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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?
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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 one gram of water one degree centigrade - specifically from 14.5 to 15.5 degrees centigrade, but we will assume it is constant over the range 0 to 37 degrees centigrade.\n\nThe following table is helpful:\n\n<table><thead><tr><th></th><th>Vol (ml)</th><th>Temp \( \left( {{}^{ \circ }\mathrm{C}}\right) \)</th><th>Calories</th></tr></thead><tr><td>Fluid 1</td><td>10</td><td>9</td><td>\( {10} \times 9 = {90} \)</td></tr><tr><td>Fluid 2</td><td>60</td><td>37</td><td>\( {60} \times {37} = {2220} \)</td></tr><tr><td>Mixture</td><td>70</td><td>\( T \)</td><td>\( {70} \times T \)</td></tr></table>\n\nThe critical step now is the conservation of energy. The calories in the mixture should be the sum of the calories in the two fluids (we assume there is no heat of mixing) so that\n\n\[ \n{70} \times T = {90} + {2220}\;T = {33}^{ \circ }\mathrm{C}\;\blacksquare \n\]
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Yes
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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\nThen for every \( x \) in \( \left\lbrack {a, b}\right\rbrack \)\n\n\[ {G}^{\prime }\left( x\right) = f\left( x\right) \]
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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 \( \left\lbrack {a, b}\right\rbrack, G\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \) . We will prove that the right hand derivative\n\n\[ {G}^{\prime + }\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow 0 + }}\frac{G\left( {x + h}\right) - G\left( x\right) }{h} = f\left( x\right) \;\text{ for }\;a \leq x < b. \]\n\nA simple modification of the argument shows that \( {G}^{\prime - }\left( x\right) = f\left( x\right) \) for \( a < x \leq b \), so that\n\n\( {G}^{\prime }\left( x\right) = f\left( x\right) \) for \( a \leq x \leq b \) .\n\nProof that \( \;{G}^{\prime + }\left( x\right) = f\left( x\right) \) . Let \( x \) and \( x + h \) be numbers in \( \left\lbrack {a, b}\right\rbrack \) with \( h > 0 \) . Refer to Figure 10.3.\n\n \n\nFigure 10.3: A. \( G\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \) . B. \( G\left( {x + h}\right) - G\left( x\right) = {\int }_{x}^{x + h}f\left( t\right) {dt} \) .
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Yes
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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 \( \left\lbrack {a, b}\right\rbrack \) \n\n\[ \nD\left( x\right) = C\text{.} \n\]
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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 \( {D}^{\prime }\left( z\right) = 0, D\left( x\right) - D\left( a\right) = 0 \) and \( D\left( x\right) = D\left( a\right) = C \) . Because \( D \) is continuous and \( D\left( x\right) = C \) for every \( x \) in \( \lbrack a, b), D\left( b\right) = C \) . End of proof.
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Yes
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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 for every \( x \) in \( \left\lbrack {a, b}\right\rbrack \)\n\n\[ \nF\left( x\right) = G\left( x\right) + C \n\]
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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\[ \n{D}^{\prime }\left( x\right) = {\left\lbrack F\left( x\right) - G\left( x\right) \right\rbrack }^{\prime } = {F}^{\prime }\left( x\right) - {G}^{\prime }\left( x\right) = 0 \n\]\n\nfor every \( x \) in \( \left\lbrack {a, b}\right\rbrack \) . By the Horizontal Graph Theorem, Theorem 10.3.1, there is a number \( C \) such that for every \( x \) in \( \left\lbrack {a, b}\right\rbrack, D\left( x\right) = C \) . Then for every \( x \) in \( \left\lbrack {a, b}\right\rbrack \)\n\n\[ \nD\left( x\right) = F\left( x\right) - G\left( x\right) = C\;\text{ and }\;F\left( x\right) = G\left( x\right) + C. \n\]\n\nEnd of proof.
