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Suppose \( {10}\mathrm{{ml}} \) of \( {9}^{ \circ }C \) water is mixed with \( {60}\mathrm{{ml}} \) of \( {37}^{ \circ }C \) water. What will be the temperature of the mixture? | We base our answer on the concept of heat content in the fluids measured with a base of zero heat content at \( 0{}^{ \circ }\mathrm{C} \) . It is helpful for this example that the specific heat of water is 1 calorie per gram-degree \( \mathrm{C} \) . The definition of a calorie is the amount of heat required to raise ... | Yes |
Theorem 10.2.1 The Fundamental Theorem of Calculus. Suppose \( f \) is a continuous function defined on an interval \( \left\lbrack {a, b}\right\rbrack \) and \( G \) is the function defined for every \( x \) in \( \left\lbrack {a, b}\right\rbrack \) by\n\n\[ G\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \]\n... | Proof of the Fundamental Theorem of Calculus. We prove the theorem for the case that \( f \) is increasing. Suppose \( f \) is an increasing and continuous function defined on \( \left\lbrack {a, b}\right\rbrack \) and for each \( x \) in \( \left\lbrack {a, b}\right\rbrack, G\left( x\right) = {\int }_{a}^{x}f\left( t\... | Yes |
Theorem 10.3.1 Horizontal Graph Theorem. If \( D \) is a continuous function defined on an interval \( \left\lbrack {a, b}\right\rbrack \) and for every number \( x \) in \( \left( {a, b}\right) \) ,\n\n\[ \n{D}^{\prime }\left( x\right) = 0 \n\] \n\nthen there is a number \( C \) such that for every number \( x \) in \... | Proof. Let \( C = D\left( a\right) \) . Suppose \( x \) is in \( \left( {a, b}\right) \) . By the Mean Value Theorem 12.1.1 there is a number \( z \) between \( a \) and \( x \) such that \n\n\[ \nD\left( x\right) - D\left( a\right) = {D}^{\prime }\left( z\right) \left( {x - a}\right) . \n\] \n\nBecause \( {D}^{\prime ... | Yes |
Theorem 10.3.2 Parallel Graph Theorem. If \( F \) and \( G \) are functions defined on an interval \( \left\lbrack {a, b}\right\rbrack \) and for every \( x \) in \( \left\lbrack {a, b}\right\rbrack \)\n\n\[ \n{F}^{\prime }\left( x\right) = {G}^{\prime }\left( x\right) \n\]\n\nthen there is a number, \( C \), such that... | Proof. Let \( D\left( x\right) = F\left( x\right) - G\left( x\right) \) for every \( x \) in \( \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ \n{D}^{\prime }\left( x\right) = {\left\lbrack F\left( x\right) - G\left( x\right) \right\rbrack }^{\prime } = {F}^{\prime }\left( x\right) - {G}^{\prime }\left( x\right) = 0 ... | Yes |
Example 10.3.1 A \( {10}\mathrm{{cc}} \) syringe has cross-sectional area \( A{\mathrm{\;{cm}}}^{2} \) ; air inside the plunger is at atmospheric pressure \( {P}_{0} \) which we assume to be 1 Atmosphere (1 Atm); the plunger is a the 10 cc mark and the neck of the syringe is blocked. The plunger is depressed a distance... | \[ {\int }_{0}^{5}A{P}_{0}\frac{s}{{10} - s}d\frac{s}{A} = {\int }_{0}^{5}{P}_{0}\frac{s}{{10} - s}{ds} \]\n\nLet\n\[ W\left( x\right) = {P}_{0}{\int }_{0}^{x}\frac{s}{{10} - s}{ds} \]\n\nThen the work done in compressing the air is \( W\left( 5\right) \) . The Fundamental Theorem of Calculus asserts that\n\[ {W}^{\pri... | Yes |
Theorem 10.4.1 Fundamental Theorem of Calculus II. Suppose \( f \) is a continuous function defined on an interval \( \left\lbrack {a, b}\right\rbrack \) and \( \mathrm{F} \) is a function defined on \( \left\lbrack {a, b}\right\rbrack \) having the property that\n\n\[ \text{for every number}t\text{in}\left\lbrack {a, ... | Proof: Suppose the hypothesis of the theorem. Let \( G \) be the function defined on \( \left\lbrack {a, b}\right\rbrack \) by\n\n\[ G\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \]\n\nFor \( x \) in \( \left\lbrack {a, b}\right\rbrack ,{G}^{\prime }\left( x\right) = f\left( x\right) \).\n\n(i)\n\nFor \( x \)... | Yes |
Evaluate \( {\int }_{0}^{1}{t}^{2}{dt} \) . | Check that\n\n\[ \text{if}\;F\left( t\right) = \frac{{t}^{3}}{3}\;\text{then}\;{F}^{\prime }\left( t\right) = {\left\lbrack \frac{{t}^{3}}{3}\right\rbrack }^{\prime } = \frac{1}{3}{\left\lbrack {t}^{3}\right\rbrack }^{\prime } = \frac{1}{3}3{t}^{2} = {t}^{2}\text{.} \]\n\nIt follows that\n\n\[ {\int }_{0}^{1}{t}^{2}{dt... | Yes |
\[ \int {e}^{\left( {t}^{2}\right) }{tdt} \] | In the second equation, identify\n\n\[ u\left( t\right) = {t}^{2},\;{G}^{\prime }\left( u\right) = {e}^{u},\;{u}^{\prime }\left( t\right) = {2t}\;\text{ and }\;G\left( u\right) = {e}^{u} \]\n\nThen\n\n\[ \int {e}^{\left( {t}^{2}\right) }{tdt} = \int \frac{1}{2}{e}^{\left( {t}^{2}\right) }{2tdt} \]\n\nArithmetic\n\n\[ =... | Yes |
Consider solving the two problems\n\n\\[ \n\\int {\\left( 1 + {t}^{4}\\right) }^{10}{t}^{3}{dt}\\;\\text{ or }\\;\\int {\\left( 1 + {t}^{4}\\right) }^{10}{t}^{2}{dt} \n\\] | Because \\( {\\left( 1 + {t}^{4}\\right) }^{10} \\) can be expanded by multiplication (using either the binomial expansion formula or by making nine multiplications) and the expanded form is a polynomial, both integrands of these two problems are polynomials and the antiderivatives of polynomials can be easily computed... | Yes |
Compute \( \int \sin \left( {\pi t}\right) {dt} \) . | Identify\n\n\[ u\left( t\right) = {\pi t},\;{G}^{\prime }\left( u\right) = \sin u,\;{u}^{\prime }\left( t\right) = \pi \;\text{ and }\;G\left( u\right) = - \cos u \]\n\nThen\n\n\[ \int \sin \left( {\pi t}\right) {dt} = \int \frac{1}{\pi }\sin \left( {\pi t}\right) {\pi dt} = \frac{1}{\pi }\int \sin \left( {\pi t}\right... | Yes |
Compute \( \int \sqrt{{5t} - 4}{dt} \) . | Identify\n\n\[ u\left( t\right) = {5t} - 4,\;{G}^{\prime }\left( u\right) = \sqrt{u} = {\left( u\right) }^{1/2},\;{u}^{\prime }\left( t\right) = 5\;\text{ and }\;G\left( u\right) = \frac{{u}^{3/2}}{3/2} \]\n\nThen\n\n\[ \int \sqrt{{5t} - 4}{dt} = \frac{1}{5}\int \sqrt{{5t} - 4}{5dt} = \frac{1}{5}\int {\left( u\left( t\... | Yes |
