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2 As x → 2−, f (x) → ∞ As x → 2+, f (x) → −∞ Hole at −3, 7 5 Horizontal asymptote: y = 1 As x → −∞, f (x) → 1+ As x → ∞, f (x) → 1− y 1 −7−6−5−4−3−2−1 y 5 4 3 2 1 −5 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3 −4 −5 5 4 3 2 1 y −5 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3 −4 −5 338 Rational Functions 10. f (x) = = 3x2 − 5x − 2 x2 − 9 (3x...
x) = x2 − x 3 − x = x(x − 1) 3 − x Domain: (−∞, 3) ∪ (3, ∞) x-intercepts: (0, 0), (1, 0) y-intercept: (0, 0) Vertical asymptote: x = 3 As x → 3−, f (x) → ∞ As x → 3+, f (x) → −∞ Slant asymptote: y = −x − 2 As x → −∞, the graph is above y = −x − 2 As x → ∞, the graph is below y = −9−8−7−7−6−5−4−3−2−1 −1 −3 −4 −5 −6 −7 −...
f (x) → −∞ As x → 3−, f (x) → ∞ As x → 3+, f (x) → −∞ Slant asymptote: y = −x As x → −∞, the graph is above y = −x As x → ∞, the graph is below y = −x 15. f (x) = x3 − 2x2 + 3x 2x2 + 2 Domain: (−∞, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Slant asymptote: y = 1 As x → −∞, the graph is below y = 1 As x → ∞, the graph...
−5 y 3 2 1 −1 1 2 3 4 5 x −1 −2 − Vertically stretch the graph of y = by a factor of 3. 3 x Reﬂect the graph of y = about the x-axis. Shift the graph of y = − up 1 unit. 3 x −6−5−4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 4.2 Graphs of Rational Functions 341 19. h(x) = −2x + 1 x Shift the graph of y = down 2 units. = −2 + 1...
, depending on x, (x − 1) may be positive (which doesn’t aﬀect the inequality) or (x − 1) could be negative (which would reverse the inequality). Instead of working by cases, we collect all of the terms on one side of the inequality with 0 on the other and make a sign diagram using the technique given on page 321 in Se...
(x). We obtain the graphs below. x−1 The ‘Intersect’ command conﬁrms that the graphs cross when x = − 1 2 and x = 0. It is clear from the calculator that the graph of y = f (x) is above the graph of y = g(x) on −∞, − 1 2 as well as on (0, ∞). According to the calculator, our solution is then −∞, − 1 ∪ [0, ∞) 2 which al...
the trip into its two parts: the initial trip downstream and the return trip upstream. For the downstream trip, all we know is that the distance traveled is 5 miles. distance downstream = rate traveling downstream · time traveling downstream 5 miles = rate traveling downstream · time traveling downstream Since the ret...
6 − R) tup = 5 E2 tdown + tup = 3 E3 Since we are ultimately after R, we need to use these three equations to get at least one equation involving only R. To that end, we solve E1 for tdown by dividing both sides3 by the quantity (6+R) to get tdown = 5 6−R. Substituting these into E3, we get:4 6+R. Similarly, we solve E...
5. 4The reader is encouraged to verify that the units in this equation are the same on both sides. To get you started, the units on the ‘3’ is ‘hours.’ 346 Rational Functions amount of work done together = rate of working together · time spent working together 1 garden = (rate of working together) · (3 hours) garden F...
), which returns the total cost of producing x items, recall that the average cost function, C(x) = C(x) computes the cost per item when x items are x produced. Suppose the cost C, in dollars, to produce x PortaBoy game systems for a local retailer is C(x) = 80x + 150, x ≥ 0. 1. Find an expression for the average cost ...
so that the average cost is less than $100 per system. Our solution, (7.5, ∞) tells us that we need to produce more than 7.5 systems to achieve this. Since it doesn’t make sense to produce half a system, our ﬁnal answer is [8, ∞). 3. When we apply Theorem 4.2 to C(x) we ﬁnd that y = 80 is a horizontal asymptote to the...
. Use a calculator to approximate (to two decimal places) the dimensions of the box which minimize the surface area. Solution. 1. We are told that the volume of the box is 1000 cubic centimeters and that x represents the width, in centimeters. From geometry, we know Volume = width × height × depth. Since the base of th...
centimeters for this to happen. 3. As x → 0+, h(x) = 1000 x2 → ∞. This means that the smaller the width x (and, in this case, depth), the larger the height h has to be in order to maintain a volume of 1000 cubic centimeters. As x → ∞, we ﬁnd h(x) → 0+, which means that in order to maintain a volume of 1000 cubic centi...
deﬁnition below. Deﬁnition 4.5. Suppose x, y and z are variable quantities. We say y varies directly with (or is directly proportional to) x if there is a constant k such that y = kx. y varies inversely with (or is inversely proportional to) x if there is a constant k such that y = k x. z varies jointly with (or is jo...
