F describes y as a function of x. The sets of x- and y-coordinates are given special names which we deﬁne below. Deﬁnition 1.7. Suppose F is a relation which describes y as a function of x. The set of the x-coordinates of the points in F is called the domain of F. The set of the y-coordinates of the points in F is called the range of F. We demonstrate ﬁnding the domain and range of functions given to us either graphically or via the roster method in the following example. Example 1.3.4. Find the domain and range of the function F = {(−3, 2), (0, 1), (4, 2), (5, 2)} and of the function G whose graph is given above on the right. Solution. The domain of F is the set of the x-coordinates of the points in F, namely {−3, 0, 4, 5} and the range of F is the set of the y-coordinates, namely {1, 2}. To determine the domain and range of G, we need to determine which x and y values occur as coordinates of points on the given graph. To ﬁnd the domain, it may be helpful to imagine collapsing the curve to the x-axis and determining the portion of the x-axis that gets covered. This is called projecting the curve to the x-axis. Before we start projecting, we need to pay attention to two 46 Relations and Functions subtle notations on the graph: the arrowhead on the lower left corner of the graph indicates that the graph continues to curve downwards to the left forever more; and the open circle at (1, 3) indicates that the point (1, 3) isn’t on the graph, but all points on the curve leading up to that point are. y project down 1 1 x −1 −1 1 x −1
project up The graph of G The graph of G We see from the ﬁgure that if we project the graph of G to the x-axis, we get all real numbers less than 1. Using interval notation, we write the domain of G as (−∞, 1). To determine the range of G, we project the curve to the y-axis as follows: y 4 3 2 1 project left project right y 4 3 2 1 −1 1 x −1 −1 1 x −1 The graph of G The graph of G Note that even though there is an open circle at (1, 3), we still include the y value of 3 in our range, since the point (−1, 3) is on the graph of G. We see that the range of G is all real numbers less than or equal to 4, or, in interval notation, (−∞, 4]. 1.3 Introduction to Functions 47 All functions are relations, but not all relations are functions. Thus the equations which described the relations in Section1.2 may or may not describe y as a function of x. The algebraic representation of functions is possibly the most important way to view them so we need a process for determining whether or not an equation of a relation represents a function. (We delay the discussion of ﬁnding the domain of a function given algebraically until Section 1.4.) Example 1.3.5. Determine which equations represent y as a function of x. 1. x3 + y2 = 1 2. x2 + y3 = 1 3. x2y = 1 − 3y Solution. For each of these equations, we solve for y and determine whether each choice of x will determine only one corresponding value of y. 1. 2. 3. x3 + y2 = 1 √ y2 = 1 − x3 y2 = y = ± 1 − x3 √ 1 − x3 extract square roots 1 − 03 = ±1, so that (0, 1) If we substitute x = 0 into our equation for y, we get y = ± and (0, −1) are on the graph of this equation. Hence, this equation does not represent y as a function of x. √ x2 + y3 = 1 3 y3 = 1 − x2 y3 = 3√ y = 3√ 1 − x2 1 − x2 For every choice of x, the equation y = 3�
� equation describes y as a function of x. 1 − x2 returns only one value of y. Hence, this x2y = 1 − 3y x2y + 3y = 1 y x2 + 3 = 1 y = 1 x2 + 3 factor For each choice of x, there is only one value for y, so this equation describes y as a function of x. We could try to use our graphing calculator to verify our responses to the previous example, but we immediately run into trouble. The calculator’s “Y=” menu requires that the equation be of the form ‘y = some expression of x’. If we wanted to verify that the ﬁrst equation in Example 1.3.5 48 Relations and Functions does not represent y as a function of x, we would need to enter two separate expressions into the calculator: one for the positive square root and one for the negative square root we found when solving the equation for y. As predicted, the resulting graph shown below clearly fails the Vertical Line Test, so the equation does not represent y as a function of x. Thus in order to use the calculator to show that x3 + y2 = 1 does not represent y as a function of x we needed to know analytically that y was not a function of x so that we could use the calculator properly. There are more advanced graphing utilities out there which can do implicit function plots, but you need to know even more Algebra to make them work properly. Do you get the point we’re trying to make here? We believe it is in your best interest to learn the analytic way of doing things so that you are always smarter than your calculator. 1.3 Introduction to Functions 49 1.3.1 Exercises In Exercises 1 - 12, determine whether or not the relation represents y as a function of x. Find the domain and range of those relations which are functions. 1. {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} 2. {(−3, 0), (1, 6), (2, −3), (4, 2), (−5, 6), (4, −9), (6, 2)} 3. {(−3, 0), (−7, 6), (5, 5), (6, 4), (4, 9),
(3, 0)} 4. {(1, 2), (4, 4), (9, 6), (16, 8), (25, 10), (36, 12),...} 5. {(x, y) \| x is an odd integer, and y is an even integer} 6. {(x, 1) \| x is an irrational number} 7. {(1, 0), (2, 1), (4, 2), (8, 3), (16, 4), (32, 5),... } 8. {..., (−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9),... } 9. {(−2, y) \| − 3 < y < 4} 10. {(x, 3) \| − 2 ≤ x < 4} 11. {x, x2 \| x is a real number} 12. {x2, x \| x is a real number} In Exercises 13 - 32, determine whether or not the relation represents y as a function of x. Find the domain and range of those relations which are functions. 13. y 4 3 2 1 14. y 4 3 2 1 −4 −3 −2 −1 1 x −1 −4 −3 −2 −1 1 x −1 50 15. 17. 19. 21. y 5 4 3 2 1 −2 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3−2−1 −1 −2 −3 −4 −5 Relations and Functions 16. 18. 20. 22. y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 y 4 3 2 1 −4 −3 −2 −5 −4 −3 −2 −1 1 2 3 x −1 −2 y 5 4 3 2 1 −5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 −5 1.3 Introduction to Functions 51 23. 25. 27. 29. y 5 4 3 2 1 −5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 −5 y 4 3 2 1 24. y 5 4 3 2 1 −1 −1 −2 −3 −4 −5 262
−1 1 2 x −2 −1 1 2 x y 4 3 2 1 28. y 4 3 2 1 −2 −1 1 2 x −2 −1 1 2 x y 2 1 30. y 2 1 −2 −1 1 2 x −3 −2 −1 1 2 3 x −1 −2 −1 −2 52 31. y 2 1 Relations and Functions 32. y 2 1 −2 −1 1 2 x −2 −1 1 2 x −1 −2 −1 −2 In Exercises 33 - 47, determine whether or not the equation represents y as a function of x. 33. y = x3 − x 36. x2 − y2 = 1 39. x = y2 + 4 42. y = √ 4 − x2 34. y = 37. y = √ x − 2 x x2 − 9 40. y = x2 + 4 43. x2 − y2 = 4 45. 2x + 3y = 4 46. 2xy = 4 35. x3y = −4 38. x = −6 41. x2 + y2 = 4 44. x3 + y3 = 4 47. x2 = y2 48. Explain why the population P of Sasquatch in a given area is a function of time t. What would be the range of this function? 49. Explain why the relation between your classmates and their email addresses may not be a function. What about phone numbers and Social Security Numbers? The process given in Example 1.3.5 for determining whether an equation of a relation represents y as a function of x breaks down if we cannot solve the equation for y in terms of x. However, that does not prevent us from proving that an equation fails to represent y as a function of x. What we really need is two points with the same x-coordinate and diﬀerent y-coordinates which both satisfy the equation so that the graph of the relation would fail the Vertical Line Test 1.1. Discuss with your classmates how you might ﬁnd such points for the relations given in Exercises 50 - 53. 50. x3 + y3 − 3xy = 0 52. y2 = x3 + 3x2 51. x4 = x2 + y2 53. (x2 + y2)2 = x3 + y3 1.3 Introduction to Functions 53 1.3.2 Answers 1. Function 2
. Not a function domain = {−3, −2, −1, 0, 1, 2,3} range = {0, 1, 4, 9} 3. Function domain = {−7, −3, 3, 4, 5, 6} range = {0, 4, 5, 6, 9} 4. Function domain = {1, 4, 9, 16, 25, 36,...} = {x \| x is a perfect square} range = {2, 4, 6, 8, 10, 12,...} = {y \| y is a positive even integer} 5. Not a function 6. Function domain = {x \| x is irrational} range = {1} 7. Function 8. Function domain = {x\|x = 2n for some whole number n} range = {y \| y is any whole number} domain = {x \| x is any integer} range = y \| y = n2 for some integer n 9. Not a function 10. Function 11. Function domain = (−∞, ∞) range = [0, ∞) domain = [−2, 4), range = {3} 12. Not a function 13. Function 14. Not a function domain = {−4, −3, −2, −1, 0, 1} range = {−1, 0, 1, 2, 3, 4} 15. Function domain = (−∞, ∞) range = [1, ∞) 17. Function domain = [2, ∞) range = [0, ∞) 19. Not a function 16. Not a function 18. Function domain = (−∞, ∞) range = (0, 4] 20. Function domain = [−5, −3) ∪ (−3, 3) range = (−2, −1) ∪ [0, 4) 54 Relations and Functions 21. Function domain = [−2, ∞) range = [−3, ∞) 23. Function domain = [−5, 4) range = [−4, 4) 25. Function domain = (−∞, ∞) range = (−∞, 4] 27. Function domain = [−2, ∞) range = (−∞, 3] 29. Function domain = (−∞, 0] ∪ (1, ∞) range = (−∞, 1] ∪ {
2} 31. Not a function 22. Not a function 24. Function domain = [0, 3) ∪ (3, 6] range = (−4, −1] ∪ [0, 4] 26. Function domain = (−∞, ∞) range = (−∞, 4] 28. Function domain = (−∞, ∞) range = (−∞, ∞) 30. Function domain = [−3, 3] range = [−2, 2] 32. Function domain = (−∞, ∞) range = {2} 33. Function 34. Function 35. Function 36. Not a function 37. Function 38. Not a function 39. Not a function 40. Function 41. Not a function 42. Function 45. Function 43. Not a function 44. Function 46. Function 47. Not a function 1.4 Function Notation 55 1.4 Function Notation In Deﬁnition 1.6, we described a function as a special kind of relation − one in which each xcoordinate is matched with only one y-coordinate. In this section, we focus more on the process by which the x is matched with the y. If we think of the domain of a function as a set of inputs and the range as a set of outputs, we can think of a function f as a process by which each input x is matched with only one output y. Since the output is completely determined by the input x and the process f, we symbolize the output with function notation: ‘f (x)’, read ‘f of x.’ In other words, f (x) is the output which results by applying the process f to the input x. In this case, the parentheses here do not indicate multiplication, as they do elsewhere in Algebra. This can cause confusion if the context is not clear, so you must read carefully. This relationship is typically visualized using a diagram similar to the one below. f x Domain (Inputs) y = f (x) Range (Outputs) The value of y is completely dependent on the choice of x. For this reason, x is often called the independent variable, or argument of f, whereas y is often called the dependent variable. As we shall see, the process of a function f is usually described using an algebraic formula. For example, suppose a function f takes a real number and performs the following two steps, in sequence 1