Compute \( \int {\left( \ln t\right) }^{3}\frac{1}{t}{dt} \) . | Identify\n\n\[ u\left( t\right) = \ln t,\;{G}^{\prime }\left( u\right) = {u}^{3},\;{u}^{\prime }\left( t\right) = \frac{1}{t}\;\text{ and }\;G\left( u\right) = \frac{{u}^{4}}{4} \]\n\nThen\n\n\[ \int {\left( \ln t\right) }^{3}\frac{1}{t}{dt} = \int {\left( u\left( t\right) \right) }^{3}{u}^{\prime }\left( t\right) {dt}... | Yes |
Example 10.5.6 Compute \( \int \tan {xdx} \) . | This is a good one. First write\n\n\[ \int \tan {xdx} = \int \frac{\sin x}{\cos x}{dx} \]\n\nThen identify\n\n\[ u\left( x\right) = \cos x,\;{G}^{\prime }\left( u\right) = \frac{1}{u},\;{u}^{\prime }\left( x\right) = - \sin x\;\text{ and }\;G\left( u\right) = \ln u \]\n\nThen\n\n\[ \int \tan x = \int \frac{\sin x}{\cos... | Yes |
How to find the volume of a potato. | A potato is pictured in Figure 11.1A with marks along an axis at \( 2\mathrm{\;{cm}} \) intervals. A cross section of the potato at position \( {10}\mathrm{\;{cm}} \) showing an area of approximately \( {17}{\mathrm{\;{cm}}}^{2} \) is pictured in Figure 11.1B. The volume of the slice between stations \( 8\mathrm{\;{cm}... | Yes |
The volume of a sphere. Consider a sphere with radius \( R \) and center at the origin of three dimensional space. We will compute the volume of the hemisphere between \( x = 0 \) and \( x = R \) . See Figure 11.2A. Suppose \( x \) is between 0 and \( R \) . The cross section of the region at \( x \) is a circle of rad... | The indefinite integral is the antiderivative of a polynomial, and\n\n\[ \n\int \pi \left( {{R}^{2} - {x}^{2}}\right) {dx} = \int \left( {\pi {R}^{2} - \pi {x}^{2}}\right) {dx} = \pi {R}^{2}x - \pi \frac{{x}^{3}}{3} + C.\n\]\n\nBy the Fundamental Theorem of Calculus II,\n\n\[ \nV = {\int }_{0}^{R}\pi \left( {{R}^{2} - ... | Yes |
Compute the volume of the triangular solid with three mutually perpendicular edges of length 2, 3, and 4, illustrated in Figure 11.2B. | At height \( z \), the cross section is a right triangle with sides \( x \) and \( y \) and the area \( A\left( z\right) = \) \( \frac{1}{2}x \times y \) . Now\n\n\[ \frac{x}{2} = \frac{4 - z}{4}\;\text{ so }\;x = \frac{2}{4}\left( {4 - z}\right) .\;\text{ Similarly,}\;y = \frac{3}{4}\left( {4 - z}\right) .\n\nTherefor... | Yes |
Find the volume of the solid \( S \) generated by rotating the region \( R \) between \( y = {x}^{2} \) and \( y = \sqrt{x} \) about the \( x \) -axis. See Figure \( {11.3}\mathrm{\;B} \) . | Solution. There are two problems here. First we compute the volume of the solid, \( {S}_{1} \), generated by rotating the region below \( y = \sqrt{x},0 \leq x \leq 1 \), about the \( x \) -axis. Then we subtract the volume of the solid, \( {S}_{2} \), generated by rotating \( y = {x}^{2},0 \leq x \leq 1 \) about the \... | Yes |
For the integration problem, \[ \int \sqrt{1 + \sqrt{x}}{dx} \] substitute \[ x = g\left( z\right) = {z}^{2}, \] and \[ {dx} = {g}^{\prime }\left( z\right) {dz} = {2zdz}. \] | Then compute \[ \int \sqrt{1 + \sqrt{{z}^{2}}}{2z};{dz} = \int \sqrt{1 + z}{2zdz} \] \[ = \int \sqrt{1 + z}\left( {{2z} + 2 - 2}\right) {dz} \] \[ = 2\int {\left( 1 + z\right) }^{3/2} - {\left( 1 + z\right) }^{1/2}{dz} \] \[ = 2\left( {\frac{{\left( 1 + z\right) }^{5/2}}{5/2} - \frac{{\left( 1 + z\right) }^{3/2}}{3/2}}... | Yes |
Compute \( \int \sqrt{1 + {x}^{2}}{dx} \) . | Let \( x = \tan z \) . Then \( {dx} = {\sec }^{2}{zdz} \) . Compute\n\n\[ \int \sqrt{1 + {\tan }^{2}z}{\sec }^{2}{zdz} = \]\n\n\[ \int {\sec }^{3}{zdz} = \int \left( {\sec z}\right) \left( {1 + {\tan }^{2}z}\right) {dz} \]\n\n\[ = \int \frac{\left( {\sec z}\right) \left( {\sec z + \tan z}\right) }{\sec z + \tan z} + \l... | Yes |
We compute the volume, \( V \), of \( S \) . | \[ V = {\int }_{0}^{a}\pi {\left( b\sqrt[4]{x}\sqrt[4]{a - x}\right) }^{2}{dx} = \pi {b}^{2}{\int }_{0}^{a}\sqrt{x}\sqrt{a - x}{dx} \]\n\nLet \( x = {az} \) ; then \( {dx} = {adz} \) and \( x = 0 \) and \( x = a \) correspond to \( z = 0 \) and \( z = 1 \) . Then\n\n\[ V = \pi {b}^{2}{\int }_{0}^{1}\sqrt{az}\sqrt{a - {... | Yes |
Compute the centroid of the right circular cone of height \( H \) and base radius \( R \) . | We picture the cone as the solid of revolution obtained by rotating the graph of \( y = \left( {R/H}\right) \times x,0 \leq x \leq H \) about the \( X \) -axis. Then \( A\left( x\right) = \pi \times {\left( \left( R/H\right) x\right) }^{2} \) and the centroid \( c \) is computed from Equation 11.11 as\n\n\[ c = \frac{{... | Yes |
Suppose a horizontal flat plate of thickness 1 and uniform density has a horizontal outline bounded by the graph of \( y = {x}^{2}, y = 1 \), and \( x = 4 \) . Where is the centroid of the plate. | Solution. There are two problems here. The first is,’What is the \( x \) -coordinate, \( \bar{x} \), of the centroid’; the second is,’What is the \( y \) -coordinate, \( \bar{y} \), of the centroid?’ The \( z \) -coordinate of the centroid is one-half the thickness, \( \bar{z} = 1/2 \) .\n\nFor the \( x \) -coordinate ... | Yes |
Consider a hemispherical region generated by rotating the graph of \( y = \sqrt{1 - {x}^{2}},0 \leq x \leq 1 \) about the \( x \) axis that is filled with a substance that has density, \( \delta \left( x\right) \), equal to \( x \) . Find the center of mass of the substance. | Solution. At position \( x \), the radius of the cross section perpendicular to the \( X \) -axis is \( \sqrt{1 - {x}^{2}} \) , the area of the cross section is \( \pi \left( {1 - {x}^{2}}\right) \) . From Equation 11.10, the\n\n\[ \text{Center of mass} = \frac{{\int }_{a}^{b}{x\delta }\left( x\right) A\left( x\right) ... | Yes |
Find the length of the graph of \( y = \ln \cos x \) on \( 0 \leq x \leq \pi /4 \) . | We write\n\n\[ f\left( x\right) = \ln \cos x\;{f}^{\prime }\left( x\right) = {\left\lbrack \ln \cos x\right\rbrack }^{\prime }\;\frac{1}{\cos x}{\left\lbrack \cos x\right\rbrack }^{\prime } = \frac{1}{\cos x}\left( {-\cos x}\right) = - \tan x \]\n\nWe leave it to you to check that Bolt out of the Blue\n\n\[ {\left\lbra... | No |