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Yes
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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 \( s/A \) to the \( {10} - s \) cc mark. Because at constant temperature, \( {PV} = \) a constant \( = {P}_{0}{V}_{0}, = {P}_{0} \), the pressure, \( {P}_{s} \), inside the syringe is \( {P}_{0}{10}/\left( {{10} - s}\right) \) . The force on the plunger is \( \left( {{P}_{s} - {P}_{0}}\right) A = A{P}_{0}(s/\left( {{10} - s}\right) \) . The work done in compressing the air from \( {10}\mathrm{{cc}} \) to \( 5\mathrm{{cc}} \) is
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\[ {\int }_{0}^{5}A{P}_{0}\frac{s}{{10} - s}d\frac{s}{A} = {\int }_{0}^{5}{P}_{0}\frac{s}{{10} - s}{ds} \]\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\[ {W}^{\prime }\left( x\right) = {P}_{0}\frac{x}{{10} - x} \]\n\nLet \( \omega \left( x\right) \) be defined by (a bolt out of the blue!)\n\[ \omega \left( x\right) = {P}_{0}\left\lbrack {-{10}\ln \left( {{10} - x}\right) - x}\right\rbrack \]\n\nExplore 10.3.1 Use derivative formulas including \( f\left( t\right) = \ln U\left( t\right) \Rightarrow {f}^{\prime }\left( t\right) = \frac{1}{U\left( t\right) }{U}^{\prime }\left( t\right) \) to show that\n\[ {\omega }^{\prime }\left( x\right) = {P}_{0}\frac{x}{{10} - x} \]\n\nThus \( {W}^{\prime }\left( x\right) = {\omega }^{\prime }\left( x\right) \) for every \( x \) in \( \left\lbrack {0,5}\right\rbrack \), and by the Parallel Graph Theorem there is a number \( C \) such that such that for every number \( x \) in \( \left\lbrack {0,5}\right\rbrack \)\n\[ W\left( x\right) = \omega \left( x\right) + C \]\n\nNow\n\[ W\left( 0\right) = {P}_{0}{\int }_{0}^{0}\frac{s}{{10} - s}{ds}\; = 0,\;\text{ and } \]\n\[ \omega \left( 0\right) = {P}_{0}\left\lbrack {-{10}\ln \left( {{10} - 0}\right) - 0}\right\rbrack = - {P}_{0}{10}\ln {10}. \]\n\nBecause \( W\left( 0\right) = \omega \left( 0\right) + C \)\n\[ 0 = - {P}_{0}{10}\ln {10} + C\;\text{ and }\;C = {P}_{0}{10}\ln {10}. \]\n\nWe can conclude that for all \( x \) in \( \left\lbrack {0,5}\right\rbrack \)\n\[ W\left( x\right) = {P}_{0}\left\lbrack {-{10}\ln \left( {{10} - x}\right) - x}\right\rbrack + {P}_{0}{10}\ln \left( {10}\right) \]\n\nThe total work done in compressing the air as\n\[ W\left( 5\right) = {P}_{0}\left\lbrack {-{10}\ln \left( {{10} - 5}\right) - 5 + {10}\ln \left( {10}\right) }\right\rbrack = {P}_{0}{1.93} \]
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Yes
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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\[ \text{for every number}t\text{in}\left\lbrack {a, b}\right\rbrack \;{F}^{\prime }\left( t\right) = f\left( t\right) \]\n\n(10.5)\n\nThen\n\n\[ {\int }_{a}^{b}f\left( t\right) {dt} = F\left( b\right) - F\left( a\right) \]\n\n(10.6)
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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 \( x \) in \( \left\lbrack {a, b}\right\rbrack ,{F}^{\prime }\left( x\right) = f\left( x\right) \).\n\n(ii)\n\nFor \( x \) in \( \left\lbrack {a, b}\right\rbrack ,{G}^{\prime }\left( x\right) = {F}^{\prime }\left( x\right) \).\n\nThere is a number \( C \) so that for \( x \) in \( \left\lbrack {a, b}\right\rbrack ,\;G\left( x\right) = F\left( x\right) + C \)\n\n(iii)\n\n\[ G\left( a\right) = 0. \]\n\n(iv)\n\n\[ G\left( a\right) = F\left( a\right) + C\text{. Therefore,}C = - F\left( a\right) \text{.} \]\n\n\[ G\left( x\right) = F\left( x\right) - F\left( a\right) . \]\n\n\[ G\left( b\right) = F\left( b\right) - F\left( a\right) . \]\n\n\[ {\int }_{a}^{b}f\left( t\right) {dt} = F\left( b\right) - F\left( a\right) . \]\n\n(v)\n\nEnd of proof.