.5 to get V = khr2. 4. Even though the problem doesn’t use the phrase ‘varies jointly’, it is implied by the fact that the current I is related to two diﬀerent quantities. Since I varies directly with V but inversely with R, we write I = kV R. 5. We write the product of the masses mM and the square of the distance as r...
volume numbers times each of the pressure numbers,11 we produce a number which is always approximately 1400. We suspect that P = 1400 V. Graphing y = 1400 along with the data gives us good reason to believe our hypotheses that P and V are, in fact, inversely related. x The graph of the data The data with y = 1400 x 3....
< 0 x4 − 4x3 + x2 − 2x − 15 x3 − 4x2 ≥ x 20. 5x3 − 12x2 + 9x + 10 x2 − 1 ≥ 3x − 1 21. Carl and Mike start a 3 mile race at the same time. If Mike ran the race at 6 miles per hour and ﬁnishes the race 10 minutes before Carl, how fast does Carl run? 22. One day, Donnie observes that the wind is blowing at 6 miles per ho...
answers to two decimal places. 28. The box for the new Sasquatch-themed cereal, ‘Crypt-Os’, is to have a volume of 140 cubic inches. For aesthetic reasons, the height of the box needs to be 1.62 times the width of the base of the box.13 Find the dimensions of the box which will minimize the surface area of the box. Wh...
.62 is a crude approximation of the so-called ‘Golden Ratio’ φ = 1+ 2 14According to www.dictionary.com, there are diﬀerent values given for this conversion. We will stick with 33.6in3. 5 √ for this problem. 4.3 Rational Inequalities and Applications 355 In Exercises 33 - 38, translate the following into mathematical e...
. 41. We know that the circumference of a circle varies directly with its radius with 2π as the constant of proportionality. (That is, we know C = 2πr.) With the help of your classmates, compile a list of other basic geometric relationships which can be seen as variations. 15Also known as the linear density. It is simp...
base of the box should be ≈ 4.12 inches, the height of the box should be ≈ 6.67 inches, and the depth of the base of the box should be ≈ 5.09 inches; minimum surface area ≈ 164.91 square inches. 29. The dimensions are ≈ 7 feet by ≈ 14 feet; minimum amount of fencing required ≈ 28 feet. 30. (a) V = πr2h (b) S = 2πr2 + ...
thoughts and gain some perspective. Chapter 1 ﬁrst introduced us to functions in Section 1.3. At that time, functions were speciﬁc kinds of relations - sets of points in the plane which passed the Vertical Line Test, Theorem 1.1. In Section 1.4, we developed the idea that functions are processes - rules which match in...
far is a combination of linear functions. 360 Further Topics in Functions Deﬁnition 5.1. Suppose f and g are two functions. The composite of g with f, denoted g ◦ f, is deﬁned by the formula (g ◦ f )(x) = g(f (x)), provided x is an element of the domain of f and f (x) is an element of the domain of g. The quantity g ◦...
of each. 4. (g ◦ f )(x) 5. (f ◦ g)(x) 6. (g ◦ h)(x) 7. (h ◦ g)(x) 8. (h ◦ h)(x) 9. (h ◦ (g ◦ f ))(x) 10. ((h ◦ g) ◦ f )(x) Solution. 1. Using Deﬁnition 5.1, (g ◦ f )(1) = g(f (1)). We ﬁnd f (1) = −3, so (g ◦ f )(1) = g(f (1)) = g(−3) = 2 5.1 Function Composition 361 2. As before, we use Deﬁnition 5.1 to write (f ◦ g)(...
� f )(x) = 2 − √ x2 − 4x + 3. To ﬁnd the domain of g ◦ f, we need to ﬁnd the elements in the domain of f whose outputs f (x) are in the domain of g. We accomplish this by following the rule set forth in Section 1.4, that is, we ﬁnd the domain before we simplify. To that end, we examine (g ◦ f )(x) = 2 − (x2 − 4x) + 3. ...
) into g ﬁrst to get (g ◦ h)(x) = g(h(x)) = g 2x − 2x x + 1 + 3 3(x + 1) x + 1 get common denominators + 2x x + 1 5x + 3 x + 1 outside in: We use the formula for g(x) ﬁrst to get (g ◦ h)(x) = g(h(x)) = 2 − h(x) + 3 = 2 − = 2 − 2x x + 1 5x + 3 x + 1 + 3 get common denominators as before To ﬁnd the domain of (g ◦ h), we ...
= 3. Squaring both sides gives us x + 3 = 9, or x = 6. Since x = 6 checks in the original equation, we know x = 6 is the only zero of the denominator. Hence, the domain of h ◦ g is [−3, 6) ∪ (6, ∞). √ √ 8. To ﬁnd (h ◦ h)(x), we substitute the function h into itself, h(h(x)). inside out: We insert the expression h(x) i...
)(x) into h ﬁrst to get (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = h √ x2 − 4x + 3 2 − + 1 x2 − 4x + − √ √ x2 − 4x + 3 x2 − 4x + 3 x2 − 4x + 3 = = outside in: We use the formula for h(x) ﬁrst to get (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = 2 ((g ◦ f )(x)) ((g ◦ f )(x)) + 1 √ x2 − 4x + − √ √ x2 − 4x + 3 x2 − 4x + 3 x2 − 4x + 3 +...