Is there an initial vertical speed of a satellite that is sufficient to insure that it will escape the Earth's gravity field without further propulsion? | The work done along an axis against a variable force \( F\left( x\right) \) is\n\n\[ W = {\int }_{a}^{b}F\left( x\right) {dx} \]\n\nThe acceleration of gravity at an altitude \( x \) is given as\n\n\[ g\left( x\right) = {9.8} \times \frac{{R}^{2}}{{\left( R + x\right) }^{2}}\;\text{ meters }/{\mathrm{{sec}}}^{2} \]\n\n... | Yes |
The Mean Value Theorem asserts that there is a tangent to the parabola \( {y}^{2} = x \) that is parallel to the secant containing \( \left( {0,0}\right) \) and \( \left( {1,1}\right) \) . Let\n\n\[ f\left( x\right) = \sqrt{x}\;0 \leq x \leq 1. \]\n\nThe function, \( f \) is continuous on the closed interval \( \left\l... | The Mean Value Theorem asserts that there is a number \( c \) such that \( {f}^{\prime }\left( c\right) = 1 \) . We solve for \( c \) in\n\n\[ {f}^{\prime }\left( c\right) = 1\;\frac{1}{2\sqrt{c}} = 1\;\frac{1}{2} = \sqrt{c}\;c = \frac{1}{4}.\;\text{ Also,}\;f\left( \frac{1}{4}\right) = \frac{1}{2} \]\n\nThe tangent to... | Yes |
Theorem 12.1.2 Rolle’s Theorem. If a function, \( f \), defined on a closed interval, \( \left\lbrack {a, b}\right\rbrack \), satisfies\n\n1. \( f\left( a\right) = f\left( b\right) \).\n\n2. \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \).\n\n3. \( {f}^{\prime }\left( c\right) \) exists for every number... | The proof is seemingly harmless. We are looking for a horizontal tangent and therefore only have to look at a high point \( \left( {u, f\left( u\right) }\right) \) and a low point \( \left( {v, f\left( v\right) }\right) \) of \( f \) on \( \left\lbrack {a, b}\right\rbrack \) .\n\nIf \( a < u < b \), as in Figure 12.2, ... | Yes |
Let \( F \) be the function defined by \( F\left( x\right) = x + {x}^{3} \) for all numbers, \( x \). We will show that no horizontal line intersects the graph of \( F \) at two distinct points, which means that \( F \) is invertible. | Observe that\n\n\[ \n{F}^{\prime }\left( x\right) = {\left\lbrack x + {x}^{3}\right\rbrack }^{\prime } = 1 + 3{x}^{2}\; > 0\;\text{ for all }x.\n\]\n\nBy Rolle’s Theorem, if some horizontal line contains two points of the graph of \( F \) then at an intervening point there is a horizontal tangent and at that point \( {... | Yes |
If the usual properties of addition, multiplication and order of the number system are assumed, one may accept the statement of the Intermediate Value Theorem 12.1.4 as an axiom and prove the statement of the Axiom of Completeness Axiom 5.2.1 (repeated above) as a theorem. | Proof. Assume the usual properties of addition, multiplication, and order of the number system and the statement of the Intermediate Value Theorem. Suppose the statement of the Completion Axiom is not true. Then there are sets of numbers, \( {S}_{1} \) and \( {S}_{2} \) such that every number is in either \( {S}_{1} \)... | Yes |
Theorem 12.2.1 Suppose \( P \) is a continuous function defined on an interval \( \left\lbrack {A, B}\right\rbrack \) and at every point \( t \) in \( \left( {A, B}\right) {P}^{\prime }\left( t\right) \) exists and \( {P}^{\prime }\left( t\right) \geq 0 \) . Then \( P \) is nondecreasing on \( \left\lbrack {A, B}\right... | Proof. Suppose \( a \) and \( b \) are numbers in \( \left\lbrack {A, B}\right\rbrack \) and \( a < b \) . Then by the Mean Value Theorem, there is a number, \( c \), between \( a \) and \( b \) such that\n\n\[ \frac{P\left( b\right) - P\left( a\right) }{b - a} = {P}^{\prime }\left( c\right) \;\text{ so that }\;P\left(... | Yes |
Theorem 12.2.2 Suppose \( P \) is a continuous function defined on an interval \( \left\lbrack {A, B}\right\rbrack \) and at every point \( t \) in \( \left( {A, B}\right) {P}^{\prime }\left( t\right) \) exists and \( {P}^{\prime }\left( t\right) > 0 \) . Then \( P \) is increasing on \( \left\lbrack {A, B}\right\rbrac... | Proof: The argument is the similar to that for Theorem 8.1.1 and is left as Exercise 12.2.3 | No |
Theorem 12.2.3 Suppose \( f \) is a function with continuous first and second derivatives throughout an interval \( \left\lbrack {a, b}\right\rbrack \) and \( c \) is a number between \( a \) and \( b \) for which \( {f}^{\prime }\left( c\right) = 0 \) . Under these conditions:\n\n1. If \( {f}^{\prime \prime }\left( c\... | Proof. We prove case 1, for \( c \) to be a local minimum of \( f \) . It will be helpful to look at the numbers on the number lines shown in Figure 12.5. Because \( {f}^{\prime \prime } \) is continuous and positive at \( c \), there is an interval, \( \left( {u, v}\right) \), containing \( c \) so that \( {f}^{\prime... | Yes |
Find a polynomial, \( p\left( t\right) = {p}_{0} + {p}_{1}t + {p}_{2}{t}^{2} + {p}_{3}{t}^{3} + {p}_{4}{t}^{4} \) that approximates the solution to\n\n\[ \n{y}^{\prime }\left( t\right) = y\left( t\right) ,\;y\left( 0\right) = 1.\n\] | Solution: We need to find \( {p}_{0},{p}_{1},{p}_{2},{p}_{3} \), and \( {p}_{4} \) . We will insist that ’ \( p\left( t\right) \) matches \( y\left( t\right) \) at \( t = 0 \) ’ meaning that\n\n\[ \np\left( 0\right) = y\left( 0\right) ,\;{p}^{\prime }\left( 0\right) = {y}^{\prime }\left( 0\right) ,\;{p}^{\left( 2\right... | Yes |
Find a polynomial, \( p\left( t\right) = {p}_{0} + {p}_{,}t + {p}_{2}{t}^{2} + {p}_{3}{t}^{3} + {p}_{4}{t}^{4} \) that approximates the solution to\n\n\[ \n{y}^{\prime \prime }\left( t\right) + y\left( t\right) = 0,\;y\left( 0\right) = 1,\;{y}^{\prime }\left( 0\right) = 0.\n\] | Solution: We need to find \( {p}_{0},{p}_{1},{p}_{2},{p}_{3} \), and \( {p}_{4} \) . We will insist that ’ \( p\left( t\right) \) match \( y\left( t\right) \) at \( t = 0 \) ’ meaning that\n\n\[ \np\left( 0\right) = y\left( 0\right) ,\;{p}^{\prime }\left( 0\right) = {y}^{\prime }\left( 0\right) ,\;{p}^{\left( 2\right) ... | Yes |