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Yes
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Evaluate \( {\int }_{0}^{1}{t}^{2}{dt} \) .
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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\[ {\int }_{0}^{1}{t}^{2}{dt} = F\left( 1\right) - F\left( 0\right) = {\left\lbrack \frac{{t}^{3}}{3}\right\rbrack }_{0}^{1} = \frac{{1}^{3}}{3} - \frac{{0}^{3}}{3} = \frac{1}{3}. \]
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Yes
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\[ \int {e}^{\left( {t}^{2}\right) }{tdt} \]
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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\[ = \frac{1}{2}\int {e}^{\left( {t}^{2}\right) }\left( {2t}\right) {dt} \]\n\nEquation 10.15\n\n\[ = \frac{1}{2}\int {e}^{u\left( t\right) }{u}^{\prime }\left( t\right) {dt} \]\n\n\[ = \frac{1}{2}{e}^{u\left( t\right) } + C \]\n\nEquation 10.20\n\n\[ = \frac{1}{2}{e}^{\left( {t}^{2}\right) } + C \]
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Yes
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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\\]
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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. The first integral can be solved without expansion, however, and is easier to compute. Identify\n\n\\[ \nu\\left( t\\right) = 1 + {t}^{4},\\;{G}^{\\prime }\\left( u\\right) = {u}^{10},\\;{u}^{\\prime \\left( t\\right) } = 4{t}^{3}\\;\\text{ and }\\;G\\left( u\\right) = \\frac{{u}^{11}}{11} \\]\n\nThen\n\n\\[ \n\\int {\\left( 1 + {t}^{4}\\right) }^{10}{t}^{3}{dt} = \\int \\frac{1}{4}{\\left( 1 + {t}^{4}\\right) }^{10}4{t}^{3}{dt}\\;\\text{Arithn} \n\\]\n\n\\[ \n= \\frac{1}{4}\\int {\\left( 1 + {t}^{4}\\right) }^{10}\\left( {4{t}^{3}}\\right) {dt} \n\\]\n\n\\[ \n= \\frac{1}{4}\\int {\\left( u\\left( t\\right) \\right) }^{10}{u}^{\\prime }\\left( t\\right) {dt} \n\\]\n\n\\[ \n= \\frac{1}{4}\\frac{{\\left( u\\left( t\\right) \\right) }^{11}}{11} + C \n\\]\n\n\\[ \n= \\frac{1}{4}\\frac{{\\left( 1 + {t}^{4}\\right) }^{11}}{11} + C\\;u\\left( t\\right) = 1 + {t}^{4} \n\\]
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Yes
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Compute \( \int \sin \left( {\pi t}\right) {dt} \) .
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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\[ \int \sin \left( {\pi t}\right) {dt} = \int \frac{1}{\pi }\sin \left( {\pi t}\right) {\pi dt} = \frac{1}{\pi }\int \sin \left( {\pi t}\right) {\pi dt} = \]\n\n\[ \frac{1}{\pi }\int \sin \left( {u\left( t\right) }\right) {u}^{\prime }\left( t\right) {dt} = \frac{1}{\pi }\left( {-\cos \left( {u\left( t\right) }\right) }\right) + C = - \frac{1}{\pi }\left( {\cos \left( {\pi t}\right) }\right) + C \]
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Yes
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Compute \( \int \sqrt{{5t} - 4}{dt} \) .
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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\[ \int \sqrt{{5t} - 4}{dt} = \frac{1}{5}\int \sqrt{{5t} - 4}{5dt} = \frac{1}{5}\int {\left( u\left( t\right) \right) }^{1/2}{u}^{\prime }\left( t\right) {dt} = \frac{1}{5}\frac{{\left( u\left( t\right) \right) }^{3/2}}{3/2} + C \]\n\n\[ = - \frac{2}{15}{\left( 5t - 4\right) }^{3/2} + C\; \]
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Yes
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Compute \( \int {\left( \ln t\right) }^{3}\frac{1}{t}{dt} \) .