)(x) indicates that we ﬁrst ﬁnd the composite h◦g and then compose that with f. From number 4, we have (h ◦ g)(x 366 Further Topics in Functions We now proceed as before. inside out: We insert the expression f (x) into h ◦ g ﬁrst to get ((h ◦ g) ◦ f )(x) = (h ◦ g)(f (x)) = (h ◦ g) x2 − 4x x2 − 4x) + 3 (x2 − 4x) + 3 x2...
iﬁed formula would indicate. Composing a function with itself, as in the case of ﬁnding (g ◦ g)(6) and (h ◦ h)(x), may seem odd. Looking at this from a procedural perspective, however, this merely indicates performing a task h and then doing it again - like setting the washing machine to do a ‘double rinse’. Composing ...
the reader to think about why the domains of these two functions are identical, too. These two facts establish the equality h ◦ (g ◦ f ) = (h ◦ g) ◦ f. A consequence of the associativity of function composition is that there is no need for parentheses when we write h ◦ g ◦ f. The second property can also be veriﬁed us...
(t))2 = 4π 3t22 = 36πt4. This formula allows us to compute the surface area directly given the time without going through the ‘middle man’ r. A useful skill in Calculus is to be able to take a complicated function and break it down into a composition of easier functions which our last example illustrates. Example 5.1.3...
function takes its input and divides it into 2, h(x) = 2 x. The claim is that G = h ◦ g ◦ f. We ﬁnd (h ◦ g ◦ f )(x) = h(g(f (x))) = h(g x2) = h x2 + 1 = 2 x2 + 1 = G(x), so we are done. 3. If we look H(x) = √ x+1√ x−1 with an eye towards building a complicated function from simpler If we deﬁne x is a simple piece of t...
g(x) = 4x2 − x 9. f (x) = 3 1 − x, g(x) = 4x x2 + 1 11. f (x) = 2x 5 − x2, g(x) = √ 4x + 1 10. f (x) = x x + 5, g(x) = 2 7 − x2 12. f (x) = √ 2x + 5, g(x) = 10x x2 + 1 In Exercises 13 - 24, use the given pair of functions to ﬁnd and simplify expressions for the following functions and state the domain of each using in...
f ◦ g)(x) 27. (g ◦ f ◦ h)(x) 28. (g ◦ h ◦ f )(x) 29. (f ◦ h ◦ g)(x) 30. (f ◦ g ◦ h)(x) In Exercises 31 - 40, write the given function as a composition of two or more non-identity functions. (There are several correct answers, so check your answer using function composition.) 31. p(x) = (2x + 3)3 33. h(x) = √ 2x − 1 35...
)(3) 47. (f ◦ g)(−3) 45. f (g(−1)) 48. (g ◦ f )(3) 46. (f ◦ f )(0) 49. g(f (−3)) 5.1 Function Composition 371 50. (g ◦ g)(−2) 51. (g ◦ f )(−2) 52. g(f (g(0))) 53. f (f (f (−1))) 54. f (f (f (f (f (1))))) 55. (g ◦ g ◦ · · · ◦ g) n times (0) In Exercises 56 - 61, use the graphs of y = f (x) and y = g(x) below to ﬁnd the ...
(0) = 1 (f ◦ g)(−1) = 1 (f ◦ f )(2) = 16 (g ◦ f )(−3) = 19 (f ◦ g) 1 2 = 4 (f ◦ f )(−2) = 16 2. For f (x) = 4 − x and g(x) = 1 − x2, (g ◦ f )(0) = −15 (f ◦ g)(−1) = 4 (f ◦ f )(2) = 2 (g ◦ f )(−3) = −48 (f ◦ g) 1 2 = 13 4 (f ◦ f )(−2) = −2 3. For f (x) = 4 − 3x and g(x) = \|x\|, (g ◦ f )(0) = 4 (f ◦ g)(−1) = 1 (f ◦ f )(2)...
f ◦ f )(−2) = 3 − √ 5 5.1 Function Composition 373 7. For f (x) = 6 − x − x2 and g(x) = x √ x + 10, (g ◦ f )(0) = 24 (f ◦ g)(−1) = 0 (f ◦ f )(2) = 6 (g ◦ f )(−3) = 0 (f ◦ g) 1 2 √ = 27−2 8 42 (f ◦ f )(−2) = −14 √ 8. For f (x) = 3 x + 1 and g(x) = 4x2 − x, (g ◦ f )(0) = 3 √ (g ◦ f )(−3f ◦ g)(−1) = 3 6 (f ◦ g) 1 2 = 3√ 1...
11 (f ◦ f )(−2) = 8 11 12. For f (x) = √ 2x + 5 and g(x) = 10x x2+1, (g ◦ f )(0) = 5 √ 3 5 (f ◦ g)(−1) is not real (f ◦ f )(2) = √ 11 (g ◦ f )(−3) is not real (f ◦ g) 1 2 = √ 13 (f ◦ f )(−2) = √ 7 13. For f (x) = 2x + 3 and g(x) = x2 − 9 (g ◦ f )(x) = 4x2 + 12x, domain: (−∞, ∞) (f ◦ g)(x) = 2x2 − 15, domain: (−∞, ∞) (...