Find a polynomial, \( p\left( t\right) = {p}_{0} + {p}_{1}t + {p}_{2}{t}^{2} + {p}_{3}{t}^{3} + {p}_{4}{t}^{4} + {p}_{5},{t}^{5} \) that approximates the solution to the logistic equation\n\n\[ \n{y}^{\prime }\left( t\right) = y\left( t\right) \left( {1 - y\left( t\right) }\right) ,\;y\left( 0\right) = 1/2.\n\] | Solution: As in Examples 12.5.1 and 12.5.2 we will match derivatives at \( t = 0 \) to find the coefficients of \( p \) .\n\n\[ \n{y}^{\prime }\left( t\right) = y\left( t\right) - {y}^{2}\left( t\right) \n\]\n\n\[ \n{y}^{\prime }\left( 0\right) = \frac{1}{4} \n\]\n\n\[ \n{y}^{\left( 2\right) }\left( t\right) = y\prime ... | Yes |
Suppose you are monitoring a rare avian population and you have the data shown. What is your best estimate for the number of adults on January 1, 2003? | Let \( t \) measure years after the year 2000, so that \( t = 0, t = 1 \), and \( t = 2 \) correspond to the years 2000,2001, and 2002, respectively, and let \( A\left( t\right) \) denote adult avian population at time \( t \) . Our problem is to estimate \( A\left( 3\right) \) .\n\nSolution 1. We know \( A\left( 1\rig... | Yes |
What is the error in\n\\[ \sin \frac{\pi }{4} \doteq \frac{\pi }{4} - \frac{1}{3!}{\\left( \frac{\pi }{4}\\right) }^{3}? \\] | The answer is\n\\[ \\text{exactly}\\;\\frac{\\sin \\left( {\\mathbf{c}}_{\\mathbf{3}}\\right) }{\\mathbf{4}!}{\\left( \frac{\\pi }{4} - \\mathbf{0}\\right) }^{\\mathbf{4}}\\text{.}\n\\]\nSome fuzz appears. We do not know the value of \\( {c}_{3} \\) . We only know that it is a number between 0 and \\( \frac{\\pi }{4} \... | Yes |
If \( \;{f}^{\prime }\left( a\right) = 0\; \) and \( \;{f}^{\prime \prime }\left( a\right) < 0\; \) then \( \left( {a, f\left( a\right) }\right) \) is a local maximum for \( f \) . | We suppose that \( {f}^{\left( 2\right) } \) is continuous so that \( {f}^{\left( 2\right) }\left( a\right) < 0 \) implies that there is an interval, \( \left( {p, q}\right) \) surrounding \( a \) so that \( {f}^{\left( 2\right) }\left( x\right) < 0 \) for all \( x \) in \( \left( {p, q}\right) \) . Suppose \( b \) is ... | Yes |
Example 12.7.3 Taylor's Theorem provides a good explanation as to why the centered difference quotient is usually a better approximation to \( {f}^{\prime }\left( a\right) \) than is the forward difference quotient. Directly from Taylor’s formula with \( b = a + h, b - a = h \) is | \[ f\left( {a + h}\right) = f\left( a\right) + {f}^{\prime }\left( a\right) h + \frac{{f}^{\left( 2\right) }\left( {c}_{1}\right) }{2!}{h}^{2} \] we can solve for \( {f}^{\prime }\left( a\right) \) and get \[ {f}^{\prime }\left( a\right) = \frac{f\left( {a + h}\right) - f\left( a\right) }{h} - \frac{{f}^{\left( 2\right... | Yes |
Property 13.1.1 A property of tangents to functions of one variable. Suppose \( f \) is a function of one variable and at a number \( a \) in its domain, \( {f}^{\prime }\left( a\right) \) exists. The graph of \( L\left( x\right) = f\left( a\right) + {f}^{\prime }\left( a\right) \left( {x - a}\right) \) is the tangent ... | \[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{\left| f\left( x\right) - L\left( x\right) \right| }{\left| x - a\right| } = \mathop{\lim }\limits_{{x \rightarrow a}}\left| \frac{f\left( x\right) - f\left( a\right) - {f}^{\prime }\left( a\right) \left( {x - a}\right) }{x - a}\right| \] \[ = \left| {\mathop{\lim }\lim... | Yes |
The graphs of \( F\left( {x, y}\right) = {x}^{2} + {y}^{2}, G\left( {x, y}\right) = {x}^{2} - {y}^{2} \) and \( H\left( {x, y}\right) = - {x}^{2} - {y}^{2} \) shown in Figure 13.10 illustrate three important options. The origin, \( \left( {0,0}\right) \), is a critical point of each of the graphs and the \( z = 0 \) pl... | For \( F \), for example,\n\n\[ F\left( {x, y}\right) = {x}^{2} + {y}^{2}\;{F}_{1}\left( {x, y}\right) = {2x}\;{F}_{2}\left( {x, y}\right) = {2y} \]\n\n\[ F\left( {0,0}\right) = 0\;{F}_{1}\left( {0,0}\right) = 0\;{F}_{2}\left( {0,0}\right) = 0 \]\n\nThe origin, \( \left( {0,0}\right) \), is a critical point of \( F \),... | Yes |
Fit a line to the data in Example Figure 13.2.3.3. | \[ {S}_{x} = 1 + 2 + 4 + 6 + 8 = {21}, \] \[ {S}_{y} = {0.5} + {0.8} + {1.0} + {1.7} + {1.8} = {5.8} \] \[ {S}_{xx} = {1}^{2} + {2}^{2} + {4}^{2} + {6}^{2} + {8}^{2} = {121}, \] \[ {S}_{xy} = 1 \cdot {0.5} + 2 \cdot {0.8} + 4 \cdot {1.0} + 6 \cdot {1.7} + 8 \cdot {1.8} = {30.7} \] \[ \Delta = n{S}_{xx} - {\left( {S}_{x... | Yes |
Find the dimensions of the largest box (rectangular solid) that will fit in a hemisphere of radius \( R \) . | Solution. Assume the hemisphere is the graph of \( z = \sqrt{{R}^{2} - {x}^{2} - {y}^{2}} \) and that the optimum box has one face in the \( x, y \) -plane and the other four corners on the hemisphere (see Figure 13.2.4.4).\n\nThe volume, \( V \) of the box is\n\n\[ V\left( {x, y}\right) = {2x} \times {2y} \times z = {... | Yes |
Example 13.3.1 The domain of the function \( F\left( {x, y}\right) = \left( {\sin {\pi x}}\right) \left( {\cos \frac{\pi }{2}y}\right) \) shown in Figure 13.12 is \( 0 \leq x \leq 1,0 \leq y \leq 1 \) . Choose \( f\left( x\right) = 0 \) and \( g\left( x\right) = 1 \), for \( 0 \leq x \leq 1 \) . The volume of the regio... | \[ {\int }_{0}^{1}{\int }_{0}^{1}\left( {\sin {\pi x}}\right) \left( {\cos \frac{\pi }{2}y}\right) {dydx} = {\int }_{0}^{1}{\left\lbrack \left( \sin \pi x\right) \left( \sin \frac{\pi }{2}y\right) \times \frac{2}{\pi }\right\rbrack }_{y = 0}^{y = 1}{dx} \] \[ = \frac{2}{\pi }{\int }_{0}^{1}\sin {\pi xdx} \] \[ = \frac{... | Yes |