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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\[ \int {\left( \ln t\right) }^{3}\frac{1}{t}{dt} = \int {\left( u\left( t\right) \right) }^{3}{u}^{\prime }\left( t\right) {dt} = \frac{{\left( u\left( t\right) \right) }^{4}}{4} + C = \frac{{\left( \ln t\right) }^{4}}{4} + C\;\blacksquare \]
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Yes
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Example 10.5.6 Compute \( \int \tan {xdx} \) .
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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\[ \int \tan x = \int \frac{\sin x}{\cos x}{dx} = - \int \frac{1}{\cos x}\left( {-\sin x}\right) {dx} = - \int \frac{1}{u\left( x\right) }{u}^{\prime }\left( x\right) {dx} \]\n\n\[ = - \ln u\left( x\right) + C = - \ln \cos x + C = \ln \sec x + C \]\n\nAlways, once an antiderivative has been computed, it can be checked by differentiation.\n\nWe check the last claim that \( \int \tan x = \ln \sec x + C \) by differentiation. We should show that \( {\left\lbrack \ln \sec x\right\rbrack }^{\prime } = \tan x. \)\n\n\[ {\left\lbrack \ln \sec x\right\rbrack }^{\prime } = \frac{1}{\sec x}{\left\lbrack \sec x\right\rbrack }^{\prime } = \frac{1}{\sec x}\sec x\tan x = \tan x \]
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Yes
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How to find the volume of a potato.
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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 \( 8\mathrm{\;{cm}} \) and \( {10}\mathrm{\;{cm}} \) is approximately \( 2\mathrm{\;{cm}} \times {17} \) \( {\mathrm{{cm}}}^{2} = {34}{\mathrm{\;{cm}}}^{3} \) . There are 9 slices, and the volume of the potato is approximately\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{9}\left( {\text{ Area of slice }k\text{ in }{\mathrm{{cm}}}^{2}}\right) \times 2\mathrm{\;{cm}}. \]\n\nThe approximation works remarkably well. Several students have computed approximate volumes and subsequently tested the volumes with liquid displacement in a beaker and found close agreement.\n\n\n\nFigure 11.1: A. A potato with \( 2\mathrm{\;{cm}} \) marks along an axis. B. A slice of the potato at position \( x = {10} \) that has area approximately \( {17}{\mathrm{\;{cm}}}^{2} \) .\n\nThe sum is also an approximation to the integral\n\n\[ {\int }_{0}^{17}A\left( x\right) {dx} \]\n\nwhere \( A\left( x\right) \) is the area of the cross section of the potato at station \( x \), and the integral may be considered to be the actual volume.
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Yes
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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 radius \( r \) and area \( \pi {r}^{2} \) . Therefore\n\n\[ \n{r}^{2} + {x}^{2} = {R}^{2},\;{r}^{2} = {R}^{2} - {x}^{2}\;\text{ and }\;A = A\left( x\right) = \pi \left( {{R}^{2} - {x}^{2}}\right) .\n\]\n\nConsequently the volume of the hemisphere is\n\n\[ \nV = {\int }_{0}^{R}\pi \left( {{R}^{2} - {x}^{2}}\right) {dx} \n\]
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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\[ \nV = {\int }_{0}^{R}\pi \left( {{R}^{2} - {x}^{2}}\right) {dx} = {\left\lbrack \pi {R}^{2}x - \pi \frac{{x}^{3}}{3}\right\rbrack }_{0}^{R} = \left( {\pi {R}^{2} \times R - \pi \frac{{R}^{3}}{3}}\right) - \left( {\pi {R}^{2} \times 0 - \pi \frac{{0}^{3}}{3}}\right) = \frac{2}{3}\pi {R}^{3}.\n\]\n\nThe volume of the sphere is twice the volume of the hemisphere.\n\n\[ \n\text{The volume of a sphere of radius}R\text{is}\;\frac{4}{3}\pi {R}^{3} \n\]
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Yes
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Compute the volume of the triangular solid with three mutually perpendicular edges of length 2, 3, and 4, illustrated in Figure 11.2B.