� f )(x) = 9x − 20, domain: (−∞, ∞) 17. For f (x) = \|x + 1\| and g(x) = √ x (g ◦ f )(x) = \|x + 1\|, domain: (−∞, ∞) √ √ (f ◦ g)(x) = \| x + 1\| = x + 1, domain: [0, ∞) (f ◦ f )(x) = \|\|x + 1\| + 1\| = \|x + 1\| + 1, domain: (−∞, ∞) 18. For f (x) = 3 − x2 and g(x) = √ x + 1 (g ◦ f )(x) = √ 4 − x2, domain: [−2, 2] (f ◦ g)(x) = 2 ...
x (f ◦ f )(x) = 9x − 4, domain: (−∞, ∞) 3, ∞ 3x+2, domain: −∞, − 2 x+3, domain: (−∞, −3) ∪ (−3, ∞) ∪ − 2 3 22. For f (x) = 3x x−1 and g(x) = x x−3 (g ◦ f )(x) = x, domain: (−∞, 1) ∪ (1, ∞) (f ◦ g)(x) = x, domain: (−∞, 3) ∪ (3, ∞) (f ◦ f )(x) = 9x 2x+1, domain: −∞, − 1, ∞) 23. For f (x) = x 2x+1 and g(x) = 2x+1 x (g ◦ ...
25. (h ◦ g ◦ f )(x) = \| √ 26. (h ◦ f ◦ g)(x) = \| − 2 √ −2x\| = √ −2x, domain: (−∞, 0] √ x, domain: [0, ∞) 27. (g ◦ f ◦ h)(x) = 28. (g ◦ h ◦ f )(x) = 29. (f ◦ h ◦ g)(x) = −2\| x\| = −2 x\| = 2 −2\|x\|, domain: {0} \| − 2x\| = √ 2\|x\|, domain: (−∞, ∞) √ x, domain: [0, ∞) 376 Further Topics in Functions 30. (f ◦ g ◦ h)(x) = −2 \|x...
g(x) = x+1 3−2x, then v(x) = (g ◦ f )(x). 40. Let f (x) = x2 and g(x) = x x2+1, then w(x) = (g ◦ f )(x). 41. F (x) = x3+6 x3−9 = (h(g(f (x))) where f (x) = x3, g(x) = x+6 √ 42. F (x) = 3 −x + 2 − 4 = k(j(f (h(g(x))))) x−9 and h(x) = √ x. 43. One possible solution is F (x) = − 1 2 (2x − 7)3 + 1 = k(j(f (h(g(x))))) wher...
(f (f (f (f (1))))) = f (f (f (f (3)))) = f (f (f (−1))) = f (f (0)) = f (1) = 3 55. (g ◦ g ◦ · · · ◦ g) (0) = 0 n times 5.1 Function Composition 377 56. (g ◦ f )(1) = 3 57. (f ◦ g)(3) = 4 58. (g ◦ f )(2) = 0 59. (f ◦ g)(0) = 4 60. (f ◦ f )(1) = 3 61. (g ◦ g)(1) = 0 62. V (x) = x3 so V (x(t)) = (t + 1)3 63. (a) R(x) =...
function g should 1. subtract 4 2. divide by 3 3 = 15 Following this procedure, we get g(x) = x−4 3. Let’s check to see if the function g does the job. If x = 5, then f (5) = 3(5) + 4 = 15 + 4 = 19. Taking the output 19 from f, we substitute it into g to get g(19) = 19−4 3 = 5, which is our original input to f. To che...
. (f ◦ g)(x) = x for all x in the domain of g then f and g are inverses of each other and the functions f and g are said to be invertible. We now formalize the concept that inverse functions exchange inputs and outputs. Theorem 5.2. Properties of Inverse Functions: Suppose f and g are inverse functions. The rangea of f...
here. 2In other words, invertible functions have exactly one inverse. 380 Further Topics in Functions f. By Theorem 5.2, the domain of g is equal to the domain of h, since both are the range of f. This means the identity function I2 applies both to the domain of h and the domain of g. Thus h = h ◦ I2 = h ◦ (f ◦ g) = (...
input, namely 2, which satisﬁes f (2) = 4. This issue is presented schematically in the picture below. x. Checking the composition yields (g◦f )(x) = g(f (x)) = f (x2 x = 2 4 We see from the diagram that since both f (−2) and f (2) are 4, it is impossible to construct a function which takes 4 back to both x = 2 and x ...
for a function to have an inverse, diﬀerent inputs must go to diﬀerent outputs, or else we will run into the same problem we did with f (x) = x2. We give this property a name. Deﬁnition 5.3. A function f is said to be one-to-one if f matches diﬀerent inputs to diﬀerent outputs. Equivalently, f is one-to-one if and onl...
x) = x2 − 2x + 4 Solution. 2. g(x) = 2x 1 − x 4. F = {(−1, 1), (0, 2), (2, 1)} 1. (a) To determine if f is one-to-one analytically, we assume f (c) = f (d) and attempt to deduce that c = d. f (c) = f (d) 1 − 2c 5 = 1 − 2d 5 1 − 2c = 1 − 2d −2c = −2d c = d Hence, f is one-to-one. (b) To check if f is one-to-one graphica...