Show that\n\n\[ \n{u}_{t}\left( {x, t}\right) = \frac{1}{{\pi }^{2}}{u}_{xx}\left( {x, t}\right) ,\;u\left( {x,0}\right) = {30} * \left( {1 - \cos {\pi x}}\right) ,\;\text{ and }\;\begin{array}{l} {u}_{x}\left( {0, t}\right) = 0 \\ {u}_{x}\left( {2, t}\right) = 0. \end{array} \n\] | Solution. First compute some partial derivatives.\n\n\[ \nu\left( {x, t}\right) = {30}\left( {1 - {e}^{-t}\cos {\pi x}}\right) \]\n\n\[ {u}_{t}\left( {x, t}\right) = {30}\left( {0 - \left( {e}^{-t}\right) \left( {-1}\right) \cos {\pi x}}\right) = {30}{e}^{-t}\cos \left( {\pi x}\right) \]\n\n\[ {u}_{x}\left( {x, t}\righ... | Yes |
Express the following sets of numbers using interval notation. | 1. The best way to proceed here is to graph the set of numbers on the number line and glean the answer from it. The inequality \( x \leq - 2 \) corresponds to the interval \( ( - \infty , - 2\rbrack \) and the inequality \( x \geq 2 \) corresponds to the interval \( \lbrack 2,\infty ) \) . Since we are looking to descr... | Yes |
Example 1.1.2. Plot the following points: \( A\left( {5,8}\right), B\left( {-\frac{5}{2},3}\right), C\left( {-{5.8}, - 3}\right), D\left( {{4.5}, - 1}\right), E\left( {5,0}\right) \) , \( F\left( {0,5}\right), G\left( {-7,0}\right), H\left( {0, - 9}\right), O\left( {0,0}\right) .{}^{10} \) | 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 ... | Yes |
Example 1.1.3. Let \( P \) be the point \( \left( {-2,3}\right) \) . Find the points which are symmetric to \( P \) about the:\n\n1. \( x \) -axis 2. \( y \) -axis 3. origin\n\nCheck 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 \( P\left( {-2,3}\right) \) .\n\n---\n\n1. To find the point symmetric about the \( x \) -axis, we replace the \( y \) -coordinate with i... | Yes |
Find and simplify the distance between \( P\left( {-2,3}\right) \) and \( Q\left( {1, - 3}\right) \) . | \[ d = \sqrt{{\left( {x}_{1} - {x}_{0}\right) }^{2} + {\left( {y}_{1} - {y}_{0}\right) }^{2}} \] \[ = \sqrt{{\left( 1 - \left( -2\right) \right) }^{2} + {\left( -3 - 3\right) }^{2}} \] \[ = \sqrt{9 + {36}} \] \[ = 3\sqrt{5} \] So the distance is \( 3\sqrt{5} \) . | Yes |
Find all of the points with \( x \) -coordinate 1 which are 4 units from the point \( \\left( {3,2}\\right) \) . | 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 \( \\left( {1, y}\\right) \). Visually,\n\n\n\nWe require that the distanc... | Yes |
Find the midpoint of the line segment connecting \( P\left( {-2,3}\right) \) and \( Q\left( {1, - 3}\right) \) . | \[ M = \left( {\frac{{x}_{0} + {x}_{1}}{2},\frac{{y}_{0} + {y}_{1}}{2}}\right) \] \[ = \left( {\frac{\left( {-2}\right) + 1}{2},\frac{3 + \left( {-3}\right) }{2}}\right) = \left( {-\frac{1}{2},\frac{0}{2}}\right) \] \[ = \left( {-\frac{1}{2},0}\right) \] The midpoint is \( \left( {-\frac{1}{2},0}\right) \). | Yes |
Example 1.1.7. If \( a \neq b \), prove that the line \( y = x \) equally divides the line segment with endpoints \( \left( {a, b}\right) \) and \( \left( {b, a}\right) \) . | Solution. To prove the claim, we use Equation 1.2 to find the midpoint\n\n\[ M = \left( {\frac{a + b}{2},\frac{b + a}{2}}\right) \]\n\n\[ = \left( {\frac{a + b}{2},\frac{a + b}{2}}\right) \]\n\nSince the \( x \) and \( y \) coordinates of this point are the same, we find that the midpoint lies on the line \( y = x \), ... | Yes |
1. \( A = \{ \left( {0,0}\right) ,\left( {-3,1}\right) ,\left( {4,2}\right) ,\left( {-3,2}\right) \} \) | 1. To graph \( A \), we simply plot all of the points which belong to \( A \), as shown below on the left. | No |
Determine whether or not \( \left( {2, - 1}\right) \) is on the graph of \( {x}^{2} + {y}^{3} = 1 \) . | We substitute \( x = 2 \) and \( y = - 1 \) into the equation to see if the equation is satisfied.\n\n\[{\left( 2\right) }^{2} + {\left( -1\right) }^{3}\overset{?}{ = }1\]\n\n\[3 \neq 1\]\n\nHence, \( \left( {2, - 1}\right) \) is not on the graph of \( {x}^{2} + {y}^{3} = 1 \) . | Yes |
Example 1.2.3. Graph \( {x}^{2} + {y}^{3} = 1 \) . | Solution. To efficiently generate points on the graph of this equation, we first solve for \( y \)\n\n\[ \n{x}^{2} + {y}^{3} = 1 \n\]\n\n\[ \n{y}^{3} = 1 - {x}^{2} \n\]\n\n\[ \n\sqrt[3]{{y}^{3}} = \sqrt[3]{1 - {x}^{2}} \n\]\n\n\[ \ny = \sqrt[3]{1 - {x}^{2}} \n\]\n\nWe now substitute a value in for \( x \), determine th... | Yes |
Find the \( x \) - and \( y \) -intercepts (if any) of the graph of \( {\left( x - 2\right) }^{2} + {y}^{2} = 1 \) . Test for symmetry. Plot additional points as needed to complete the graph. | Solution. To look for \( x \) -intercepts, we set \( y = 0 \) and solve\n\n\[{\left( x - 2\right) }^{2} + {y}^{2} = 1\]\n\n\[{\left( x - 2\right) }^{2} + {0}^{2} = 1\]\n\n\[{\left( x - 2\right) }^{2} = 1\]\n\n\[\sqrt{{\left( x - 2\right) }^{2}} = \sqrt{1}\;\text{extract square roots}\]\n\n\[x - 2 = \pm 1\]\n\n\[x = 2 \... | Yes |
Example 1.3.1. Which of the following relations describe \( y \) as a function of \( x \) ?\n\n1. \( {R}_{1} = \{ \left( {-2,1}\right) ,\left( {1,3}\right) ,\left( {1,4}\right) ,\left( {3, - 1}\right) \} \) 2. \( {R}_{2} = \{ \left( {-2,1}\right) ,\left( {1,3}\right) ,\left( {2,3}\right) ,\left( {3, - 1}\right) \} \) | Solution. A quick scan of the points in \( {R}_{1} \) reveals that the \( x \) -coordinate 1 is matched with two different \( y \) -coordinates: namely 3 and 4. Hence in \( {R}_{1}, y \) is not a function of \( x \) . On the other hand, every \( x \) -coordinate in \( {R}_{2} \) occurs only once which means each \( x \... | Yes |
Use the Vertical Line Test to determine which of the following relations describes \( y \) as a function of \( x \) . | 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 \( x \) ... | Yes |
Use the Vertical Line Test to determine which of the following relations describes \( y \) as a function of \( x \) . | Both \( {S}_{1} \) and \( {S}_{2} \) are slight modifications to the relation \( S \) in the previous example whose graph we determined passed the Vertical Line Test. In both \( {S}_{1} \) and \( {S}_{2} \), it is the addition of the point \( \left( {1,2}\right) \) which threatens to cause trouble. In \( {S}_{1} \), th... | Yes |