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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\nTherefore\n\n\[ A\left( z\right) = \frac{1}{2}\frac{2}{4}\left( {4 - z}\right) \frac{3}{4}\left( {4 - z}\right) = \frac{3}{16}{\left( 4 - z\right) }^{2},\n\nand\n\n\[ V = {\int }_{0}^{4}A\left( z\right) {dz} = {\int }_{0}^{4}\frac{3}{16}{\left( 4 - z\right) }^{2}{dz}\; = \;\frac{3}{16}{\int }_{0}^{4}{\left( 4 - z\right) }^{2}{dz}.\n\nCheck that \( {\left\lbrack -\frac{{\left( 4 - z\right) }^{3}}{3}\right\rbrack }^{\prime } = {\left( 4 - z\right) }^{2} \) . Then\n\n\[ \frac{3}{16}{\int }_{0}^{4}{\left( 4 - z\right) }^{2}{dz} = \frac{3}{16}{\left\lbrack -\frac{{\left( 4 - z\right) }^{3}}{3}\right\rbrack }_{0}^{4} = \frac{3}{16}\left( {-0 - \left( {-\frac{{\left( 4\right) }^{3}}{3}}\right) }\right) = 4.\]
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Yes
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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 \( {11.3}\mathrm{\;B} \) .
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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 \( x \) -axis.\n\n\[ \text{Volume of}\;{S}_{1}\;{\int }_{0}^{1}\pi {\left( \sqrt{x}\right) }^{2}{dx} = \pi {\int }_{0}^{1}{xdx} = \pi {\left\lbrack \frac{{x}^{2}}{2}\right\rbrack }_{0}^{1} = \frac{\pi }{2} \]\n\n\[ \text{Volume of}\;{S}_{2}\;{\int }_{0}^{1}\pi {\left( {x}^{2}\right) }^{2}{dx} = \pi {\int }_{0}^{1}{x}^{4}{dx} = \pi {\left\lbrack \frac{{x}^{5}}{5}\right\rbrack }_{0}^{1} = \frac{\pi }{5} \]\n\n\[ \text{Volume of}\;S\;\frac{\pi }{2} - \frac{\pi }{5} = \frac{3\pi }{10} \]\nAlternatively, we can compute the area of the ’washer’ at \( x \) which is\n\n\[ A\left( x\right) = \pi {\left( \sqrt{x}\right) }^{2} - \pi {\left( {x}^{2}\right) }^{2}\; = \;\pi \left( {x - {x}^{4}}\right) . \]\n\nThen\n\n\[ \text{Volume of}S = {\int }_{0}^{1}\pi {\left( x - {x}^{4}\right) }^{2}{dx} = \pi {\left\lbrack \frac{{x}^{2}}{2} - \frac{{x}^{5}}{5}\right\rbrack }_{0}^{1} = \pi \left( {\frac{1}{2} - \frac{1}{5}}\right) = \frac{3\pi }{10}\text{-} \]
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Yes
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For the integration problem, \[ \int \sqrt{1 + \sqrt{x}}{dx} \] substitute \[ x = g\left( z\right) = {z}^{2}, \] and \[ {dx} = {g}^{\prime }\left( z\right) {dz} = {2zdz}. \]
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Then compute \[ \int \sqrt{1 + \sqrt{{z}^{2}}}{2z};{dz} = \int \sqrt{1 + z}{2zdz} \] \[ = \int \sqrt{1 + z}\left( {{2z} + 2 - 2}\right) {dz} \] \[ = 2\int {\left( 1 + z\right) }^{3/2} - {\left( 1 + z\right) }^{1/2}{dz} \] \[ = 2\left( {\frac{{\left( 1 + z\right) }^{5/2}}{5/2} - \frac{{\left( 1 + z\right) }^{3/2}}{3/2}}\right) + C. \] One can remember that \( x = {z}^{2} \), and assume \( z = \sqrt{x} \) and write \[ \int \sqrt{1 + \sqrt{x}}{dx} = 2\left( {\frac{{\left( 1 + \sqrt{x}\right) }^{5/2}}{5/2} - \frac{{\left( 1 + \sqrt{x}\right) }^{3/2}}{3/2}}\right) + C. \]
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Yes
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Compute \( \int \sqrt{1 + {x}^{2}}{dx} \) .