c2 − 2c = d2 − 2d c2 − d2 − 2c + 2d = 0 (c + d)(c − d) − 2(c − d) = 0 (c − d)((c + d) − 2) = 0 c − d = 0 or c = d or factor by grouping We get c = d as one possibility, but we also get the possibility that c = 2 − d. This suggests that f may not be one-to-one. Taking d = 0, we get c = 0 or c = 2. With f (0) = 4 and f ...
the chapter. The function g(x) = 2x 1−x is a bit trickier since x occurs in two places. When one evaluates g(x) for a speciﬁc value of x, which is ﬁrst, the 2x or the 1 − x? We can imagine functions more complicated than these so we need to develop a general methodology to attack this problem. Theorem 5.2 tells us equ...
x 1 − 2x) + + x + 1 2 We now check that f ◦ f −1 (x) = x for all x in the range of f which is also all real numbers. (Recall that the domain of f −1) is the range of f.) f ◦ f −1 (x) = f (f −1(x)) = = = 1 − 2f −1(x + 5x − 1 5 5x = 5 = x To check our answer graphically, we graph y = f (x) and y = f −1(x) on the same set...
all x in the range of g. From the graph of g in Example 5.2.1, we have that the range of g is (−∞, −2) ∪ (−2, ∞). This matches the domain we get from the formula g−1(x) = x x+2, as it should. g ◦ g−1 (x) = g g−1(x 2x (x + 2) − x 2x x + 2) (x + 2) x + 2 clear denominators 1 − = x Graphing y = g(x) and y = g−1(x) on the...
it looks like we’ll run into the same trouble as before, but when we check the composition, the domain restriction on g saves the day. We get g−1 ◦ g (x) = g−1(g(x)) = g−1 x2 = x2 = \|x\| = x, since x ≥ 0. Checking g ◦ g−1 (x) = g g−1(x) = √ x)2 = x. Graphing6 g and g−1 on the same set of axes shows that they are reﬂect...
(j(x)) (x2 − 2x + 4) − 3 x2 − 2x + 1 (x − 1)2 = j−1 x2 − 2x + x − 1\| = 1 − (−(x − 1)) = x since x ≤ 1 Checking j ◦ j−1, we get j ◦ j−1 (x) = j j−1(x) x − 3 x − 32 − − 32 − 7Here, we use the Quadratic Formula to solve for y. For ‘completeness,’ we note you can (and should!) also consider solving for y by ‘completing’ th...
� x2 + 2x + 1 − 1 (x + 1)2 − 1 = = = = \|x + 1 since x ≥ −1 Graphically, everything checks out as well, provided that we remember the domain restriction on k−1 means we take the right half of the parabola. y 2 y = k(x) 1 −2 −1 1 2 x −1 y = k−1(x) −2 Our last example of the section gives an application of inverse functio...
the weekly sales as its output. Hence, p−1(220) = 20 means 20 systems will be sold in a week if the price is set at $220 per system. 3 = −1.5 500−2x 3. We compute P ◦ p−1 (x) = P p−1(x) = P 500−2x − 150. After a hefty amount of Elementary Algebra,8 we obtain P ◦ p−1 (x) = − 2 3 x2 +220x− 40450. To understand what this...
1 and vice-versa. 1. f (x) = 6x − 2 3. f (x) = 5. f (x) = x − 2 3 + 4 √ 3x − 1 + 5 7. f (x. f (x) = 5 3x − 1 2. f (x) = 42 − x 4. f (x) = 1 − 6. f (x) = 2 − 4 + 3x 5 √ x − 5 8. f (x) = 1 − 2 √ 2x + 5 √ 10. f (x) = 3 − 3 x − 2 11. f (x) = x2 − 10x, x ≥ 5 12. f (x) = 3(x + 4)2 − 5, x ≤ −4 13. f (x) = x2 − 6x + 5, x ≤ 3 1...
Opis per week is P (x) = −15x2 + 350x − 2000, for 0 ≤ x ≤ 30. Find P ◦ p−1 (x) and determine what price per dOpi would yield the maximum proﬁt. What is the maximum proﬁt? How many dOpis need to be produced and sold to achieve the maximum proﬁt? 5.2 Inverse Functions 395 26. Show that the Fahrenheit to Celsius conversio...
in Functions 5.2.2 Answers 1. f −1(x) = x + 2 6 3. f −1(x) = 3x − 10 5. f −1(x) = 1 3 (x − 5)2 + 1 3, x ≥ 5 7. f −1(x) = 1 9 (x + 4)2 + 1, x ≥ −4 9. f −1(x) = 1 3 x5 + 1 √ 3 11. f −1(x) = 5 + x + 25 2. f −1(x) = 42 − x 4. f −1(x. f −1(x) = (x − 2)2 + 5, x ≤ 2 8. f −1(x) = 1 8 (x − 1)2 − 5 2, x ≤ 1 10. f −1(x) = −(x − ...
can produce and sell p−1(275) = 11.6 dOpis. Since we cannot sell part of a system, we need to adjust the price to sell either 11 dOpis or 12 dOpis. We ﬁnd p(11) = 285 and p(12) = 270, which means we set the price per dOpi at either $285 or $270, respectively. The proﬁts at these prices are P ◦ p−1 (285) = 35 and P ◦ p...