. multiply by 3 2. add 4 If we choose 5 as our input, in step 1 we multiply by 3 to get (5)(3) = 15. In step 2, we add 4 to our result from step 1 which yields 15 + 4 = 19. Using function notation, we would write f (5) = 19 to indicate that the result of applying the process f to the input 5 gives the output 19. In general, if we use x for the input, applying step 1 produces 3x. Following with step 2 produces 3x + 4 as our ﬁnal output. Hence for an input x, we get the output f (x) = 3x + 4. Notice that to check our formula for the case x = 5, we replace the occurrence of x in the formula for f (x) with 5 to get f (5) = 3(5) + 4 = 15 + 4 = 19, as required. 56 Relations and Functions Example 1.4.1. Suppose a function g is described by applying the following steps, in sequence 1. add 4 2. multiply by 3 Determine g(5) and ﬁnd an expression for g(x). Solution. Starting with 5, step 1 gives 5 + 4 = 9. Continuing with step 2, we get (3)(9) = 27. To ﬁnd a formula for g(x), we start with our input x. Step 1 produces x + 4. We now wish to multiply this entire quantity by 3, so we use a parentheses: 3(x + 4) = 3x + 12. Hence, g(x) = 3x + 12. We can check our formula by replacing x with 5 to get g(5) = 3(5) + 12 = 15 + 12 = 27. Most of the functions we will encounter in College Algebra will be described using formulas like the ones we developed for f (x) and g(x) above. Evaluating formulas using this function notation is a key skill for success in this and many other Math courses. Example 1.4.2. Let f (x) = −x2 + 3x + 4 1. Find and simplify the following. (a) f (−1), f (0), f (2) (b) f (2x), 2f (x) (c) f (x + 2), f (x) + 2, f (x) + f (2) 2
. Solve f (x) = 4. Solution. 1. (a) To ﬁnd f (−1), we replace every occurrence of x in the expression f (x) with −1 f (−1) = −(−1)2 + 3(−1) + 4 = −(1) + (−3) + 4 = 0 Similarly, f (0) = −(0)2 + 3(0) + 4 = 4, and f (2) = −(2)2 + 3(2) + 4 = −4 + 6 + 4 = 6. (b) To ﬁnd f (2x), we replace every occurrence of x with the quantity 2x f (2x) = −(2x)2 + 3(2x) + 4 = −(4x2) + (6x) + 4 = −4x2 + 6x + 4 The expression 2f (x) means we multiply the expression f (x) by 2 2f (x) = 2 −x2 + 3x + 4 = −2x2 + 6x + 8 1.4 Function Notation 57 (c) To ﬁnd f (x + 2), we replace every occurrence of x with the quantity x + 2 f (x + 2) = −(x + 2)2 + 3(x + 2) + 4 = − x2 + 4x + 4 + (3x + 6) + 4 = −x2 − 4x − 4 + 3x + 6 + 4 = −x2 − x + 6 To ﬁnd f (x) + 2, we add 2 to the expression for f (x) f (x) + 2 = −x2 + 3x + 4 + 2 = −x2 + 3x + 6 From our work above, we see f (2) = 6 so that f (x) + f (2) = −x2 + 3x + 4 + 6 = −x2 + 3x + 10 2. Since f (x) = −x2 + 3x + 4, the equation f (x) = 4 is equivalent to −x2 + 3x + 4 = 4. Solving we get −x2 + 3x = 0, or x(−x + 3) = 0. We get x = 0 or x = 3, and we can verify these answers by checking that f (0
) = 4 and f (3) = 4. A few notes about Example 1.4.2 are in order. First note the diﬀerence between the answers for f (2x) and 2f (x). For f (2x), we are multiplying the input by 2; for 2f (x), we are multiplying the output by 2. As we see, we get entirely diﬀerent results. Along these lines, note that f (x + 2), f (x) + 2 and f (x) + f (2) are three diﬀerent expressions as well. Even though function notation uses parentheses, as does multiplication, there is no general ‘distributive property’ of function notation. Finally, note the practice of using parentheses when substituting one algebraic expression into another; we highly recommend this practice as it will reduce careless errors. Suppose now we wish to ﬁnd r(3) for r(x) = 2x x2−9. Substitution gives r(3) = 2(3) (3)2 − 9 = 6 0, which is undeﬁned. (Why is this, again?) The number 3 is not an allowable input to the function r; in other words, 3 is not in the domain of r. Which other real numbers are forbidden in this formula? We think back to arithmetic. The reason r(3) is undeﬁned is because substitution results in a division by 0. To determine which other numbers result in such a transgression, we set the denominator equal to 0 and solve x2 − 9 = 0 x2 = 9 √ x2 = x = ±3 √ 9 extract square roots 58 Relations and Functions As long as we substitute numbers other than 3 and −3, the expression r(x) is a real number. Hence, we write our domain in interval notation1 as (−∞, −3) ∪ (−3, 3) ∪ (3, ∞). When a formula for a function is given, we assume that the function is valid for all real numbers which make arithmetic sense when substituted into the formula. This set of numbers is often called the implied domain2 of the function. At this stage, there are only two mathematical sins we need to avoid: division by 0 and extracting even roots of negative numbers. The following example illustrates these concepts. Example 1.4.3. Find the domain3 of
the following functions. 1. g(x) = √ 4 − 3x 3. f (x) = 2 4x x − 3 1 − 5. r(t) = 4 √ t + 3 6 − Solution. √ 2. h(x) = 5 4 − 3x 4. F (x) = √ 4 2x + 1 x2 − 1 6. I(x) = 3x2 x 1. The potential disaster for g is if the radicand4 is negative. To avoid this, we set 4 − 3x ≥ 0. 3, the expression From this, we get 3x ≤ 4 or x ≤ 4 4 − 3x ≥ 0, and the formula g(x) returns a real number. Our domain is −∞, 4 3 3. What this shows is that as long as x ≤ 4. 2. The formula for h(x) is hauntingly close to that of g(x) with one key diﬀerence − whereas the expression for g(x) includes an even indexed root (namely a square root), the formula for h(x) involves an odd indexed root (the ﬁfth root). Since odd roots of real numbers (even negative real numbers) are real numbers, there is no restriction on the inputs to h. Hence, the domain is (−∞, ∞). 3. In the expression for f, there are two denominators. We need to make sure neither of them is 0. To that end, we set each denominator equal to 0 and solve. For the ‘small’ denominator, we get x − 3 = 0 or x = 3. For the ‘large’ denominator 1See the Exercises for Section 1.1. 2or, ‘implicit domain’ 3The word ‘implied’ is, well, implied. 4The ‘radicand’ is the expression ‘inside’ the radical. 1.4 Function Notation 59 1 − 4x x − 3 = 0 1 = (1)(x − 3) = 4x x − 3 4x x − 3 x − 3 = 4x −3 = 3x −1 = x (x − 3) clear denominators So we get two real numbers which make denominators 0, namely x = −1 and x = 3. Our domain is all real numbers except −1 and 3: (−
∞, −1) ∪ (−1, 3) ∪ (3, ∞). 4. In ﬁnding the domain of F, we notice that we have two potentially hazardous issues: not only do we have a denominator, we have a fourth (even-indexed) root. Our strategy is to determine the restrictions imposed by each part and select the real numbers which satisfy both conditions. To satisfy the fourth root, we require 2x + 1 ≥ 0. From this we get 2x ≥ −1 or x ≥ − 1 2. Next, we round up the values of x which could cause trouble in the denominator by setting the denominator equal to 0. We get x2 − 1 = 0, or x = ±1. Hence, in order for a real number x to be in the domain of F, x ≥ − 1 2 but x = ±1. In interval notation, this set is − 1 2, 1 ∪ (1, ∞). 5. Don’t be put oﬀ by the ‘t’ here. It is an independent variable representing a real number, just like x does, and is subject to the same restrictions. As in the previous problem, we have double danger here: we have a square root and a denominator. To satisfy the square root, we need a non-negative radicand so we set t + 3 ≥ 0 to get t ≥ −3. Setting the denominator equal to zero gives 6 − t + 3 = 6. Squaring both sides gives t + 3 = 36, or t = 33. Since we squared both sides in the course of solving this equation, we need to check our answer.5 Sure enough, when t = 33, 6 − 36 = 0, so t = 33 will cause problems in the denominator. At last we can ﬁnd the domain of r: we need t ≥ −3, but t = 33. Our ﬁnal answer is [−3, 33) ∪ (33, ∞). t + 3 = 0, or. It’s tempting to simplify I(x) = 3x2 x = 3x, and, since there are no longer any denominators, claim that there are no longer any restrictions. However, in simplifying I(x), we are assuming x = 0, since 0 0 is undeﬁned.6 Proceeding as before, we �
�nd the domain of I to be all real numbers except 0: (−∞, 0) ∪ (0, ∞). It is worth reiterating the importance of ﬁnding the domain of a function before simplifying, as evidenced by the function I in the previous example. Even though the formula I(x) simpliﬁes to 5Do you remember why? Consider squaring both sides to ‘solve’ 6More precisely, the fraction 0 0 is an ‘indeterminant form’. Calculus is required tame such beasts. √ t + 1 = −2. 60 Relations and Functions 3x, it would be inaccurate to write I(x) = 3x without adding the stipulation that x = 0. It would be analogous to not reporting taxable income or some other sin of omission. 1.4.1 Modeling with Functions The importance of Mathematics to our society lies in its value to approximate, or model real-world phenomenon. Whether it be used to predict the high temperature on a given day, determine the hours of daylight on a given day, or predict population trends of various and sundry real and mythical beasts,7 Mathematics is second only to literacy in the importance humanity’s development.8 It is important to keep in mind that anytime Mathematics is used to approximate reality, there are always limitations to the model. For example, suppose grapes are on sale at the local market for $1.50 per pound. Then one pound of grapes costs $1.50, two pounds of grapes cost $3.00, and so forth. Suppose we want to develop a formula which relates the cost of buying grapes to the amount of grapes being purchased. Since these two quantities vary from situation to situation, we assign them variables. Let c denote the cost of the grapes and let g denote the amount of grapes purchased. To ﬁnd the cost c of the grapes, we multiply the amount of grapes g by the price $1.50 dollars per pound to get c = 1.5g In order for the units to be correct in the formula, g must be measured in pounds of grapes in which case the computed value of c is measured in dollars. Since we’re interested in ﬁnding the cost c given an amount g, we think of g as the independent variable and c as the dependent variable. Using the language of function notation, we write c(g) = 1.5g where g is
the amount of grapes purchased (in pounds) and c(g) is the cost (in dollars). For example, c(5) represents the cost, in dollars, to purchase 5 pounds of grapes. In this case, c(5) = 1.5(5) = 7.5, so it would cost $7.50. If, on the other hand, we wanted to ﬁnd the amount of grapes we can purchase for $5, we would need to set c(g) = 5 and solve for g. In this case, c(g) = 1.5g, so solving c(g) = 5 is equivalent to solving 1.5g = 5 Doing so gives g = 5 1.5 = 3.3. This means we can purchase exactly 3.3 pounds of grapes for $5. Of course, you would be hard-pressed to buy exactly 3.3 pounds of grapes,9 and this leads us to our next topic of discussion, the applied domain10 of a function. Even though, mathematically, c(g) = 1.5g has no domain restrictions (there are no denominators and no even-indexed radicals), there are certain values of g that don’t make any physical sense. For example, g = −1 corresponds to ‘purchasing’ −1 pounds of grapes.11 Also, unless the ‘local market’ mentioned is the State of California (or some other exporter of grapes), it also doesn’t make much sense for g = 500,000,000, either. So the reality of the situation limits what g can be, and 7See Sections 2.5, 11.1, and 6.5, respectively. 8In Carl’s humble opinion, of course... 9You could get close... within a certain speciﬁed margin of error, perhaps. 10or, ‘explicit domain’ 11Maybe this means returning a pound of grapes? 1.4 Function Notation 61 these limits determine the applied domain of g. Typically, an applied domain is stated explicitly. In this case, it would be common to see something like c(g) = 1.5g, 0 ≤ g ≤ 100, meaning the number of pounds of grapes purchased is limited from 0 up to 100. The upper bound here, 100 may represent the inventory of the market, or some other limit as set by local policy or law