Find the domain and range of the function \( F = \{ \left( {-3,2}\right) ,\left( {0,1}\right) ,\left( {4,2}\right) ,\left( {5,2}\right) \} \) 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 ... | Yes |
Example 1.3.5. Determine which equations represent \( y \) as a function of \( x \) . | Solution. For each of these equations, we solve for \( y \) and determine whether each choice of \( x \) will determine only one corresponding value of \( y \) .\n\n1.\n\n\[ \n{x}^{3} + {y}^{2} = 1 \n\] \n\n\[ \n{y}^{2} = 1 - {x}^{3} \n\] \n\n\[ \n\sqrt{{y}^{2}} = \sqrt{1 - {x}^{3}}\;\text{extract square roots} \n\] \n... | Yes |
Suppose a function \( g \) is described by applying the following steps, in sequence\n\n1. add 4\n\n2. multiply by 3\n\nDetermine \( g\\left( 5\\right) \) and find an expression for \( g\\left( x\\right) \) . | Solution. Starting with 5, step 1 gives \( 5 + 4 = 9 \) . Continuing with step 2, we get \( \\left( 3\\right) \\left( 9\\right) = {27} \) . To find a formula for \( g\\left( x\\right) \\), 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 pare... | Yes |
1. (a) \( f\left( {-1}\right), f\left( 0\right), f\left( 2\right) \) | To find \( f\left( {-1}\right) \), we replace every occurrence of \( x \) in the expression \( f\left( x\right) \) with -1\n\n\[ f\left( {-1}\right) = - {\left( -1\right) }^{2} + 3\left( {-1}\right) + 4 \]\n\n\[ = - \left( 1\right) + \left( {-3}\right) + 4 \]\n\n\[ = 0 \]\n\nSimilarly, \( f\left( 0\right) = - {\left( 0... | Yes |
Find the domain \( {}^{3} \) of the following functions.\n\n1. \( g\left( x\right) = \sqrt{4 - {3x}} \) 2. \( h\left( x\right) = \sqrt[5]{4 - {3x}} \) | 1. The potential disaster for \( g \) is if the radicand \( {}^{4} \) is negative. To avoid this, we set \( 4 - {3x} \geq 0 \) . From this, we get \( {3x} \leq 4 \) or \( x \leq \frac{4}{3} \) . What this shows is that as long as \( x \leq \frac{4}{3} \), the expression \( 4 - {3x} \geq 0 \), and the formula \( g\left(... | Yes |
1. Find and interpret \( h\left( {10}\right) \) and \( h\left( {60}\right) \) . | 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 \leq t ... | Yes |
1. Find \( \left( {f + g}\right) \left( {-1}\right) \) | To find \( \left( {f + g}\right) \left( {-1}\right) \) we first find \( f\left( {-1}\right) = 8 \) and \( g\left( {-1}\right) = 4 \) . By definition, we have that \( \left( {f + g}\right) \left( {-1}\right) = f\left( {-1}\right) + g\left( {-1}\right) = 8 + 4 = {12}. \) | Yes |
Find and simplify the difference quotients for the following functions 1. \( f\left( x\right) = {x}^{2} - x - 2 \) | To find \( f\left( {x + h}\right) \), we replace every occurrence of \( x \) in the formula \( f\left( x\right) = {x}^{2} - x - 2 \) with the quantity \( \left( {x + h}\right) \) to get \[ f\left( {x + h}\right) = {\left( x + h\right) }^{2} - \left( {x + h}\right) - 2 \] \[ = {x}^{2} + {2xh} + {h}^{2} - x - h - 2\text{... | Yes |
1. Find and interpret \( C\left( 0\right) \) . | We substitute \( x = 0 \) into the formula for \( C\left( x\right) \) and get \( C\left( 0\right) = {100}\left( 0\right) + {2000} = {2000} \) . 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 m... | Yes |
Example 1.6.1. Graph \( f\left( x\right) = {x}^{2} - x - 6 \) . | Solution. To graph \( f \), we graph the equation \( y = f\left( x\right) \) . 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\left( x\right)... | Yes |
Graph: \( f\left( x\right) = \left\{ \begin{matrix} 4 - {x}^{2} & \text{ if } & x < 1 \\ x - 3, & \text{ if } & x \geq 1 \end{matrix}\right. \) | 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 \( f\left( x\right) = 0 \) . The twist is that we have two formulas for \( f\left( x\right) \) . For \( x < 1 \), we use the formula \( f\left( x\ri... | Yes |
Determine analytically if the following functions are even, odd, or neither even nor odd. Verify your result with a graphing calculator. | The first step in all of these problems is to replace \( x \) with \( - x \) and simplify.\n\n1.\n\n\[ f\left( x\right) = \frac{5}{2 - {x}^{2}} \]\n\n\[ f\left( {-x}\right) = \frac{5}{2 - {\left( -x\right) }^{2}} \]\n\n\[ f\left( {-x}\right) = \frac{5}{2 - {x}^{2}} \]\n\n\[ f\left( {-x}\right) = f\left( x\right) \]\n\n... | Yes |
1. Find the domain of \( f \). | 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 \( \left\lbrack {-4,4}\right\rbrack \) . | Yes |
Example 1.6.5. Let \( f\left( x\right) = \frac{15x}{{x}^{2} + 3} \) . Use a graphing calculator to approximate the intervals on which \( f \) is increasing and those on which it is decreasing. Approximate all extrema. | Solution. Entering this function into the calculator gives\n\n \n\nUsing the Minimum and Maximum features, we ge... | Yes |
Find the points on the graph of \( y = {\left( x - 3\right) }^{2} \) which are closest to the origin. Round your answers to two decimal places. | Solution. Suppose a point \( \left( {x, y}\right) \) is on the graph of \( y = {\left( x - 3\right) }^{2} \) . Its distance to the origin \( \left( {0,0}\right) \) is given by\n\n\[ d = \sqrt{{\left( x - 0\right) }^{2} + {\left( y - 0\right) }^{2}} \]\n\n\[ = \sqrt{{x}^{2} + {y}^{2}} \]\n\n\[ = \sqrt{{x}^{2} + {\left\l... | Yes |
Theorem 1.2. Vertical Shifts. Suppose \( f \) is a function and \( k \) is a positive number. | - To graph \( y = f\\left( x\\right) + k \), shift the graph of \( y = f\\left( x\\right) \) up \( k \) units by adding \( k \) to the \( y \)-coordinates of the points on the graph of \( f \). \n\n- To graph \( y = f\\left( x\\right) - k \), shift the graph of \( y = f\\left( x\\right) \) down \( k \) units by subtrac... | Yes |