ﬁne f (x) = n x functionally as the inverse of g(x) = xn with the stipulation that when n is even, the domain of g is restricted to [0, ∞). From what we know about g(x) = xn from Section 3.1 along with Theorem 5.3, we can produce √ x by reﬂecting the graphs of g(x) = xn across the line y = x. Below are the the graphs o...
following. If n is odd, then by deﬁnition n √ √ xy)n = xy. Given that n is the unique real number such that ( n = xy, √ √ √ it must be the case that n xy is the unique non-negative real x n y. If n is even, then n √ √ xy)n = xy. Also note that since n is even, n number such that ( n y are also non-negative √ √ √ and h...
= x. = x √ −12 However, if we substitute x = −1 and apply Deﬁnition 5.5, we ﬁnd (−1)2/3 = 3 = (−1)2 = 1 so that (−1)2/33/2 = x. If we take the time to rewrite x2/33/2 = 13 = 1. We see in this case that x2/33/2 = 13/2 = √ with radicals, we see 13 2 x2/33/2 √ x23/2 3 = = 3 √ 3 x2 √ = 3 3 x = √ 3 x3 = \|x\| 2Otherwise we’d...
some of the new features of these graphs. Example 5.3.1. For the following functions, state their domains and create sign diagrams. Check your answer graphically using your calculator. √ 1. f (x) = 3x 3 2 − x 3. h(x) = 3 8x x + 1 Solution. 2. g(x. k(x) = √ 2x x2 − 1 1. As far as domain is concerned, f (x) has no denom...
only this portion of the number line. To ﬁnd the zeros of r we set √ √ = 24 from which r(x) = 0 and solve 2 − 4 x + 3 = 0. We get 4 we obtain x + 3 = 16 or x = 13. Since we raised both sides of an equation to an even power, we need to check to see if x = 13 is an extraneous solution.7 We ﬁnd x = 13 does check since √ ...
of h, we set h(x) = 0. To 8x solve 3 x+1 = 0. We get 8x = 0, or x = 0. Below is the resulting sign diagram and corresponding graph. From the graph, it appears as though x = −1 is a vertical asymptote. Carrying out an analysis as x → −1 as in Section 4.2 conﬁrms this. (We leave the details to the reader.) Near x = 0, w...
otes, one at x = −1 and one at x = 1. The gap in the graph between the asymptotes is because of the gap in the domain of k. Concerning end behavior, there appear to be two horizontal asymptotes, y = 2 and y = −2. To see why this is the case, we think of x → ±∞. The radicand of the denominator x2 − 1 ≈ x2, and as such, ...
feature. 5.3 Other Algebraic Functions 403 1. x2/3 < x4/3 − 6 2. 3(2 − x)1/3 ≤ x(2 − x)−2/3 Solution. 1. To solve x2/3 < x4/3 − 6, we get 0 on one side and attempt to solve x4/3 − x2/3 − 6 > 0. We set r(x) = x4/3 − x2/3 − 6 and note that since the denominators in the exponents are 3, they correspond to cube roots, whi...
) = x2/3 and g(x) = x4/3 − 6. The solution to x2/3 < x4/3 − 6 corresponds to the inequality f (x) < g(x), which means we are looking for the x values for which the graph of f is below the graph of g. Using the ‘Intersect’ command we conﬁrm10 that the graphs cross at x = ±3 3. We see that the graph of f is below the gra...
)2 = 0. After cubing both sides, and subsequently taking square roots, we get 2 − x = 0, or x = 2. Hence, the domain of r is (−∞, 2) ∪ (2, ∞). To ﬁnd the zeros of r, we set r(x) = 0. There are two school of thought on how to proceed and we demonstrate both. Factoring Approach. From r(x) = 3(2 − x)1/3 − x(2 − x)−2/3, we...
2 − x)2/3 1 3 − − = = = 3(2 − x)1/3(2 − x)2/3 (2 − x)2/3 3 + 2 3(2 − x) (2 − x)2/3 3(2 − x)3/3 (2 − x)2/3 3(2 − x)1 (2 − x)2/3 3(2 − x) − x (2 − x)2/3 6 − 4x (2 − x)2/3 As before, when we set r(x) = 0 we obtain x = 3 2. x (2 − x)2/3 x (2 − x)2/3 x (2 − x)2/3 = = = − − common denominator since 3√ √ u3 = ( 3 u)3 = u We n...
before heading oﬀ road directly towards the Outpost. Determine a reasonable applied domain for the problem. 2. Use your calculator to graph y = C(x) on its domain. What is the minimum cost? How far along Route 117 should the cable be run before turning oﬀ of the road? Solution. 1. The cost is broken into two parts: th...
graph and identify any vertical or horizontal asymp- totes, ‘unusual steepness’ or cusps. 1. f (x) = √ 1 − x2 √ 3. f (x) = x 1 − x2 5. f (x) = 4 16x x2 − 9 7. f (x) = x 2 3 (x − 7) 1 3 9. f (x) = x(x + 5)(x − 4) 2. f (x) = 4. f (x) = x √ x2 − 1 √ x2 − 1 6. f (x) = 3√ 5x x3 + 8 3 2 (x − 7) 1 3 8. f (x) = x 10. f (x) = ...