. Even with this restriction, our model has its limitations. As we saw above, it is virtually impossible to buy exactly 3.3 pounds of grapes so that our cost is exactly $5. In this case, being sensible shoppers, we would most likely ‘round down’ and purchase 3 pounds of grapes or however close the market scale can read to 3.3 without being over. It is time for a more sophisticated example. Example 1.4.4. The height h in feet of a model rocket above the ground t seconds after lift-oﬀ is given by h(t) = −5t2 + 100t, 0, if 0 ≤ t ≤ 20 t > 20 if 1. Find and interpret h(10) and h(60). 2. Solve h(t) = 375 and interpret your answers. Solution. 1. We ﬁrst note that the independent variable here is t, chosen because it represents time. Secondly, the function is broken up into two rules: one formula for values of t between 0 and 20 inclusive, and another for values of t greater than 20. Since t = 10 satisﬁes the inequality 0 ≤ t ≤ 20, we use the ﬁrst formula listed, h(t) = −5t2 + 100t, to ﬁnd h(10). We get h(10) = −5(10)2 + 100(10) = 500. Since t represents the number of seconds since lift-oﬀ and h(t) is the height above the ground in feet, the equation h(10) = 500 means that 10 seconds after lift-oﬀ, the model rocket is 500 feet above the ground. To ﬁnd h(60), we note that t = 60 satisﬁes t > 20, so we use the rule h(t) = 0. This function returns a value of 0 regardless of what value is substituted in for t, so h(60) = 0. This means that 60 seconds after lift-oﬀ, the rocket is 0 feet above the ground; in other words, a minute after lift-oﬀ, the rocket has already returned to Earth. 2. Since the function h is deﬁned in pieces, we need to solve h(t) = 375 in pieces. For 0 ≤ t ≤ 20, h(t) = −5t2
+ 100t, so for these values of t, we solve −5t2 + 100t = 375. Rearranging terms, we get 5t2 − 100t + 375 = 0, and factoring gives 5(t − 5)(t − 15) = 0. Our answers are t = 5 and t = 15, and since both of these values of t lie between 0 and 20, we keep both solutions. For t > 20, h(t) = 0, and in this case, there are no solutions to 0 = 375. In terms of the model rocket, solving h(t) = 375 corresponds to ﬁnding when, if ever, the rocket reaches 375 feet above the ground. Our two answers, t = 5 and t = 15 correspond to the rocket reaching this altitude twice – once 5 seconds after launch, and again 15 seconds after launch.12 12What goes up... 62 Relations and Functions The type of function in the previous example is called a piecewise-deﬁned function, or ‘piecewise’ function for short. Many real-world phenomena, income tax formulas13 for example, are modeled by such functions. By the way, if we wanted to avoid using a piecewise function in Example 1.4.4, we could have used h(t) = −5t2 + 100t on the explicit domain 0 ≤ t ≤ 20 because after 20 seconds, the rocket is on the ground and stops moving. In many cases, though, piecewise functions are your only choice, so it’s best to understand them well. Mathematical modeling is not a one-section topic. It’s not even a one-course topic as is evidenced by undergraduate and graduate courses in mathematical modeling being oﬀered at many universities. Thus our goal in this section cannot possibly be to tell you the whole story. What we can do is get you started. As we study new classes of functions, we will see what phenomena they can be used to model. In that respect, mathematical modeling cannot be a topic in a book, but rather, must be a theme of the book. For now, we have you explore some very basic models in the Exercises because you need to crawl to walk to run. As we learn more about functions, we’ll help you build your own models and get you on your way to applying Mathematics to your world. 13See the Internal Revenue Service’s website 1
.4 Function Notation 1.4.2 Exercises 63 In Exercises 1 - 10, ﬁnd an expression for f (x) and state its domain. 1. f is a function that takes a real number x and performs the following three steps in the order given: (1) multiply by 2; (2) add 3; (3) divide by 4. 2. f is a function that takes a real number x and performs the following three steps in the order given: (1) add 3; (2) multiply by 2; (3) divide by 4. 3. f is a function that takes a real number x and performs the following three steps in the order given: (1) divide by 4; (2) add 3; (3) multiply by 2. 4. f is a function that takes a real number x and performs the following three steps in the order given: (1) multiply by 2; (2) add 3; (3) take the square root. 5. f is a function that takes a real number x and performs the following three steps in the order given: (1) add 3; (2) multiply by 2; (3) take the square root. 6. f is a function that takes a real number x and performs the following three steps in the order given: (1) add 3; (2) take the square root; (3) multiply by 2. 7. f is a function that takes a real number x and performs the following three steps in the order given: (1) take the square root; (2) subtract 13; (3) make the quantity the denominator of a fraction with numerator 4. 8. f is a function that takes a real number x and performs the following three steps in the order given: (1) subtract 13; (2) take the square root; (3) make the quantity the denominator of a fraction with numerator 4. 9. f is a function that takes a real number x and performs the following three steps in the order given: (1) take the square root; (2) make the quantity the denominator of a fraction with numerator 4; (3) subtract 13. 10. f is a function that takes a real number x and performs the following three steps in the order given: (1) make the quantity the denominator of a fraction with numerator 4; (2) take the square root; (3
) subtract 13. In Exercises 11 - 18, use the given function f to ﬁnd and simplify the following: f (3) f (4x) f (x − 4) f (−1) 4f (x) f (x) − 4 f 3 2 f (−x) f x2 64 Relations and Functions 11. f (x) = 2x + 1 13. f (x) = 2 − x2 15. f (x) = x x − 1 17. f (x) = 6 12. f (x) = 3 − 4x 14. f (x) = x2 − 3x + 2 16. f (x) = 2 x3 18. f (x) = 0 In Exercises 19 - 26, use the given function f to ﬁnd and simplify the following: f (2) 2f (a) f 2 a 19. f (x) = 2x − 5 21. f (x) = 2x2 − 1 2x + 1 23. f (x) = 25. f (x) = √ x 2 f (−2) f (a + 2) f (a) 2 f (2a) f (a) + f (2) f (a + h) 20. f (x) = 5 − 2x 22. f (x) = 3x2 + 3x − 2 24. f (x) = 117 26. f (x) = 2 x In Exercises 27 - 34, use the given function f to ﬁnd f (0) and solve f (x) = 0 27. f (x) = 2x − 1 29. f (x) = 2x2 − 6 31. f (x) = √ x + 4 28. f (x) = 3 − 2 5 x 30. f (x) = x2 − x − 12 √ 32. f (x) = 1 − 2x 33. f (x) = 3 4 − x   35. Let f (x) =  (a) f (−4) (d) f (3.001) 34. f (x) = 3x2 − 12x 4 − x2 √ x + 5 9 − x2 −x + 5 x ≤ −3 if if −3 < x ≤ 3 x > 3
if Compute the following function values. (b) f (−3) (e) f (−3.001) (c) f (3) (f) f (2) 1.4 Function Notation 65 36. Let f (x) =    √ x2 1 − x2 x x ≤ −1 if if −1 < x ≤ 1 x > 1 if Compute the following function values. (a) f (4) (d) f (0) (b) f (−3) (e) f (−1) (c) f (1) (f) f (−0.999) In Exercises 37 - 62, ﬁnd the (implied) domain of the function. 37. f (x) = x4 − 13x3 + 56x2 − 19 38. f (x) = x2 + 4 39. f (x) = 41. f (x) = 43. f (x) = 45. f (x) = x − 2 x + 1 2x x2 + 3 x + 4 x2 − 36 √ 3 − x 47. f (x) = 9x √ x + 3 49. f (x) = √ 6x − 2 √ 51. f (x) = 3 6x − 2 53. f (x) = √ 6x − 2 x2 − 36 55. s(t) = t t − 8 57. b(θ) = 59. α(y 61. T (t) = √ t − 8 5 − t 40. f (x) = 3x x2 + x − 2 42. f (x) = 44. f (x) = 46. f (x) = 48. f (x) = 2x x2 − 3 x − 2 x − 2 √ 2x + 5 √ 7 − x x2 + 1 50. f (x) = √ 6 6x − 2 6 √ 6x − 2 52. f (x) = 54. f (x) = 56. Q(r) = 58. A(x) = 60. g(v) = 62. u(w) = 4 − √ 3 6x − 2 x2 + 36 √ r r − 8 √ 1 4 − 1 v2 66 Relations and Functions 63.
The area A enclosed by a square, in square inches, is a function of the length of one of its sides x, when measured in inches. This relation is expressed by the formula A(x) = x2 for x > 0. Find A(3) and solve A(x) = 36. Interpret your answers to each. Why is x restricted to x > 0? 64. The area A enclosed by a circle, in square meters, is a function of its radius r, when measured in meters. This relation is expressed by the formula A(r) = πr2 for r > 0. Find A(2) and solve A(r) = 16π. Interpret your answers to each. Why is r restricted to r > 0? 65. The volume V enclosed by a cube, in cubic centimeters, is a function of the length of one of its sides x, when measured in centimeters. This relation is expressed by the formula V (x) = x3 for x > 0. Find V (5) and solve V (x) = 27. Interpret your answers to each. Why is x restricted to x > 0? 66. The volume V enclosed by a sphere, in cubic feet, is a function of the radius of the sphere r, 3 r3 for r > 0. Find when measured in feet. This relation is expressed by the formula V (r) = 4π V (3) and solve V (r) = 32π 3. Interpret your answers to each. Why is r restricted to r > 0? 67. The height of an object dropped from the roof of an eight story building is modeled by: h(t) = −16t2 + 64, 0 ≤ t ≤ 2. Here, h is the height of the object oﬀ the ground, in feet, t seconds after the object is dropped. Find h(0) and solve h(t) = 0. Interpret your answers to each. Why is t restricted to 0 ≤ t ≤ 2? 68. The temperature T in degrees Fahrenheit t hours after 6 AM is given by T (t) = − 1 2 t2 + 8t + 3 for 0 ≤ t ≤ 12. Find and interpret T (0), T (6) and T (12). 69. The function C(x) = x2 − 10x + 27 models the cost, in hundreds of dollars, to produce x thousand pens. Find and interpret C(0), C(
2) and C(5). 70. Using data from the Bureau of Transportation Statistics, the average fuel economy F in miles per gallon for passenger cars in the US can be modeled by F (t) = −0.0076t2 + 0.45t + 16, 0 ≤ t ≤ 28, where t is the number of years since 1980. Use your calculator to ﬁnd F (0), F (14) and F (28). Round your answers to two decimal places and interpret your answers to each. 71. The population of Sasquatch in Portage County can be modeled by the function P (t) = 150t t+15, where t represents the number of years since 1803. Find and interpret P (0) and P (205). Discuss with your classmates what the applied domain and range of P should be. 72. For n copies of the book Me and my Sasquatch, a print on-demand company charges C(n) dollars, where C(n) is determined by the formula C(n) =    15n 13.50n 12n if if if 1 ≤ n ≤ 25 25 < n ≤ 50 n > 50 (a) Find and interpret C(20). 1.4 Function Notation 67 (b) How much does it cost to order 50 copies of the book? What about 51 copies? (c) Your answer to 72b should get you thinking. Suppose a bookstore estimates it will sell 50 copies of the book. How many books can, in fact, be ordered for the same price as those 50 copies? (Round your answer to a whole number of books.) 73. An on-line comic book retailer charges shipping costs according to the following formula S(n) = 1.5n + 2.5 0 if if 1 ≤ n ≤ 14 n ≥ 15 where n is the number of comic books purchased and S(n) is the shipping cost in dollars. (a) What is the cost to ship 10 comic books? (b) What is the signiﬁcance of the formula S(n) = 0 for n ≥ 15? 74. The cost C (in dollars) to talk m minutes a month on a mobile phone plan is modeled by C(m) = 25 25 + 0.1(m − 1000) 0 ≤ m ≤ 1000 if if m > 1000 (a) How much does it cost to