Theorem 1.3. Horizontal Shifts. Suppose \( f \) is a function and \( h \) is a positive number. | - To graph \( y = f\\left( {x + h}\\right) \), shift the graph of \( y = f\\left( x\\right) \) left \( h \) units by subtracting \( h \) from the \( x \) -coordinates of the points on the graph of \( f \) .\n\n- To graph \( y = f\\left( {x - h}\\right) \), shift the graph of \( y = f\\left( x\\right) \) right \( h \) u... | Yes |
1. Graph \( f\left( x\right) = \sqrt{x} \) . Plot at least three points. | ## Solution.\n\n1. Owing to the square root, the domain of \( f \) is \( x \geq 0 \), or \( \lbrack 0,\infty ) \) . We choose perfect squares to build our table and graph below. From the graph we verify the domain of \( f \) is \( \lbrack 0,\infty ) \) and the range of \( f \) is also \( \lbrack 0,\infty ) \) .\n\n<tab... | Yes |
Theorem 1.4. Reflections. Suppose \( f \) is a function. | - To graph \( y = - f\left( x\right) \), reflect the graph of \( y = f\left( x\right) \) across the \( x \) -axis by multiplying the \( y \) -coordinates of the points on the graph of \( f \) by -1 .\n\n- To graph \( y = f\left( {-x}\right) \), reflect the graph of \( y = f\left( x\right) \) across the \( y \) -axis by... | Yes |
Example 1.7.2. Let \( f\left( x\right) = \sqrt{x} \) . Use the graph of \( f \) from Example 1.7.1 to graph the following functions. Also, state their domains and ranges.\n\n1. \( g\left( x\right) = \sqrt{-x} \) 2. \( j\left( x\right) = \sqrt{3 - x} \) 3. \( m\left( x\right) = 3 - \sqrt{x} \) | ## Solution.\n\n1. The mere sight of \( \sqrt{-x} \) usually causes alarm, if not panic. When we discussed domains in Section 1.4, we clearly banished negatives from the radicands of even roots. However, we must remember that \( x \) is a variable, and as such, the quantity \( - x \) isn’t always negative. For example,... | Yes |
Theorem 1.5. Vertical Scalings. Suppose \( f \) is a function and \( a > 0 \) . To graph \( y = {af}\left( x\right) \) , multiply all of the \( y \) -coordinates of the points on the graph of \( f \) by \( a \) . We say the graph of \( f \) has been vertically scaled by a factor of \( a \) . | - If \( a > 1 \), we say the graph of \( f \) has undergone a vertical stretching (expansion, dilation) by a factor of \( a \) .\n\n- If \( 0 < a < 1 \), we say the graph of \( f \) has undergone a vertical shrinking (compression, contraction) by a factor of \( \frac{1}{a} \) . | Yes |
Let \( f\left( x\right) = \sqrt{x} \) . Use the graph of \( f \) from Example 1.7.1 to graph the following functions. Also, state their domains and ranges.\n\n1. \( g\left( x\right) = 3\sqrt{x} \) | Solution.\n\n1. First we note that the domain of \( g \) is \( \lbrack 0,\infty ) \) for the usual reason. Next, we have \( g\left( x\right) = {3f}\left( x\right) \) so by Theorem 1.5, we obtain the graph of \( g \) by multiplying all of the \( y \) -coordinates of the points on the graph of \( f \) by 3 . The result i... | Yes |
Theorem 1.7. Transformations. Suppose \( f \) is a function. If \( A \neq 0 \) and \( B \neq 0 \), then to graph\n\n\[ g\left( x\right) = {Af}\left( {{Bx} + H}\right) + K \] | 1. Subtract \( H \) from each of the \( x \) -coordinates of the points on the graph of \( f \) . This results in a horizontal shift to the left if \( H > 0 \) or right if \( H < 0 \) .\n\n2. Divide the \( x \) -coordinates of the points on the graph obtained in Step 1 by \( B \) . This results in a horizontal scaling,... | Yes |
Example 1.7.5. Let \( f\left( x\right) = {x}^{2} \) . Find and simplify the formula of the function \( g\left( x\right) \) whose graph is the result of \( f \) undergoing the following sequence of transformations.\n\n1. Vertical shift up 2 units\n\n2. Reflection across the \( x \) -axis\n\n3. Horizontal shift right 1 u... | Solution. We build up to a formula for \( g\left( x\right) \) using intermediate functions as we’ve seen in previous examples. We let \( {g}_{1} \) take care of our first step. Theorem 1.2 tells us \( {g}_{1}\left( x\right) = f\left( x\right) + 2 = {x}^{2} + 2 \) . Next, we reflect the graph of \( {g}_{1} \) about the ... | Yes |
Find the slope of the line containing the following pairs of points, if it exists. Plot each pair of points and the line containing them. | Solution. In each of these examples, we apply the slope formula, Equation 2.1.\n\n1. \( m = \frac{4 - 0}{2 - 0} = \frac{4}{2} = 2 \) \n\n2. \( m = \frac{4 - 2}{3 - \left( {-1}\right) } = \frac{2}{4} = \frac{1}{2} \)\... | Yes |
1. Find the slope of the line containing the points \( \left( {6,{24}}\right) \) and \( \left( {{10},{32}}\right) \) . | 1. For the slope, we have \( m = \frac{{32} - {24}}{{10} - 6} = \frac{8}{4} = 2 \) . | Yes |
Example 2.1.3. Write the equation of the line containing the points \( \left( {-1,3}\right) \) and \( \left( {2,1}\right) \) . | Solution. In order to use Equation 2.2 we need to find the slope of the line in question so we use Equation 2.1 to get \( m = \frac{\Delta y}{\Delta x} = \frac{1 - 3}{2 - \left( {-1}\right) } = - \frac{2}{3} \) . We are spoiled for choice for a point \( \left( {{x}_{0},{y}_{0}}\right) \) . We’ll use \( \left( {-1,3}\ri... | Yes |
Graph the following functions. Identify the slope and \( y \) -intercept. | 1. To graph \( f\left( x\right) = 3 \), we graph \( y = 3 \). This is a horizontal line \( \left( {m = 0}\right) \) through \( \left( {0,3}\right) \). | No |
1. Find and interpret \( C\left( {10}\right) \) . | To find \( C\left( {10}\right) \), we replace every occurrence of \( x \) with 10 in the formula for \( C\left( x\right) \) to get \( C\left( {10}\right) = {80}\left( {10}\right) + {150} = {950} \) . Since \( x \) represents the number of PortaBoys produced, and \( C\left( x\right) \) represents the cost, in dollars, \... | Yes |