x 2 3 (x − 3)− 2 3 ≥ 0 30. − 4 3√ 32. 3 (x − 2)− 4 3 + 8 9 x(x − 2)− 7 3 ≥ 0 x3 + 3x2 − 6x − 8 > x + 1 3 4 (x − 3)− 2 4 (x − 3) 1 3 < 0 33. 1 3 x 34. x− 1 3 + 3 4 x− 1 3 − x− 4 3 (x − 3)− 2 3 (x − 3)− 5 3 (x2 − 3x + 2) ≥ 0 35. 2 3 (x + 4) 3 5 (x − 2)− 1 3 + 3 5 (x + 4)− 2 5 (x − 2) 2 3 ≥ 0 36. Rework Example 5.3.3 so ...
wind chill temperature. Round your answer to two decimal places. (b) Suppose the air temperature is 37◦F and the wind chill temperature is 30◦F. Find the wind speed. Round your answer to two decimal places. 39. As a follow-up to Exercise 38, suppose the air temperature is 28◦F. (a) Use the formula from Exercise 38 to ...
According to Einstein’s Theory of Special Relativity, the observed mass m of an object is a function of how fast the object is traveling. Speciﬁcally, m(x) = mr 1 − x2 c2 where m(0) = mr is the mass of the object at rest, x is the speed of the object and c is the speed of light. (a) Find the applied domain of the func...
) With the help of your classmates, generalize parts (a) and (b) to two cases: v2 > v1 and v2 < v1. We will discuss the case of v1 = v2 in Exercise 32 in Section 6.5. 45. Verify the Quotient Rule for Radicals in Theorem 5.6. 46. Show that 3 2 x 2 3 = x for all x ≥ 0. √ 47. Show that 3 2 is an irrational number by ﬁrst ...
16x x2 − 9 Domain: (−3, 0] ∪ (3, ∞) 0 (+) (+) −3 0 3 Vertical asymptotes: x = −3 and x = 3 Horizontal asymptote: y = 0 Unusual steepness at x = 0 No cusps 6. f (x) = 3√ 5x x3 + 8 Domain: (−∞, −2) ∪ (−2, ∞) (+) (−) 0 (+) −2 0 Vertical asymptote x = −2 Horizontal asymptote y = 5 No unusual steepness or cusps 7. f (x) = ...
1 −1 1 3 5 7 9 11 x −2 −3 −5−4−3−2−5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 √ 12. g(x) = − √ 13. g(x) = 4 x − 1 − 2 √ 14. g(x5 −3 −1 1 3 5 7 x y −1 −2 1 3 5 7 9 11 13 15 17 19 21 x y 5 4 3 2 1 −1 7 8 23 x 14Using Calculus it can be shown that y = x + 1 is a slant asymptote of this graph. 414 Further Topics in Functions √ 15. ...
πr2 √ 2 = π2r6+90000 r, r > 0 (c) The calculator gives the absolute minimum at the point ≈ (4.07, 90.23). This means the radius should be (approximately) 4.07 centimeters and the height should be 5.76 centimeters to give a minimum surface area of 90.23 square centimeters. 38. (a) W ≈ 37.55◦F. (b) V ≈ 9.84 miles per ho...
Chewbacca’s path - in other words, where Fritzy catches Chewbacca. (b) y = 1 2x − 2 6 x3 + 1 3. Using the techniques from Chapter 4, we ﬁnd as x → 0+, y → ∞ which means, in this case, Fritzy’s pursuit never ends; he never catches Chewbacca. This makes sense since Chewbacca has a head start and is running faster than F...
= −100 or x = −1000, the function f (x) = 2x takes on values like f (−100) = 2−100 = 1 2100 or f (−1000) = 2−1000 = 1 21000. In other words, as x → −∞, 2x ≈ 1 very big (+) ≈ very small (+) √ √ √ So as x → −∞, 2x → 0+. This is represented graphically using the x-axis (the line y = 0) as a horizontal asymptote. On the ﬂ...
1 −2 is not a real number. In general, if x is any rational number with an even denominator, then (−2)x is not deﬁned, so we must restrict our attention to bases b ≥ 0. What about b = 0? The function f (x) = 0x is undeﬁned for x ≤ 0 because we cannot divide by 0 and 00 is an indeterminant form. For x > 0, 0x = 0 so th...
of f (x) by reﬂecting it across the y-axis. We get = 2−x = f (−x), where f (x) = 2x. Thinking back to Section 1.7, = 2−1x 3−2−1 1 2 3 x y = f (x) = 2x reﬂect across y-axis −−−−−−−−−−−−→ multiply each x-coordinate by −1 −3−2−1 1 2 3 y = g(x) = 2−x = 1 2 x x We see that the domain and range of g match that of f, namely ...