talk 750 minutes per month with this plan? (b) How much does it cost to talk 20 hours a month with this plan? (c) Explain the terms of the plan verbally. 75. In Section 1.1.1 we deﬁned the set of integers as Z = {..., −3, −2, −1, 0, 1, 2, 3,...}.14 The greatest integer of x, denoted by x, is deﬁned to be the largest integer k with k ≤ x. (a) Find 0.785, 117, −2.001, and π + 6 (b) Discuss with your classmates how x may be described as a piecewise deﬁned function. HINT: There are inﬁnitely many pieces! (c) Is a + b = a + b always true? What if a or b is an integer? Test some values, make a conjecture, and explain your result. 76. We have through our examples tried to convince you that, in general, f (a + b) = f (a) + f (b). It has been our experience that students refuse to believe us so we’ll try again with a diﬀerent approach. With the help of your classmates, ﬁnd a function f for which the following properties are always true. (a) f (0) = f (−1 + 1) = f (−1) + f (1) 14The use of the letter Z for the integers is ostensibly because the German word zahlen means ‘to count.’ 68 Relations and Functions (b) f (5) = f (2 + 3) = f (2) + f (3) (c) f (−6) = f (0 − 6) = f (0) − f (6) (d) f (a + b) = f (a) + f (b) regardless of what two numbers we give you for a and b. How many functions did you ﬁnd that failed to satisfy the conditions above? Did f (x) = x2 work? What about f (x) =? Did you ﬁnd an attribute x or f (x) = 3x + 7 or f (x) = common to those functions that did succeed? You should have, because there is only one extremely special family of functions that actually works here
. Thus we return to our previous statement, in general, f (a + b) = f (a) + f (b). 1 x √ 1.4 Function Notation 1.4.3 Answers 1. f (x) = 2x+3 4 Domain: (−∞, ∞) 69 2. f (x) = 2(x+3) 4 = x+3 2 Domain: (−∞, ∞) 3. f (x Domain: (−∞, ∞) √ 4. f (x) = 2x + 3 Domain: − 3 2, ∞ 5. f (x) = 2(x + 3) = √ 2x + 6 Domain: [−3, ∞) 7. f (x) = 4√ x−13 Domain: [0, 169) ∪ (169, ∞) 6. f (x) = 2 √ x + 3 Domain: [−3, ∞) 8. f (x) = 4√ x−13 Domain: (13, ∞) 9. f (x) = 4√ x − 13 Domain: (0, ∞) 11. For f (x) = 2x + 1 10. f (x) = 4 x − 13 = 2√ x − 13 Domain: (0, ∞) f (3) = 7 f (−1) = −1 f 3 2 = 4 f (4x) = 8x + 1 4f (x) = 8x + 4 f (−x) = −2x + 1 f (x − 4) = 2x − 7 f (x) − 4 = 2x − 3 f x2 = 2x2 + 1 12. For f (x) = 3 − 4x f (3) = −9 f (−1) = 7 f 3 2 = −3 f (4x) = 3 − 16x 4f (x) = 12 − 16x f (−x) = 4x + 3 f (x − 4) = 19 − 4x f (x) − 4 = −4x − 1 f x2 = 3 − 4x2 70 Relations and Functions 13. For f (x) = 2 − x2 f (3) = −7 f (−14x) = 2 − 16x2 4f (x) = 8 − 4x2 f (−x) = 2 −
x2 f (x − 4) = −x2 + 8x − 14 f (x) − 4 = −x2 − 2 f x2 = 2 − x4 14. For f (x) = x2 − 3x + 2 f (3) = 2 f (−14x) = 16x2 − 12x + 2 4f (x) = 4x2 − 12x + 8 f (−x) = x2 + 3x + 2 f (x − 4) = x2 − 11x + 30 f (x) − 4 = x2 − 3x − 2 f x2 = x4 − 3x2 + 2 15. For f (x) = x x−1 f (3) = 3 2 f (4x) = 4x 4x−1 f (x − 4) = x−4 x−5 16. For f (x) = 2 x3 f (3) = 2 27 f (4x) = 1 32x3 f (x − 4) = 2 (x−4)3 = 2 x3−12x2+48x−64 17. For f (x) = 6 f (3) = 6 f (4x) = 6 f (−1) = 1 2 4f (x) = 4x x−1 f (x) − 4 = x x−1 − 4 = 4−3x x−1 f (−1) = −2 4f (x) = 8 x3 f (x) − 4 = 2 x3 − 4 = 2−4x3 x3 f (−1) = 6 4f (x) = 24 f (x − 4) = 6 f (x (−x) = x x+1 f x2 = x2 x2−1 f 3 2 = 16 27 f (−x) = − 2 x3 f x2 = 2 x6 f 3 2 = 6 f (−x) = 6 f x2 = 6 1.4 Function Notation 71 18. For f (x) = 0 f (3) = 0 f (−1) = 0 f (4x) = 0 4f (x) = 0 f (x − 4) = 0 f (x) − 4 = −4 f 3 2 = 0 f (−x) = 0 f x2 = 0 19. For f (x) = 2x − 5 f (2) = −1 f (−2
) = −9 f (2a) = 4a − 5 2f (a) = 4a − 10 f (a + 2) = 2a − 1 f (a) + f (2) = 2a − −5a a 20. For f (x) = 5 − 2x f (a) 2 = 2a−5 2 f (a + h) = 2a + 2h − 5 f (2) = 1 f (−2) = 9 f (2a) = 5 − 4a 2f (a) = 10 − 4a f (a + 2) = 1 − 2a f (a) + f (2) = 6 − 2a f 2 a = 5 − 4 a = 5a−4 a 21. For f (x) = 2x2 − 1 f (a) 2 = 5−2a 2 f (a + h) = 5 − 2a − 2h f (2) = 7 f (−2) = 7 f (2a) = 8a2 − 1 2f (a) = 4a2 − 2 f (a + 2) = 2a2 + 8a + 7 f (a) + f (2) = 2a2 + 6 f 2 a = 8 a2 − 1 = 8−a2 a2 f (a) 2 = 2a2−1 2 f (a + h) = 2a2 + 4ah + 2h2 − 1 72 Relations and Functions 22. For f (x) = 3x2 + 3x − 2 f (2) = 16 f (−2) = 4 f (2a) = 12a2 + 6a − 2 2f (a) = 6a2 + 6a − 4 f (a+2) = 3a2 +15a+16 f (a) + f (2) = 3a2 + 3a + 14 f 2 a = 12 a2 + 6 a − 2 = 12+6a−2a2 a2 23. For f (x) = √ 2x + 1 f (a) 2 = 3a2+3a−2 2 f (a + h) = 3a2 + 6ah + 3h2 + 3a + 3h − 2 f (2) = √ 5 f (−2) is not real f (2a) = √ 4a + 1 2f (a)
= 2 √ 2a + 1 f (a + 2) = √ 2a + 5 f (a)+f (2) = √ 2a + 1+ √ 5 f (a) 2 = √ 2a+1 2 f (a+h) = √ 2a + 2h + +4 a 24. For f (x) = 117 f (2) = 117 f (−2) = 117 f (2a) = 117 2f (a) = 234 f (a + 2) = 117 f (a) + f (2) = 234 f 2 a = 117 25. For f (x) = x 2 f (2) = 1 2f (aa) 2 = 117 2 f (a + h) = 117 f (−2) = −1 f (2a) = a f (a + 2) = a+2 2 f (a) 2 = a 4 f (a) + f (2) = a 2 + 1 = a+2 2 f (a + h) = a+h 2 1.4 Function Notation 73 26. For f (x) = 2 x f (2) = 1 2f (a (−2) = −1 f (a + 2) = 2 a+2 f (a) 2 = 1 a f (2a) = 1 a f (a) + f (2) = 2 a + 1 = a+2 2 f (a + h) = 2 a+h 27. For f (x) = 2x − 1, f (0) = −1 and f (x) = 0 when x = 1 2 28. For f (x) = 3 − 2 5 x, f (0) = 3 and f (x) = 0 when x = 15 2 29. For f (x) = 2x2 − 6, f (0) = −6 and f (x) = 0 when x = ± √ 3 30. For f (x) = x2 − x − 12, f (0) = −12 and f (x) = 0 when x = −3 or x = 4 31. For f (x) = 32. For f (x) = √ √ x + 4, f (0) = 2 and f (x) = 0 when x = −4 1 − 2x, f (0) = 1 and f (x)
= 0 when x = 1 2 33. For f (x) = 3 4−x, f (0) = 3 4 and f (x) is never equal to 0 34. For f (x) = 3x2−12x 4−x2, f (0) = 0 and f (x) = 0 when x = 0 or x = 4 35. (a) f (−4) = 1 (b) f (−3) = 2 (d) f (3.001) = 1.999 (e) f (−3.001) = 1.999 36. (a) f (4) = 4 (d) f (0) = 1 37. (−∞, ∞) 39. (−∞, −1) ∪ (−1, ∞) 41. (−∞, ∞) (b) f (−3) = 9 (e) f (−1) = 1 (c) f (3) = 0 √ (f) f (2) = 5 (c) f (1) = 0 (f) f (−0.999) ≈ 0.0447 38. (−∞, ∞) 40. (−∞, −2) ∪ (−2, 1) ∪ (1, ∞) 42. (−∞, − √ √ 3) ∪ (− √ 3, 3) ∪ ( √ 3, ∞) 43. (−∞, −6) ∪ (−6, 6) ∪ (6, ∞) 44. (−∞, 2) ∪ (2, ∞) 45. (−∞, 3] 46. − 5 2, ∞ 74 47. [−3, ∞) 49. 1 3, ∞ 51. (−∞, ∞) 53. 1 3, 6 ∪ (6, ∞) 55. (−∞, 8) ∪ (8, ∞) 57. (8, ∞) 59. (−∞, 8) ∪ (8, ∞) 61. [0, 5) ∪ (5, ∞) Relations and Functions 48. (−∞, 7] 50. 1 523, ∞) 54. (−∞, ∞) 56. [0, 8) ∪ (8, ∞) 58. [7, 9] 60. −�
�, −, 1 2 ∪ 1 2, ∞ 62. [0, 25) ∪ (25, ∞) 63. A(3) = 9, so the area enclosed by a square with a side of length 3 inches is 9 square inches. The solutions to A(x) = 36 are x = ±6. Since x is restricted to x > 0, we only keep x = 6. This means for the area enclosed by the square to be 36 square inches, the length of the side needs to be 6 inches. Since x represents a length, x > 0. 64. A(2) = 4π, so the area enclosed by a circle with radius 2 meters is 4π square meters. The solutions to A(r) = 16π are r = ±4. Since r is restricted to r > 0, we only keep r = 4. This means for the area enclosed by the circle to be 16π square meters, the radius needs to be 4 meters. Since r represents a radius (length), r > 0. 65. V (5) = 125, so the volume enclosed by a cube with a side of length 5 centimeters is 125 cubic centimeters. The solution to V (x) = 27 is x = 3. This means for the volume enclosed by the cube to be 27 cubic centimeters, the length of the side needs to 3 centimeters. Since x represents a length, x > 0. 66. V (3) = 36π, so the volume enclosed by a sphere with radius 3 feet is 36π cubic feet. The is r = 2. This means for the volume enclosed by the sphere to be 32π 3 solution to V (r) = 32π 3 cubic feet, the radius needs to 2 feet. Since r represents a radius (length), r > 0. 67. h(0) = 64, so at the moment the object is dropped oﬀ the building, the object is 64 feet oﬀ of the ground. The solutions to h(t) = 0 are t = ±2. Since we restrict 0 ≤ t ≤ 2, we only keep t = 2. This means 2 seconds after the object is dropped oﬀ the building, it is 0 feet oﬀ the ground. Said diﬀerently, the object hits the ground after 2 seconds. The restriction 0 ≤ t ≤ 2 restricts the time to be between the moment the object is released and the moment it hits the
ground. 68. T (0) = 3, so at 6 AM (0 hours after 6 AM), it is 3◦ Fahrenheit. T (6) = 33, so at noon (6 hours after 6 AM), the temperature is 33◦ Fahrenheit. T (12) = 27, so at 6 PM (12 hours after 6 AM), it is 27◦ Fahrenheit. 1.4 Function Notation 75 69. C(0) = 27, so to make 0 pens, it costs15 $2700. C(2) = 11, so to make 2000 pens, it costs $1100. C(5) = 2, so to make 5000 pens, it costs $2000. 70. F (0) = 16.00, so in 1980 (0 years after 1980), the average fuel economy of passenger cars in the US was 16.00 miles per gallon. F (14) = 20.81, so in 1994 (14 years after 1980), the average fuel economy of passenger cars in the US was 20.81 miles per gallon. F (28) = 22.64, so in 2008 (28 years after 1980), the average fuel economy of passenger cars in the US was 22.64 miles per gallon. 71. P (0) = 0 which means in 1803 (0 years after 1803), there are no Sasquatch in Portage County. 22 ≈ 139.77, so in 2008 (205 years after 1803), there were between 139 and 140 P (205) = 3075 Sasquatch in Portage County. 72. (a) C(20) = 300. It costs $300 for 20 copies of the book. (b) C(50) = 675, so it costs $675 for 50 copies of the book. C(51) = 612, so it costs $612 for 51 copies of the book. (c) 56 books. 73. (a) S(10) = 17.5, so it costs $17.50 to ship 10 comic books. (b) There is free shipping on orders of 15 or more comic books. 74. (a) C(750) = 25, so it costs $25 to talk 750 minutes per month with this plan. (b) Since 20 hours = 1200 minutes, we substitute m = 1200 and get C(1200) = 45. It costs $45 to talk 20 hours per month with this plan. (c) It costs $
25 for up to 1000 minutes and 10 cents per minute for each minute over 1000 minutes. 75. (a) 0.785 = 0, 117 = 117, −2.001 = −3, and π + 6 = 9 15This is called the ‘ﬁxed’ or ‘start-up’ cost. We’ll revisit this concept on page 82. 76 Relations and Functions 1.5 Function Arithmetic In the previous section we used the newly deﬁned function notation to make sense of expressions such as ‘f (x) + 2’ and ‘2f (x)’ for a given function f. It would seem natural, then, that functions should have their own arithmetic which is consistent with the arithmetic of real numbers. The following deﬁnitions allow us to add, subtract, multiply and divide functions using the arithmetic we already know for real numbers. Function Arithmetic Suppose f and g are functions and x is in both the domain of f and the domain of g.a The sum of f and g, denoted f + g, is the function deﬁned by the formula (f + g)(x) = f (x) + g(x) The diﬀerence of f and g, denoted f − g, is the function deﬁned by the formula (f − g)(x) = f (x) − g(x) The product of f and g, denoted f g, is the function deﬁned by the formula (f g)(x) = f (x)g(x) The quotient of f and g, denoted, is the function deﬁned by the formula f g f g (x) = f (x) g(x), provided g(x) = 0. aThus x is an element of the intersection of the two domains. In other words, to add two functions, we add their outputs; to subtract two functions, we subtract their outputs, and so on. Note that while the formula (f + g)(x) = f (x) + g(x) looks suspiciously like some kind of distributive property, it is nothing of the sort; the addition on the left hand side of the equation is function addition, and we are using this equation to deﬁne the output of the new function f + g as the sum of