1. Find a linear function which fits this data. Use the weekly sales \( x \) as the independent variable and the price \( p \) as the dependent variable. | 1. We recall from Section 1.4 the meaning of ’independent’ and ’dependent’ variable. Since \( x \) is to be the independent variable, and \( p \) the dependent variable, we treat \( x \) as the input variable and \( p \) as the output variable. Hence, we are looking for a function of the form \( p\left( x\right) = {mx}... | Yes |
1. Find and simplify an expression for the weekly revenue \( R\left( x\right) \) as a function of weekly sales \( x \) . | 1. Since \( R = {xp} \), we substitute \( p\left( x\right) = - {1.5x} + {250} \) from Example 2.1.6 to get \( R\left( x\right) = x( - {1.5x} + \) \( {250}) = - {1.5}{x}^{2} + {250x} \) . Since we determined the price-demand function \( p\left( x\right) \) is restricted to \( 0 \leq x \leq {166}, R\left( x\right) \) is ... | Yes |
Theorem 2.1. Properties of Absolute Value: Let \( a, b \) and \( x \) be real numbers and let \( n \) be an integer. \( {}^{a} \) Then\n\n- Product Rule: \( \left| {ab}\right| = \left| a\right| \left| b\right| \) | The proofs of the Product and Quotient Rules in Theorem 2.1 boil down to checking four cases: when both \( a \) and \( b \) are positive; when they are both negative; when one is positive and the other is negative; and when one or both are zero.\n\nFor example, suppose we wish to show that \( \left| {ab}\right| = \left... | Yes |
1. \( \left| {{3x} - 1}\right| = 6 \) | The equation \( \left| {{3x} - 1}\right| = 6 \) is of the form \( \left| x\right| = c \) for \( c > 0 \), so by the Equality Properties, \( \left| {{3x} - 1}\right| = 6 \) is equivalent to \( {3x} - 1 = 6 \) or \( {3x} - 1 = - 6 \) . Solving the former, we arrive at \( x = \frac{7}{3} \) , and solving the latter, we ge... | Yes |
Graph each of the following functions.\n\n1. \( f\left( x\right) = \left| x\right| \) 2. \( g\left( x\right) = \left| {x - 3}\right| \) 3. \( h\left( x\right) = \left| x\right| - 3 \) 4. \( i\left( x\right) = 4 - 2\left| {{3x} + 1}\right| \)\n\nFind the zeros of each function and the \( x \) - and \( y \) -intercepts o... | ## Solution.\n\n1. To find the zeros of \( f \), we set \( f\left( x\right) = 0 \) . We get \( \left| x\right| = 0 \), which, by Theorem 2.1 gives us \( x = 0 \) . Since the zeros of \( f \) are the \( x \) -coordinates of the \( x \) -intercepts of the graph of \( y = f\left( x\right) \), we get \( \left( {0,0}\right)... | Yes |
Graph the following functions starting with the graph of \( f\left( x\right) = \left| x\right| \) and using transformations. | Solution. We begin by graphing \( f\left( x\right) = \left| x\right| \) and labeling three points, \( \left( {-1,1}\right) ,\left( {0,0}\right) \) and \( \left( {1,1}\right) \) .\n\n\n\n\( f\left( x\right) = \left| x... | Yes |
1. \( f\left( x\right) = \frac{\left| x\right| }{x} \) | We first note that, due to the fraction in the formula of \( f\left( x\right), x \neq 0 \) . Thus the domain is \( \left( {-\infty ,0}\right) \cup \left( {0,\infty }\right) \) . To find the zeros of \( f \), we set \( f\left( x\right) = \frac{\left| x\right| }{x} = 0 \) . This last equation implies \( \left| x\right| =... | Yes |
Graph the following functions starting with the graph of \( f\left( x\right) = {x}^{2} \) and using transformations. Find the vertex, state the range and find the \( x \) - and \( y \) -intercepts, if any exist.\n\n1. \( g\left( x\right) = {\left( x + 2\right) }^{2} - 3 \) | Since \( g\left( x\right) = {\left( x + 2\right) }^{2} - 3 = f\left( {x + 2}\right) - 3 \), Theorem 1.7 instructs us to first subtract 2 from each of the \( x \) -values of the points on \( y = f\left( x\right) \) . This shifts the graph of \( y = f\left( x\right) \) to the left 2 units and moves \( \left( {-2,4}\right... | Yes |
Theorem 2.2. Vertex Formula for Quadratics in Standard Form: For the quadratic function \( f\left( x\right) = a{\left( x - h\right) }^{2} + k \), where \( a, h \) and \( k \) are real numbers with \( a \neq 0 \), the vertex of the graph of \( y = f\left( x\right) \) is \( \left( {h, k}\right) \) . | To see why the formula in Theorem 2.2 produces the vertex, consider the graph of the equation \( y = a{\left( x - h\right) }^{2} + k \) . When we substitute \( x = h \), we get \( y = k \), so \( \left( {h, k}\right) \) is on the graph. If \( x \neq h \), then \( x - h \neq 0 \) so \( {\left( x - h\right) }^{2} \) is a... | Yes |
Convert the functions below from general form to standard form. Find the vertex, axis of symmetry and any \( x \) - or \( y \) -intercepts. Graph each function and determine its range.\n\n1. \( f\left( x\right) = {x}^{2} - {4x} + 3 \) | ## Solution.\n\n1. To convert from general form to standard form, we complete the square. \( {}^{7} \) First, we verify that the coefficient of \( {x}^{2} \) is 1 . Next, we find the coefficient of \( x \), in this case -4, and take half of it to get \( \frac{1}{2}\left( {-4}\right) = - 2 \) . This tells us that our ta... | Yes |
1. Determine the weekly profit function \( P\left( x\right) \) . | To find the profit function \( P\left( x\right) \), we subtract\n\n\[ P\left( x\right) = R\left( x\right) - C\left( x\right) = \left( {-{1.5}{x}^{2} + {250x}}\right) - \left( {{80x} + {150}}\right) = - {1.5}{x}^{2} + {170x} - {150}. \]\n\nSince the revenue function is valid when \( 0 \leq x \leq {166}, P \) is also res... | Yes |
Much to Donnie's surprise and delight, he inherits a large parcel of land in Ashtabula County from one of his (e)strange(d) relatives. The time is finally right for him to pursue his dream of farming alpaca. He wishes to build a rectangular pasture, and estimates that he has enough money for 200 linear feet of fencing ... | Solution. It is always helpful to sketch the problem situation, so we do so below.\n\n\n\nWe are tasked to find the dimensions of the pasture which would give a maximum area. We let \( w \) denote the width of the pa... | Yes |
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