Of all of the bases for exponential functions, two occur the most often in scientiﬁc circles. The ﬁrst, base 10, is often called the common base. The second base is an irrational number, e ≈ 2.718, called the natural base. We will more formally discuss the origins of this number in Section 6.5. For now, it is enough t...
horizontal 4 asymptote remains y = 0. Finally, we restrict the domain to [0, ∞) to ﬁt with the applied domain given to us. We have the result below. and 1, 4 5 x x y 2 (0, 1) −3−2−1 1 2 3 x H.A. y = 0 y = f (x) = 4 5 x y 30 (0, 25) 20 15 10 5 vertical scale by a factor of 25 −−−−−−−−−−−−−−−−−−−−−→ multiply each y-coor...
y = T (t) using transformations, we start with the basic function, f (t) = et. As we have already remarked, e ≈ 2.718 > 1 so the graph of f is an increasing exponential with y-intercept (0, 1) and horizontal asymptote y = 0. The points −1, e−1 ≈ (−1, 0.37) and (1, e) ≈ (1, 2.72) are also on the graph. Since the formul...
this in greater detail in Section 6.5. 422 Exponential and Logarithmic Functions y 7 6 5 4 3 2 (0, 1) y 180 160 140 120 100 80 60 40 20 H.A. y = 70 −3−2−1 1 2 3 t H.A. y = 0 y = f (t) = et −−−−−−−−−−−−→ y = T (t) 2 4 6 8 10 12 14 16 18 20 t 3. From the graph, we see that the horizontal asymptote is y = 70. It is worth...
is loge(x) and is usually written ln(x). Since logs are deﬁned as the inverses of exponential functions, we can use Theorems 5.2 and 5.3 to tell us about logarithmic functions. For example, we know that the domain of a log function is the range of an exponential function, namely (0, ∞), and that the range of a log fun...
of f and x = 0 is a vertical asymptote of the graph of f. f is one-to-one, continuous and smooth ba = c if and only if logb(c) = a. That is, logb(c) is the exponent you put on b to obtain c. logb (bx) = x for all x and blogb(x) = x for all x > 0 If b > 1: If 0 < b < 1: – f is always increasing – As x → 0+, f (x) → −∞ ...
0.001) Solution. 2. log2 1 8 6. 2log2(8) 3. log√ 5(25) 7. 117− log117(6) 3√ e2 4. ln 1. The number log3(81) is the exponent we put on 3 to get 81. As such, we want to write 81 as a power of 3. We ﬁnd 81 = 34, so that log3(81) = 4., we need rewrite 1 2. To ﬁnd log2 8 as a power of 2. We ﬁnd 1 1 8 8 = 1 √ 23 = 2−3, so lo...
of exponents followed by Theorem 6.2 to get9 117− log117(6) = 1 117log117(6) = 1 6 9It is worth a moment of your time to think your way through why 117log117(6) = 6. By deﬁnition, log117(6) is the exponent we put on 117 to get 6. What are we doing with this exponent? We are putting it on 117. By deﬁnition we get 6. In...
. Subtracting 3 gives us x = −3 as our asymptote, then multplying by −1 gives us the vertical asymptote x = 3. Next, we multiply the y-coordinates by 2 which results in a vertical stretch by a factor of 2, then we ﬁnish by subtracting 1 from the y-coordinates which shifts the graph down 1 unit. We leave it to the reade...
ned.) Our last example solidiﬁes this and reviews all of the material in the section. Example 6.1.5. Let f (x) = 2x−1 − 3. 1. Graph f using transformations and state the domain and range of f. 2. Explain why f is invertible and ﬁnd a formula for f −1(x). 3. Graph f −1 using transformations and state the domain and rang...
Test so f is one-to-one, hence invertible. To ﬁnd a formula for f −1(x), we normally set y = f (x), interchange the x and y, then proceed to solve for y. Doing so in this situation leads us to the equation x = 2y−1 − 3. We have yet to discuss how to solve this kind of equation, so we will attempt to ﬁnd the formula fo...
of f, and the range of f −1 is (−∞, ∞), which matches the domain of f. y 4 3 2 1 −3−2−1 −1 −2 −2−1 −1 −2 −(x) = log2(x) −−−−−−−−−−−−→ y = f −1(x) = log2(x + 3) + 1 4. We now verify that f (x) = 2x−1 − 3 and f −1(x) = log2(x + 3) + 1 satisfy the composition requirement for inverses. For all real numbers x, 428 Exponent...
. 1 3 −2 = 9 7. e0 = 1 10. log3 1 81 = −4 2. 5−3 = 1 125 5. 4 25 −1/2 = 5 2 8. log5(25) = 2 11. log 4 3 3 4 = −1 13. log(0.1) = −1 14. ln(e) = 1 In Exercises 16 - 42, evaluate the expression. 16. log3(27) 19. log6 1 36 22. log 1 5 (625) 25. log 1 1000000 28. log4(8) 31. log36 √ 4 36 34. log36 36216 37. log 3√ 105 40. l...
57. f (x) = ln(−2x3 − x2 + 13x − 6) 48. f (x) = log x2 + 9x + 18 4x − 20 50. f (x) = ln(4x − 20) + ln x2 + 9x + 18 52. f (x) = 4 54. f (x) = ln( √ 56. f (x) = log4(x) √ x − 4 − 3) −1 − x (x) log 1 2 In Exercises 58 - 63, sketch the graph of y = g(x) by starting with the graph of y = f (x) and using transformations. Tr...

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