the real number outputs from f and g. Example 1.5.1. Let f (x) = 6x2 − 2x and g(x) = 3 − 1 x. 1. Find (f + g)(−1) 2. Find (f g)(2) 3. Find the domain of g − f then ﬁnd and simplify a formula for (g − f )(x). 1.5 Function Arithmetic 77 4. Find the domain of g f then ﬁnd and simplify a formula for g f (x). Solution. 1. To ﬁnd (f + g)(−1) we ﬁrst ﬁnd f (−1) = 8 and g(−1) = 4. By deﬁnition, we have that (f + g)(−1) = f (−1) + g(−1) = 8 + 4 = 12. 2. To ﬁnd (f g)(2), we ﬁrst need f (2) and g(2). Since f (2) = 20 and g(2) = 5 2, our formula yields (f g)(2) = f (2)g(2) = (20) 5 2 = 50. 3. One method to ﬁnd the domain of g−f is to ﬁnd the domain of g and of f separately, then ﬁnd the intersection of these two sets. Owing to the denominator in the expression g(x) = 3 − 1 x, we get that the domain of g is (−∞, 0) ∪ (0, ∞). Since f (x) = 6x2 − 2x is valid for all real numbers, we have no further restrictions. Thus the domain of g − f matches the domain of g, namely, (−∞, 0) ∪ (0, ∞). A second method is to analyze the formula for (g − f )(x) before simplifying and look for the usual domain issues. In this case, (g − f )(x) = g(x) − f (x) = 3 − 1 x − 6x2 − 2x, so we ﬁnd, as before, the domain is (−∞, 0) ∪ (0, ∞). Moving along, we need to simplify a formula for (g − f )(x
). In this case, we get common denominators and attempt to reduce the resulting fraction. Doing so, we get (g − f )(x) = g(x) − f (x) = 3 − 1 x − 6x2 − 2x = 3 − 1 x − 6x2 + 2x get common denominators = = = + − − 1 x 6x3 x 2x2 3x x x 3x − 1 − 6x3 − 2x2 x −6x3 − 2x2 + 3x − 1 x 4. As in the previous example, we have two ways to approach ﬁnding the domain of g f. First, we can ﬁnd the domain of g and f separately, and ﬁnd the intersection of these two sets. In f (x), we are introducing a new denominator, namely f (x), so we addition, since need to guard against this being 0 as well. Our previous work tells us that the domain of g is (−∞, 0) ∪ (0, ∞) and the domain of f is (−∞, ∞). Setting f (x) = 0 gives 6x2 − 2x = 0 (x) = g(x) g f 78 Relations and Functions 3. As a result, the domain of g f is all real numbers except x = 0 and x = 1 3, or or x = 0, 1 (−∞, 0) ∪ 0, 1 3 ∪ 1 3, ∞. Alternatively, we may proceed as above and analyze the expression simplifying. In this case, g f (x) = g(x) f (x) before g f (x) = g(x) f (x) = 3 − 1 x 6x2 − 2x We see immediately from the ‘little’ denominator that x = 0. To keep the ‘big’ denominator away from 0, we solve 6x2 − 2x = 0 and get x = 0 or x = 1 3. Hence, as before, we ﬁnd the domain of g f to be (−∞, 0) ∪ 0, 1 ∪ 1 3 Next, we ﬁnd and simplify a formula for (x). 3, ∞. g f g f (x(x) f (x) 3 − 1 x 6x2 − 2x 3 − 1 x
· x x x 3 − 6x2 − 2x 1 x (6x2 − 2x) x 3x − 1 (6x2 − 2x) x 3x − 1 2x2(3x − 1) 1 (3x − 1) 2x2 (3x − 1) 1 2x2 simplify compound fractions factor cancel Please note the importance of ﬁnding the domain of a function before simplifying its expression. In number 4 in Example 1.5.1 above, had we waited to ﬁnd the domain of g f until after simplifying, we’d just have the formula 1 2x2 to go by, and we would (incorrectly!) state the domain as (−∞, 0)∪(0, ∞), since the other troublesome number, x = 1 3, was canceled away.1 1We’ll see what this means geometrically in Chapter 4. 1.5 Function Arithmetic 79 Next, we turn our attention to the diﬀerence quotient of a function. Deﬁnition 1.8. Given a function f, the diﬀerence quotient of f is the expression f (x + h) − f (x) h We will revisit this concept in Section 2.1, but for now, we use it as a way to practice function notation and function arithmetic. For reasons which will become clear in Calculus, ‘simplifying’ a diﬀerence quotient means rewriting it in a form where the ‘h’ in the deﬁnition of the diﬀerence quotient cancels from the denominator. Once that happens, we consider our work to be done. Example 1.5.2. Find and simplify the diﬀerence quotients for the following functions 1. f (x) = x2 − x − 2 Solution. 2. g(x) = 3 2x + 1 3. r(x) = √ x 1. To ﬁnd f (x + h), we replace every occurrence of x in the formula f (x) = x2 − x − 2 with the quantity (x + h) to get f (x + h) = (x + h)2 − (x + h) − 2 = x2 + 2xh + h2 − x − h − 2. So the
diﬀerence quotient is f (x + h) − f (x) h = = = = = x2 + 2xh + h2 − x − h − 2 − x2 − x − 2 h x2 + 2xh + h2 − x − h − 2 − x2 + x + 2 h 2xh + h2 − h h h (2x + h − 1) h h (2x + h − 1) h factor cancel = 2x + h − 1. 80 Relations and Functions 2. To ﬁnd g(x + h), we replace every occurrence of x in the formula g(x) = 3 2x+1 with the quantity (x + h) to get which yields g(x + h) − g(x) h g(x + h) = = 3 2(x + h) + 1 3 2x + 2h + 1, 3 2x + 2h + 1 h 3 2x + 2h + 1 h − − 3 2x + 1 3 2x + 1 · (2x + 2h + 1)(2x + 1) (2x + 2h + 1)(2x + 1) 3(2x + 1) − 3(2x + 2h + 1) h(2x + 2h + 1)(2x + 1) 6x + 3 − 6x − 6h − 3 h(2x + 2h + 1)(2x + 1) −6h h(2x + 2h + 1)(2x + 1) −6h h(2x + 2h + 1)(2x + 1) −6 (2x + 2h + 1)(2x + 1). = = = = = = = Since we have managed to cancel the original ‘h’ from the denominator, we are done. 3. For r(x) = √ x, we get r(x + h) = √ x + h so the diﬀerence quotient is r(x + h) − r(x In order to cancel the ‘h’ from the denominator, we rationalize the numerator by multiplying by its conjugate.2 2Rationalizing the numerator!? How’s that for a twist! 1.5 Function Arithmetic 81 r(x + h) − r(x Multiply
by the conjugate. Diﬀerence of Squares. √ x + h2 )2 x (x + h Since we have removed the original ‘h’ from the denominator, we are done. As mentioned before, we will revisit diﬀerence quotients in Section 2.1 where we will explain them geometrically. For now, we want to move on to some classic applications of function arithmetic from Economics and for that, we need to think like an entrepreneur.3 Suppose you are a manufacturer making a certain product.4 Let x be the production level, that is, the number of items produced in a given time period. It is customary to let C(x) denote the function which calculates the total cost of producing the x items. The quantity C(0), which represents the cost of producing no items, is called the ﬁxed cost, and represents the amount of money required to begin production. Associated with the total cost C(x) is cost per item, or average cost, denoted C(x) and read ‘C-bar’ of x. To compute C(x), we take the total cost C(x) and divide by the number of items produced x to get C(x) = C(x) x On the retail end, we have the price p charged per item. To simplify the dialog and computations in this text, we assume that the number of items sold equals the number of items produced. From a 3Not really, but “entrepreneur” is the buzzword of the day and we’re trying to be trendy. 4Poorly designed resin Sasquatch statues, for example. Feel free to choose your own entrepreneurial fantasy. 82 Relations and Functions retail perspective, it seems natural to think of the number of items sold, x, as a function of the price charged, p. After all, the retailer can easily adjust the price to sell more product. In the language of functions, x would be the dependent variable and p would be the independent variable or, using function notation, we have a function x(p). While we will adopt this convention later in the text,5 we will hold with tradition at this point and consider the price p as a function of the number of items sold, x. That is, we regard x as the independent variable and p as the dependent variable and speak of the price-demand function, p(x). Hence
, p(x) returns the price charged per item when x items are produced and sold. Our next function to consider is the revenue function, R(x). The function R(x) computes the amount of money collected as a result of selling x items. Since p(x) is the price charged per item, we have R(x) = xp(x). Finally, the proﬁt function, P (x) calculates how much money is earned after the costs are paid. That is, P (x) = (R − C)(x) = R(x) − C(x). We summarize all of these functions below. Summary of Common Economic Functions Suppose x represents the quantity of items produced and sold. The price-demand function p(x) calculates the price per item. The revenue function R(x) calculates the total money collected by selling x items at a price p(x), R(x) = x p(x). The cost function C(x) calculates the cost to produce x items. The value C(0) is called the ﬁxed cost or start-up cost. The average cost function C(x) = C(x) x Here, we necessarily assume x > 0. calculates the cost per item when making x items. The proﬁt function P (x) calculates the money earned after costs are paid when x items are produced and sold, P (x) = (R − C)(x) = R(x) − C(x). It is high time for an example. Example 1.5.3. Let x represent the number of dOpi media players (‘dOpis’6) produced and sold in a typical week. Suppose the cost, in dollars, to produce x dOpis is given by C(x) = 100x + 2000, for x ≥ 0, and the price, in dollars per dOpi, is given by p(x) = 450 − 15x for 0 ≤ x ≤ 30. 1. Find and interpret C(0). 2. Find and interpret C(10). 3. Find and interpret p(0) and p(20). 4. Solve p(x) = 0 and interpret the result. 5. Find and simplify expressions for the revenue function R(x) and the proﬁt function P (x). 6. Find and interpret R(0) and P (0). 7
. Solve P (x) = 0 and interpret the result. 5See Example 5.2.4 in Section 5.2. 6Pronounced ‘dopeys’... 1.5 Function Arithmetic 83 Solution. 1. We substitute x = 0 into the formula for C(x) and get C(0) = 100(0) + 2000 = 2000. This means to produce 0 dOpis, it costs $2000. In other words, the ﬁxed (or start-up) costs are $2000. The reader is encouraged to contemplate what sorts of expenses these might be. 2. Since C(x) = C(x) x, C(10) = C(10) 10 = 3000 10 = 300. This means when 10 dOpis are produced, the cost to manufacture them amounts to $300 per dOpi. 3. Plugging x = 0 into the expression for p(x) gives p(0) = 450 − 15(0) = 450. This means no dOpis are sold if the price is $450 per dOpi. On the other hand, p(20) = 450 − 15(20) = 150 which means to sell 20 dOpis in a typical week, the price should be set at $150 per dOpi. 4. Setting p(x) = 0 gives 450 − 15x = 0. Solving gives x = 30. This means in order to sell 30 dOpis in a typical week, the price needs to be set to $0. What’s more, this means that even if dOpis were given away for free, the retailer would only be able to move 30 of them.7 5. To ﬁnd the revenue, we compute R(x) = xp(x) = x(450 − 15x) = 450x − 15x2. Since the formula for p(x) is valid only for 0 ≤ x ≤ 30, our formula R(x) is also restricted to 0 ≤ x ≤ 30. For the proﬁt, P (x) = (R − C)(x) = R(x) − C(x). Using the given formula for C(x) and the derived formula for R(x), we get P (x) = 450x − 15x2−(100x+2000) = −15x2+350x−2000. As
before, the validity of this formula is for 0 ≤ x ≤ 30 only. 6. We ﬁnd R(0) = 0 which means if no dOpis are sold, we have no revenue, which makes sense. Turning to proﬁt, P (0) = −2000 since P (x) = R(x)−C(x) and P (0) = R(0)−C(0) = −2000. This means that if no dOpis are sold, more money ($2000 to be exact!) was put into producing the dOpis than was recouped in sales. In number 1, we found the ﬁxed costs to be $2000, so it makes sense that if we sell no dOpis, we are out those start-up costs. 7. Setting P (x) = 0 gives −15x2 + 350x − 2000 = 0. Factoring gives −5(x − 10)(3x − 40) = 0 so x = 10 or x = 40 3. What do these values mean in the context of the problem? Since P (x) = R(x) − C(x), solving P (x) = 0 is the same as solving R(x) = C(x). This means that the solutions to P (x) = 0 are the production (and sales) ﬁgures for which the sales revenue exactly balances the total production costs. These are the so-called ‘break even’ points. The solution x = 10 means 10 dOpis should be produced (and sold) during the week to recoup the cost of production. For x = 40 3 = 13.3, things are a bit more complicated. Even though x = 13.3 satisﬁes 0 ≤ x ≤ 30, and hence is in the domain of P, it doesn’t make sense in the context of this problem to produce a fractional part of a dOpi.8 Evaluating P (13) = 15 and P (14) = −40, we see that producing and selling 13 dOpis per week makes a (slight) proﬁt, whereas producing just one more puts us back into the red. While breaking even is nice, we ultimately would like to ﬁnd what production level (and price) will result in the largest proﬁt, and we’ll do just
that... in Section 2.3. 7Imagine that! Giving something away for free and hardly anyone taking advantage of it... 8We’ve seen this sort of thing before in Section 1.4.1. 84 1.5.1 Exercises Relations and Functions In Exercises 1 - 10, use the pair of functions f and g to ﬁnd the following values if they exist. (f + g)(2) (f g) 1 2 (f − g)(−1) f g (0) (g − f )(1) g f (−2) 1. f (x) = 3x + 1 and g(x) = 4 − x 2. f (x) = x2 and g(x) = −2x + 1 3. f (x) = x2 − x and g(x) = 12 − x2 4. f (x) = 2x3 and g(x) = −x2 − 2x − 3 √ 5. f (x) = x + 3 and g(x) = 2x − 1 √ 6. f (x) = 4 − x and g(x) = √ x + 2 7. f (x) = 2x and g(x) = 9. f (x) = x2 and g(x) = 1 2x + 1 1 x2 8. f (x) = x2 and g(x) = 3 2x − 3 10. f (x) = x2 + 1 and g(x) = 1 x2 + 1 In Exercises 11 - 20, use the pair of functions f and g to ﬁnd the domain of the indicated function then ﬁnd and simplify an expression for it. (f + g)(x) (f − g)(x) (f g)(x) f g (x) 11. f (x) = 2x + 1 and g(x) = x − 2 12. f (x) = 1 − 4x and g(x) = 2x − 1 13. f (x) = x2 and g(x) = 3x − 1 14. f (x) = x2 − x and g(x) = 7x 15. f (x) = x2 − 4 and g(x) = 3x + 6 16. f (x) = −x2
+ x + 6 and g(x) = x2 − 9 17. f (x) = x 2 and g(x) = 19. f (x) = x and g(x) = 2 x √ x + 1 18. f (x) = x − 1 and g(x) = 1 x − 1 20. f (x) = √ x − 5 and g(x) = f (x) = √ x − 5 In Exercises 21 - 45, ﬁnd and simplify the diﬀerence quotient f (x + h) − f (x) h for the given function. 21. f (x) = 2x − 5 23. f (x) = 6 25. f (x) = −x2 + 2x − 1 22. f (x) = −3x + 5 24. f (x) = 3x2 − x 26. f (x) = 4x2 1.5 Function Arithmetic 85 27. f (x) = x − x2 28. f (x) = x3 + 1 29. f (x) = mx + b where m = 0 30. f (x) = ax2 + bx + c where a = 0 31. f (x) = 33. f (x) = 2 x 1 x2 35. f (x) = 37. f (x) = 1 4x − 3 x x − 9 √ √ √ 39. f (x) = 41. f (x) = 43. f (x) = x − 9 −4x + 5 ax + b, where a = 0. 32. f (x) = 34. f (x) = 36. f (x 3x x + 1 38. f (x) = 40. f (x) = 42. f (x) = x2 2x + 1 √ 2x + 1 √ 4 − x 44. f (x) = x √ x √ 45. f (x) = 3 x. HINT: (a − b) a2 + ab + b2 = a3 − b3 In Exercises 46 - 50, C(x) denotes the cost to produce x items and p(x) denotes the price-demand function in the given economic scenario. In each Exercise, do the following: Find and interpret C(0). Find
and interpret C(10). Find and interpret p(5) Find and simplify R(x). Find and simplify P (x). Solve P (x) = 0 and interpret. 46. The cost, in dollars, to produce x “I’d rather be a Sasquatch” T-Shirts is C(x) = 2x + 26, x ≥ 0 and the price-demand function, in dollars per shirt, is p(x) = 30 − 2x, 0 ≤ x ≤ 15. 47. The cost, in dollars, to produce x bottles of 100% All-Natural Certiﬁed Free-Trade Organic Sasquatch Tonic is C(x) = 10x + 100, x ≥ 0 and the price-demand function, in dollars per bottle, is p(x) = 35 − x, 0 ≤ x ≤ 35. 48. The cost, in cents, to produce x cups of Mountain Thunder Lemonade at Junior’s Lemonade Stand is C(x) = 18x + 240, x ≥ 0 and the price-demand function, in cents per cup, is p(x) = 90 − 3x, 0 ≤ x ≤ 30. 49. The daily cost, in dollars, to produce x Sasquatch Berry Pies C(x) = 3x + 36, x ≥ 0 and the price-demand function, in dollars per pie, is p(x) = 12 − 0.5x, 0 ≤ x ≤ 24. 86 Relations and Functions 50. The monthly cost, in hundreds of dollars, to produce x custom built electric scooters is C(x) = 20x + 1000, x ≥ 0 and the price-demand function, in hundreds of dollars per scooter, is p(x) = 140 − 2x, 0 ≤ x ≤ 70. In Exercises 51 - 62, let f be the function deﬁned by f = {(−3, 4), (−2, 2), (−1, 0), (0, 1), (1, 3), (2, 4), (3, −1)} and let g be the function deﬁned g = {(−3, −2), (−2, 0), (−1, −4), (0, 0), (1, −3), (2, 1), (3, 2)}. Compute the indicated value if it exists. 51. (f +
g)(−3) 54. (g + f )(1) 57. 60. f g g f (−2) (−1) 52. (f − g)(2) 55. (g − f )(3) 58. 61. f g g f (−1) (3) 53. (f g)(−1) 56. (gf )(−3) 59. 62. f g g f (2) (−3) 1.5 Function Arithmetic 87 1.5.2 Answers 1. For f (x) = 3x + 1 and g(x) = 4 − x (f + g)(2) = 9 (f g) 1 2 = 35 4 (f − g)(−1) = −7 (g − f )(1) = −1 f g (0) = 1 4 g f (−2) = − 6 5 2. For f (x) = x2 and g(x) = −2x + 1 (f + g)(2) = 1 (f − g)(−1) = −2 (g − f )(1) = −2 (f g) 1 2 = 0 f g (0) = 0 g f (−2) = 5 4 3. For f (x) = x2 − x and g(x) = 12 − x2 (f + g)(2) = 10 (f − g)(−1) = −9 (g − f )(1) = 11 (f g) 1 2 = − 47 16 f g (0) = 0 g f (−2) = 4 3 4. For f (x) = 2x3 and g(x) = −x2 − 2x − 3 (f + g)(2) = 5 (f g) 1 2 = − 17 16 (f − g)(−1) = 0 (g − f )(1) = −8 f g (0) = 0 g f (−2) = 3 16 5. For f (x) = √ x + 3 and g(x) = 2x − 1 (f + g)(2) = 3 + √ 5 (f − g)(−1) = 3 + √ 2 (f g) 1 2 = 0 f g √ (0) = − 3 (g − f )(1) = −1 g f (−2) = −5 6. For f (x
) = √ 4 − x and g(x) = √ (f + g)(2f − g)(−1) = −1 + √ 5 (f g) 1 2 = √ 35 2 f g √ 2 (0) = (g − f )(1) = 0 g f (−2) = 0 88 Relations and Functions 7. For f (x) = 2x and g(x) = 1 2x+1 (f + g)(2) = 21 5 (f g) 1 2 = 1 2 (f − g)(−1) = −1 f g (0) = 0 (g − f )(1) = − 5 3 g f (−2) = 1 12 8. For f (x) = x2 and g(x) = 3 2x−3 (f + g)(2) = 7 (f g) 1 2 = − 3 8 (f − g)(−1) = 8 5 f g (0) = 0 (g − f )(1) = −4 g f (−2) = − 3 28 9. For f (x) = x2 and g(x) = 1 x2 (f + g)(2) = 17 4 (f g) 1 2 = 1 (f − g)(−1) = 0 f g (0) is undeﬁned. (g − f )(1) = 0 g f (−2) = 1 16 10. For f (x) = x2 + 1 and g(x) = 1 x2+1 (f + g)(2) = 26 5 (f g) 1 2 = 1 (f − g)(−1) = 3 2 f g (0) = 1 (g − f )(1) = − 3 2 g f (−2) = 1 25 11. For f (x) = 2x + 1 and g(x) = x − 2 (f + g)(x) = 3x − 1 Domain: (−∞, ∞) (f g)(x) = 2x2 − 3x − 2 Domain: (−∞, ∞) 12. For f (x) = 1 − 4x and g(x) = 2x − 1 (f + g)(x) = −2x Domain: (−∞, ∞) (f g)(x) = −8x2 + 6x
− 1 Domain: (−∞, ∞) (f − g)(x) = x + 3 Domain: (−∞, ∞) f g (x) = 2x+1 x−2 Domain: (−∞, 2) ∪ (2, ∞) (f − g)(x) = 2 − 6x Domain: (−∞, ∞) f (x) = 1−4x g 2x−1 Domain: −∞, 1 2 ∪ 1 2, ∞ 1.5 Function Arithmetic 89 13. For f (x) = x2 and g(x) = 3x − 1 (f + g)(x) = x2 + 3x − 1 Domain: (−∞, ∞) (f g)(x) = 3x3 − x2 Domain: (−∞, ∞) 14. For f (x) = x2 − x and g(x) = 7x (f + g)(x) = x2 + 6x Domain: (−∞, ∞) (f g)(x) = 7x3 − 7x2 Domain: (−∞, ∞) (f − g)(x) = x2 − 3x + 1 Domain: (−∞, ∞) f (x) = x2 3x−1 g Domain: −∞, 1 3 ∪ 1 3, ∞ (f − g)(x) = x2 − 8x Domain: (−∞, ∞) f g (x) = x−1 7 Domain: (−∞, 0) ∪ (0, ∞) 15. For f (x) = x2 − 4 and g(x) = 3x + 6 (f + g)(x) = x2 + 3x + 2 Domain: (−∞, ∞) (f g)(x) = 3x3 + 6x2 − 12x − 24 Domain: (−∞, ∞) 16. For f (x) = −x2 + x + 6 and g(x) = x2 − 9 (f − g)(x) = x2 − 3x − 10 Domain: (−∞, ∞) f g (x) = x−2 3 Domain: (−∞, −2) ∪ (−2, ∞) (f + g)(x) = x − 3 Domain: (−
∞, ∞) (f g)(x) = −x4 + x3 + 15x2 − 9x − 54 Domain: (−∞, ∞) (f − g)(x) = −2x2 + x + 15 Domain: (−∞, ∞) f g (x) = − x+2 x+3 Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 17. For f (x) = x 2 and g(x) = 2 x (f + g)(x) = x2+4 2x Domain: (−∞, 0) ∪ (0, ∞) (f g)(x) = 1 Domain: (−∞, 0) ∪ (0, ∞) (f − g)(x) = x2−4 2x Domain: (−∞, 0) ∪ (0, ∞) f g (x) = x2 4 Domain: (−∞, 0) ∪ (0, ∞) 90 Relations and Functions 18. For f (x) = x − 1 and g(x) = 1 x−1 (f + g)(x) = x2−2x+2 x−1 Domain: (−∞, 1) ∪ (1, ∞) (f − g)(x) = x2−2x x−1 Domain: (−∞, 1) ∪ (1, ∞) (f g)(x) = 1 Domain: (−∞, 1) ∪ (1, ∞) f g (x) = x2 − 2x + 1 Domain: (−∞, 1) ∪ (1, ∞) √ x + 1 19. For f (x) = x and g(x) = √ (f + g)(x) = x + Domain: [−1, ∞) x + 1 (f − g)(x) = x − Domain: [−1, ∞) √ x + 1 (f g)(x) = x √ x + 1 Domain: [−1, ∞) 20. For f (x) = √ x − 5 and g(x) = f (x) = √ x − 5 f g (x) = x√ Domain: (−1, ∞) x+1 √
(f + g)(x) = 2 Domain: [5, ∞) x − 5 (f g)(x) = x − 5 Domain: [5, ∞) 21. 2 23. 0 25. −2x − h + 2 27. −2x − h + 1 29. m 31. 33. 35. −2 x(x + h) −(2x + h) x2(x + h)2 −4 (4x − 3)(4x + 4h − 3) (f − g)(x) = 0 Domain: [5, ∞) f g (x) = 1 Domain: (5, ∞) 22. −3 24. 6x + 3h − 1 26. 8x + 4h 28. 3x2 + 3xh + h2 30. 2ax + ah + b 32. 34. 36. 3 (1 − x − h)(1 − x) −2 (x + 5)(x + h + 5) 3 (x + 1)(x + h + 1) 1.5 Function Arithmetic 91 37. −9 (x − 9)(x + h − 9) 39. √ 41. √ 434 −4x − 4h + 5 + √ −4x + 5 a ax + ah + b + √ ax + b 38. 2x2 + 2xh + 2x + h (2x + 1)(2x + 2h + 1) 40. √ 42. √ 2 2x + 2h + 1 + √ 2x + 1 − 44. 3x2 + 3xh + h2 (x + h)3/2 + x3/2 45. 46. 47. 48. 1 (x + h)2/3 + (x + h)1/3x1/3 + x2/3 C(0) = 26, so the ﬁxed costs are $26. C(10) = 4.6, so when 10 shirts are produced, the cost per shirt is $4.60. p(5) = 20, so to sell 5 shirts, set the price at $20 per shirt. R(x) = −2x2 + 30x, 0 ≤ x ≤ 15 P (x) = −2x2 + 28x − 26, 0 ≤ x ≤ 15 P (x) = 0 when x =
1 and x = 13. These are the ‘break even’ points, so selling 1 shirt or 13 shirts will guarantee the revenue earned exactly recoups the cost of production. C(0) = 100, so the ﬁxed costs are $100. C(10) = 20, so when 10 bottles of tonic are produced, the cost per bottle is $20. p(5) = 30, so to sell 5 bottles of tonic, set the price at $30 per bottle. R(x) = −x2 + 35x, 0 ≤ x ≤ 35 P (x) = −x2 + 25x − 100, 0 ≤ x ≤ 35 P (x) = 0 when x = 5 and x = 20. These are the ‘break even’ points, so selling 5 bottles of tonic or 20 bottles of tonic will guarantee the revenue earned exactly recoups the cost of production. C(0) = 240, so the ﬁxed costs are 240¢ or $2.40. C(10) = 42, so when 10 cups of lemonade are made, the cost per cup is 42¢. p(5) = 75, so to sell 5 cups of lemonade, set the price at 75¢ per cup. R(x) = −3x2 + 90x, 0 ≤ x ≤ 30 P (x) = −3x2 + 72x − 240, 0 ≤ x ≤ 30 P (x) = 0 when x = 4 and x = 20. These are the ‘break even’ points, so selling 4 cups of lemonade or 20 cups of lemonade will guarantee the revenue earned exactly recoups the cost of production. 92 49. Relations and Functions C(0) = 36, so the daily ﬁxed costs are $36. C(10) = 6.6, so when 10 pies are made, the cost per pie is $6.60. p(5) = 9.5, so to sell 5 pies a day, set the price at $9.50 per pie. R(x) = −0.5x2 + 12x, 0 ≤ x ≤ 24 P (x) = −0.5x2 + 9x − 36, 0 ≤ x ≤ 24 P (x) = 0 when x = 6 and x = 12. These are the ‘break even’ points, so selling
6 pies or 12 pies a day will guarantee the revenue earned exactly recoups the cost of production. 50. C(0) = 1000, so the monthly ﬁxed costs are 1000 hundred dollars, or $100,000. C(10) = 120, so when 10 scooters are made, the cost per scooter is 120 hundred dollars, or $12,000. p(5) = 130, so to sell 5 scooters a month, set the price at 130 hundred dollars, or $13,000 per scooter. R(x) = −2x2 + 140x, 0 ≤ x ≤ 70 P (x) = −2x2 + 120x − 1000, 0 ≤ x ≤ 70 P (x) = 0 when x = 10 and x = 50. These are the ‘break even’ points, so selling 10 scooters or 50 scooters a month will guarantee the revenue earned exactly recoups the cost of production. 51. (f + g)(−3) = 2 52. (f − g)(2) = 3 53. (f g)(−1) = 0 54. (g + f )(1) = 0 55. (g − f )(3) = 3 56. (gf )(−3) = −8 57. 60. f g g f (−2) does not exist (−1) does not exist 58. 61. f g g f (−1) = 0 (3) = −2 59. 62. f g g f (2) = 4 (−3) = − 1 2 1.6 Graphs of Functions 93 1.6 Graphs of Functions In Section 1.3 we deﬁned a function as a special type of relation; one in which each x-coordinate was matched with only one y-coordinate. We spent most of our time in that section looking at functions graphically because they were, after all, just sets of points in the plane. Then in Section 1.4 we described a function as a process and deﬁned the notation necessary to work with functions algebraically. So now it’s time to look at functions graphically again, only this time we’ll do so with the notation deﬁned in Section 1.4. We start with what should not be a surprising connection. The Fundamental Graphing Principle for Functions The graph of a function f is the set of
points which satisfy the equation y = f (x). That is, the point (x, y) is on the graph of f if and only if y = f (x). Example 1.6.1. Graph f (x) = x2 − x − 6. Solution. To graph f, we graph the equation y = f (x). To this end, we use the techniques outlined in Section 1.2.1. Speciﬁcally, we check for intercepts, test for symmetry, and plot additional points as needed. To ﬁnd the x-intercepts, we set y = 0. Since y = f (x), this means f (x) = 0. f (x) = x2 − x − 6 0 = x2 − x − 6 0 = (x − 3)(x + 2) factor x − 3 = 0 or x + 2 = 0 x = −2, 3 So we get (−2, 0) and (3, 0) as x-intercepts. To ﬁnd the y-intercept, we set x = 0. Using function notation, this is the same as ﬁnding f (0) and f (0) = 02 − 0 − 6 = −6. Thus the y-intercept is (0, −6). As far as symmetry is concerned, we can tell from the intercepts that the graph possesses none of the three symmetries discussed thus far. (You should verify this.) We can make a table analogous to the ones we made in Section 1.2.1, plot the points and connect the dots in a somewhat pleasing fashion to get the graph below on the right. x −3 −2 −1 0 1 2 3 4 f (x) 6 0 (x, f (x)) (−3, 6) (−2, 0) −4 (−1, −4) (0, −6) −6 (1, −6) −6 (2, −4) −4 (3, 0) 0 (4, 63−2−1 −1 1 2 3 4 x −2 −3 −4 −5 −6 94 Relations and Functions Graphing piecewise-deﬁned functions is a bit more of a challenge. Example 1.6.2. Graph: f (x) = 4 − x2 x − 3, if x < 1 if x ≥ 1 Solution. We
proceed as before – ﬁnding intercepts, testing for symmetry and then plotting additional points as needed. To ﬁnd the x-intercepts, as before, we set f (x) = 0. The twist is that we have two formulas for f (x). For x < 1, we use the formula f (x) = 4 − x2. Setting f (x) = 0 gives 0 = 4 − x2, so that x = ±2. However, of these two answers, only x = −2 ﬁts in the domain x < 1 for this piece. This means the only x-intercept for the x < 1 region of the x-axis is (−2, 0). For x ≥ 1, f (x) = x − 3. Setting f (x) = 0 gives 0 = x − 3, or x = 3. Since x = 3 satisﬁes the inequality x ≥ 1, we get (3, 0) as another x-intercept. Next, we seek the y-intercept. Notice that x = 0 falls in the domain x < 1. Thus f (0) = 4 − 02 = 4 yields the y-intercept (0, 4). As far as symmetry is concerned, you can check that the equation y = 4 − x2 is symmetric about the y-axis; unfortunately, this equation (and its symmetry) is valid only for x < 1. You can also verify y = x − 3 possesses none of the symmetries discussed in the Section 1.2.1. When plotting additional points, it is important to keep in mind the restrictions on x for each piece of the function. The sticking point for this function is x = 1, since this is where the equations change. When x = 1, we use the formula f (x) = x − 3, so the point on the graph (1, f (1)) is (1, −2). However, for all values less than 1, we use the formula f (x) = 4 − x2. As we have discussed earlier in Section 1.2, there is no real number which immediately precedes x = 1 on the number line. Thus for the values x = 0.9, x = 0.99, x = 0.999, and so on, we ﬁnd the corresponding y values using the formula f (x) = 4 − x2. Making
a table as before, we see that as the x values sneak up to x = 1 in this fashion, the f (x) values inch closer and closer1 to 4 − 12 = 3. To indicate this graphically, we use an open circle at the point (1, 3). Putting all of this information together and plotting additional points, we get (x, f (x)) x f (x) (0.9, 3.19) 0.9 3.19 0.99 ≈ 3.02 (0.99, 3.02) 0.999 ≈ 3.002 (0.999, 3.002) y 4 3 2 1 −3 −2 −1 −1 1 2 3 x −2 −3 −4 1We’ve just stepped into Calculus here! 1.6 Graphs of Functions 95 In the previous two examples, the x-coordinates of the x-intercepts of the graph of y = f (x) were found by solving f (x) = 0. For this reason, they are called the zeros of f. Deﬁnition 1.9. The zeros of a function f are the solutions to the equation f (x) = 0. In other words, x is a zero of f if and only if (x, 0) is an x-intercept of the graph of y = f (x). Of the three symmetries discussed in Section 1.2.1, only two are of signiﬁcance to functions: symmetry about the y-axis and symmetry about the origin.2 Recall that we can test whether the graph of an equation is symmetric about the y-axis by replacing x with −x and checking to see if an equivalent equation results. If we are graphing the equation y = f (x), substituting −x for x results in the equation y = f (−x). In order for this equation to be equivalent to the original equation y = f (x) we need f (−x) = f (x). In a similar fashion, we recall that to test an equation’s graph for symmetry about the origin, we replace x and y with −x and −y, respectively. Doing this substitution in the equation y = f (x) results in −y = f (−x). Solving the latter equation for y gives y = −f (−x). In order for this equation to be equivalent to
the original equation y = f (x) we need −f (−x) = f (x), or, equivalently, f (−x) = −f (x). These results are summarized below. Testing the Graph of a Function for Symmetry The graph of a function f is symmetric about the y-axis if and only if f (−x) = f (x) for all x in the domain of f. about the origin if and only if f (−x) = −f (x) for all x in the domain of f. For reasons which won’t become clear until we study polynomials, we call a function even if its graph is symmetric about the y-axis or odd if its graph is symmetric about the origin. Apart from a very specialized family of functions which are both even and odd,3 functions fall into one of three distinct categories: even, odd, or neither even nor odd. Example 1.6.3. Determine analytically if the following functions are even, odd, or neither even nor odd. Verify your result with a graphing calculator. 1. f (x) = 5 2 − x2 2. g(x) = 5x 2 − x2 3. h(x) = 5. j(x) = x2 − 5x 2 − x3 x 100 − 1 4. i(x) = 6. p(x) = 5x 2x − x3 x + 3 if x < 0 if x ≥ 0 −x + 3, Solution. The ﬁrst step in all of these problems is to replace x with −x and simplify. 2Why are we so dismissive about symmetry about the x-axis for graphs of functions? 3Any ideas? 96 1. 2. Relations and Functions f (x) = f (−x) = f (−x) = 5 2 − x2 5 2 − (−x)2 5 2 − x2 f (−x) = f (x) Hence, f is even. The graphing calculator furnishes the following. This suggests4 that the graph of f is symmetric about the y-axis, as expected. g(x) = g(−x) = g(−x) = 5x 2 − x2 5(−x) 2 − (−x)2 −5x 2 − x2 It doesn’t appear that g(−x) is equivalent to g(x). To prove
this, we check with an x value. After some trial and error, we see that g(1) = 5 whereas g(−1) = −5. This proves that g is not even, but it doesn’t rule out the possibility that g is odd. (Why not?) To check if g is odd, we compare g(−x) with −g(x) −g(x) = − 5x 2 − x2 −5x 2 − x2 = −g(x) = g(−x) Hence, g is odd. Graphically, 4‘Suggests’ is about the extent of what it can do. 1.6 Graphs of Functions 97 3. 4. The calculator indicates the graph of g is symmetric about the origin, as expected. h(x) = h(−x) = h(−x) = 5x 2 − x3 5(−x) 2 − (−x)3 −5x 2 + x3 Once again, h(−x) doesn’t appear to be equivalent to h(x). We check with an x value, for example, h(1) = 5 but h(−1) = − 5 3. This proves that h is not even and it also shows h is not odd. (Why?) Graphically, The graph of h appears to be neither symmetric about the y-axis nor the origin. i(x) = i(−x) = i(−x) = 5x 2x − x3 5(−x) 2(−x) − (−x)3 −5x −2x + x3 The expression i(−x) doesn’t appear to be equivalent to i(x). However, after checking some x values, for example x = 1 yields i(1) = 5 and i(−1) = 5, it appears that i(−x) does, in fact, equal i(x). However, while this suggests i is even, it doesn’t prove it. (It does, however, prove 98 Relations and Functions i is not odd.) To prove i(−x) = i(x), we need to manipulate our expressions for i(x) and i(−x) and show that they are equivalent. A clue as to how to proceed is in the numerators: in the formula for i(x), the numerator is 5x and in
i(−x) the numerator is −5x. To re-write i(x) with a numerator of −5x, we need to multiply its numerator by −1. To keep the value of the fraction the same, we need to multiply the denominator by −1 as well. Thus i(x) = = = 5x 2x − x3 (−1)5x (−1) (2x − x3) −5x −2x + x3 Hence, i(x) = i(−x), so i is even. The calculator supports our conclusion. 5. j(x) = x2 − x 100 j(−x) = (−x)2 − j(−x) = x2 + x 100 − 1 −x 100 − 1 − 1 The expression for j(−x) doesn’t seem to be equivalent to j(x), so we check using x = 1 to get j(1) = − 1 100. This rules out j being even. However, it doesn’t rule out j being odd. Examining −j(x) gives 100 and j(−1) = 1 j(x) = x2 − −j(x) = − x 100 x2 − −j(x) = −x2 + + 1 − 1 − 1 x 100 x 100 The expression −j(x) doesn’t seem to match j(−x) either. Testing x = 2 gives j(2) = 149 50 and j(−2) = 151 50, so j is not odd, either. The calculator gives: 1.6 Graphs of Functions 99 The calculator suggests that the graph of j is symmetric about the y-axis which would imply that j is even. However, we have proven that is not the case. 6. Testing the graph of y = p(x) for symmetry is complicated by the fact p(x) is a piecewisedeﬁned function. As always, we handle this by checking the condition for symmetry by checking it on each piece of the domain. We ﬁrst consider the case when x < 0 and set about ﬁnding the correct expression for p(−x). Even though p(x) = x + 3 for x < 0, p(−x) = −x + 3 here. The reason for this is that since x < 0, −x >
0 which means to ﬁnd p(−x), we need to use the other formula for p(x), namely p(x) = −x+3. Hence, for x < 0, p(−x) = −(−x)+3 = x + 3 = p(x). For x ≥ 0, p(x) = −x + 3 and we have two cases. If x > 0, then −x < 0 so p(−x) = (−x) + 3 = −x + 3 = p(x). If x = 0, then p(0) = 3 = p(−0). Hence, in all cases, p(−x) = p(x), so p is even. Since p(0) = 3 but p(−0) = p(0) = 3 = −3, we also have p is not odd. While graphing y = p(x) is not onerous to do by hand, it is instructive to see how to enter this into our calculator. By using some of the logical commands,5 we have: The calculator bears shows that the graph appears to be symmetric about the y-axis. There are two lessons to be learned from the last example. The ﬁrst is that sampling function values at particular x values is not enough to prove that a function is even or odd − despite the fact that j(−1) = −j(1), j turned out not to be odd. Secondly, while the calculator may suggest mathematical truths, it is the Algebra which proves mathematical truths.6 5Consult your owner’s manual, instructor, or favorite video site! 6Or, in other words, don’t rely too heavily on the machine! 100 Relations and Functions 1.6.1 General Function Behavior The last topic we wish to address in this section is general function behavior. As you shall see in the next several chapters, each family of functions has its own unique attributes and we will study them all in great detail. The purpose of this section’s discussion, then, is to lay the foundation for that further study by investigating aspects of function behavior which apply to all functions. To start, we will examine the concepts of increasing, decreasing and constant. Before deﬁning the concepts algebraically, it is instructive to ﬁrst look at them graphically. Consider the graph of the function f below. (6,
5.5) (−2, 4.54 −3 −2 −1 −1 1 2 3 4 5 6 7 x (−4, −3) −2 −3 −4 −5 −6 −7 −8 −9 (4, −6) (3, −8) (5, −6) The graph of y = f (x) Reading from left to right, the graph ‘starts’ at the point (−4, −3) and ‘ends’ at the point (6, 5.5). If we imagine walking from left to right on the graph, between (−4, −3) and (−2, 4.5), we are walking ‘uphill’; then between (−2, 4.5) and (3, −8), we are walking ‘downhill’; and between (3, −8) and (4, −6), we are walking ‘uphill’ once more. From (4, −6) to (5, −6), we ‘level oﬀ’, and then resume walking ‘uphill’ from (5, −6) to (6, 5.5). In other words, for the x values between −4 and −2 (inclusive), the y-coordinates on the graph are getting larger, or increasing, as we move from left to right. Since y = f (x), the y values on the graph are the function values, and we say that the function f is increasing on the interval [−4, −2]. Analogously, we say that f is decreasing on the interval [−2, 3] increasing once more on the interval [3, 4], constant on [4, 5], and ﬁnally increasing once again on [5, 6]. It is extremely important to notice that the behavior (increasing, decreasing or constant) occurs on an interval on the x-axis. When we say that the function f is increasing 1.6 Graphs of Functions 101 on [−4, −2] we do not mention the actual y values that f attains along the way. Thus, we report where the behavior occurs, not to what extent the behavior occurs.7 Also notice that we do not say that a function is increasing, decreasing or constant at a single x value. In fact, we would run into serious trouble in our previous example if we tried to do so because x
= −2 is contained in an interval on which f was increasing and one on which it is decreasing. (There’s more on this issue – and many others – in the Exercises.) We’re now ready for the more formal algebraic deﬁnitions of what it means for a function to be increasing, decreasing or constant. Deﬁnition 1.10. Suppose f is a function deﬁned on an interval I. We say f is: increasing on I if and only if f (a) < f (b) for all real numbers a, b in I with a < b. decreasing on I if and only if f (a) > f (b) for all real numbers a, b in I with a < b. constant on I if and only if f (a) = f (b) for all real numbers a, b in I. It is worth taking some time to see that the algebraic descriptions of increasing, decreasing and constant as stated in Deﬁnition 1.10 agree with our graphical descriptions given earlier. You should look back through the examples and exercise sets in previous sections where graphs were given to see if you can determine the intervals on which the functions are increasing, decreasing or constant. Can you ﬁnd an example of a function for which none of the concepts in Deﬁnition 1.10 apply? Now let’s turn our attention to a few of the points on the graph. Clearly the point (−2, 4.5) does not have the largest y value of all of the points on the graph of f − indeed that honor goes to (6, 5.5) − but (−2, 4.5) should get some sort of consolation prize for being ‘the top of the hill’ between x = −4 and x = 3. We say that the function f has a local maximum8 at the point (−2, 4.5), because the y-coordinate 4.5 is the largest y-value (hence, function value) on the curve ‘near’9 x = −2. Similarly, we say that the function f has a local minimum10 at the point (3, −8), since the y-coordinate −8 is the smallest function value near x = 3. Although it is tempting to say that local extrema11 occur when the function changes from increasing to decreasing or vice versa
, it is not a precise enough way to deﬁne the concepts for the needs of Calculus. At the risk of being pedantic, we will present the traditional deﬁnitions and thoroughly vet the pathologies they induce in the Exercises. We have one last observation to make before we proceed to the algebraic deﬁnitions and look at a fairly tame, yet helpful, example. If we look at the entire graph, we see that the largest y value (the largest function value) is 5.5 at x = 6. In this case, we say the maximum12 of f is 5.5; similarly, the minimum13 of f is −8. 7The notions of how quickly or how slowly a function increases or decreases are explored in Calculus. 8Also called ‘relative maximum’. 9We will make this more precise in a moment. 10Also called a ‘relative minimum’. 11‘Maxima’ is the plural of ‘maximum’ and ‘mimima’ is the plural of ‘minimum’. ‘Extrema’ is the plural of ‘extremum’ which combines maximum and minimum. 12Sometimes called the ‘absolute’ or ‘global’ maximum. 13Again, ‘absolute’ or ‘global’ minimum can be used. 102 Relations and Functions We formalize these concepts in the following deﬁnitions. Deﬁnition 1.11. Suppose f is a function with f (a) = b. We say f has a local maximum at the point (a, b) if and only if there is an open interval I containing a for which f (a) ≥ f (x) for all x in I. The value f (a) = b is called ‘a local maximum value of f ’ in this case. We say f has a local minimum at the point (a, b) if and only if there is an open interval I containing a for which f (a) ≤ f (x) for all x in I. The value f (a) = b is called ‘a local minimum value of f ’ in this case. The value b is called the maximum of f if b ≥ f (x) for all x in the domain of f. The value b is called the minimum of f if b ≤ f (x) for
all x in the domain of f. It’s important to note that not every function will have all of these features. Indeed, it is possible to have a function with no local or absolute extrema at all! (Any ideas of what such a function’s graph would have to look like?) We shall see examples of functions in the Exercises which have one or two, but not all, of these features, some that have instances of each type of extremum and some functions that seem to defy common sense. In all cases, though, we shall adhere to the algebraic deﬁnitions above as we explore the wonderful diversity of graphs that functions provide us. Here is the ‘tame’ example which was promised earlier. It summarizes all of the concepts presented in this section as well as some from previous sections so you should spend some time thinking deeply about it before proceeding to the Exercises. Example 1.6.4. Given the graph of y = f (x) below, answer all of the following questions. y (0, 3) 4 3 2 1 (−2, 0) (2, 0) −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 (−4, −3) (4, −3) 1.6 Graphs of Functions 103 1. Find the domain of f. 2. Find the range of f. 3. List the x-intercepts, if any exist. 4. List the y-intercepts, if any exist. 5. Find the zeros of f. 7. Determine f (2). 6. Solve f (x) < 0. 8. Solve f (x) = −3. 9. Find the number of solutions to f (x) = 1. 10. Does f appear to be even, odd, or neither? 11. List the intervals on which f is increasing. 12. List the intervals on which f is decreasing. 13. List the local maximums, if any exist. 14. List the local minimums, if any exist. 15. Find the maximum, if it exists. 16. Find the minimum, if it exists. Solution. 1. To ﬁnd the domain of f, we proceed as in Section 1.3. By projecting the graph to the x-axis, we see that the portion of the x-axis which corresponds to a point on the graph is everything
from −4 to 4, inclusive. Hence, the domain is [−4, 4]. 2. To ﬁnd the range, we project the graph to the y-axis. We see that the y values from −3 to 3, inclusive, constitute the range of f. Hence, our answer is [−3, 3]. 3. The x-intercepts are the points on the graph with y-coordinate 0, namely (−2, 0) and (2, 0). 4. The y-intercept is the point on the graph with x-coordinate 0, namely (0, 3). 5. The zeros of f are the x-coordinates of the x-intercepts of the graph of y = f (x) which are x = −2, 2. 6. To solve f (x) < 0, we look for the x values of the points on the graph where the y-coordinate is less than 0. Graphically, we are looking for where the graph is below the x-axis. This happens for the x values from −4 to −2 and again from 2 to 4. So our answer is [−4, −2) ∪ (2, 4]. 7. Since the graph of f is the graph of the equation y = f (x), f (2) is the y-coordinate of the point which corresponds to x = 2. Since the point (2, 0) is on the graph, we have f (2) = 0. 8. To solve f (x) = −3, we look where y = f (x) = −3. We ﬁnd two points with a y-coordinate of −3, namely (−4, −3) and (4, −3). Hence, the solutions to f (x) = −3 are x = ±4. 9. As in the previous problem, to solve f (x) = 1, we look for points on the graph where the y-coordinate is 1. Even though these points aren’t speciﬁed, we see that the curve has two points with a y value of 1, as seen in the graph below. That means there are two solutions to f (x) = 1. 104 Relations and Functions y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 10. The graph appears to
be symmetric about the y-axis. This suggests14 that f is even. 11. As we move from left to right, the graph rises from (−4, −3) to (0, 3). This means f is increasing on the interval [−4, 0]. (Remember, the answer here is an interval on the x-axis.) 12. As we move from left to right, the graph falls from (0, 3) to (4, −3). This means f is decreasing on the interval [0, 4]. (Remember, the answer here is an interval on the x-axis.) 13. The function has its only local maximum at (0, 3) so f (0) = 3 is the local minimum value. 14. There are no local minimums. Why don’t (−4, −3) and (4, −3) count? Let’s consider the point (−4, −3) for a moment. Recall that, in the deﬁnition of local minimum, there needs to be an open interval I which contains x = −4 such that f (−4) < f (x) for all x in I diﬀerent from −4. But if we put an open interval around x = −4 a portion of that interval will lie outside of the domain of f. Because we are unable to fulﬁll the requirements of the deﬁnition for a local minimum, we cannot claim that f has one at (−4, −3). The point (4, −3) fails for the same reason − no open interval around x = 4 stays within the domain of f. 15. The maximum value of f is the largest y-coordinate which is 3. 16. The minimum value of f is the smallest y-coordinate which is −3. With few exceptions, we will not develop techniques in College Algebra which allow us to determine the intervals on which a function is increasing, decreasing or constant or to ﬁnd the local maximums and local minimums analytically; this is the business of Calculus.15 When we have need to ﬁnd such beasts, we will resort to the calculator. Most graphing calculators have ‘Minimum’ and ‘Maximum’ features which can be used to approximate these values, as we now demonstrate. 14but does not prove 15Although, truth be told, there is only one step of Cal

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