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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\[ = \int \frac{\left( {\sec z}\right) \left( {\sec z + \tan z}\right) }{\sec z + \tan z} + \left( {\int \tan z\sec z\tan {zdz}}\right) \]\n\n\[ u = \tan z,\;{v}^{\prime } = \sec z\tan z \]\n\n\[ = \int \frac{{\sec }^{2}z + \sec z\tan z}{\tan z + \sec z}{dz} + \left( {\sec z\tan z-\int \sec z{\sec }^{2}{zdz}}\right) \]\n\n\[ = \ln \left( {\tan z + \sec z}\right) + \sec z\tan z - \int {\sec }^{3}{zdz} \]\n\n\[ 2\int {\sec }^{3}{zdz} = \ln \left( {\tan z + \sec z}\right) + \sec z\tan z + C \]\n\n\[ \int \sqrt{1 + {x}^{2}}{dx} = \frac{1}{2}\left( {\ln \left( {x + \sqrt{1 + {x}^{2}}}\right) + x\sqrt{1 + {x}^{2}}}\right) + C \]
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Yes
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We compute the volume, \( V \), of \( S \) .
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\[ V = {\int }_{0}^{a}\pi {\left( b\sqrt[4]{x}\sqrt[4]{a - x}\right) }^{2}{dx} = \pi {b}^{2}{\int }_{0}^{a}\sqrt{x}\sqrt{a - x}{dx} \]\n\nLet \( x = {az} \) ; then \( {dx} = {adz} \) and \( x = 0 \) and \( x = a \) correspond to \( z = 0 \) and \( z = 1 \) . Then\n\n\[ V = \pi {b}^{2}{\int }_{0}^{1}\sqrt{az}\sqrt{a - {az}}{adz} = \pi {b}^{2}{a}^{2}{\int }_{0}^{1}\sqrt{z}\sqrt{1 - z}{dz} \]\n\nThe skies darken and lightning abounds.\n\n\[ \int \sqrt{\mathrm{z}}\sqrt{1 - \mathrm{z}}\mathrm{{dz}} = \frac{1}{4}\arcsin \sqrt{\mathrm{z}} - \frac{1}{4}\sqrt{\mathrm{z}}\sqrt{1 - \mathrm{z}}\left( {1 - 2\mathrm{z}}\right) + \mathrm{C} \]\n\n(11.7)\n\nThen\n\n\[ V = \pi {b}^{2}\frac{{a}^{2}}{4}{\left\lbrack \arcsin \sqrt{z} - \sqrt{z}\sqrt{1 - z}\left( 1 - 2z\right) \right\rbrack }_{0}^{1} = {a}^{2}{b}^{2}\frac{{\pi }^{2}}{8} \]
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Yes
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Compute the centroid of the right circular cone of height \( H \) and base radius \( R \) .
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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\[ c = \frac{{\int }_{0}^{H}{x\pi } \times {\left( \left( R/H\right) x\right) }^{2}{dx}}{{\int }_{0}^{H}\pi \times {\left( \left( R/H\right) x\right) }^{2}{dx}} = \frac{{\int }_{0}^{H}{x}^{3}{dx}}{{\int }_{0}^{H}{x}^{2}{dx}} = \frac{{H}^{4}/4}{{H}^{3}/3} = \frac{3}{4}H \]\n\nThe centroid, \( c \), is \( 3/4 \) the distance from the vertex of the cone and \( 1/4 \) the distance from the base of the cone.
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Yes
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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.
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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 of the centroid, the cross sectional area of the region in a plane perpendicular to the \( X \) -axis at a position \( x \) is \( {x}^{2} \times 1 = {x}^{2} \) for \( 0 \leq x \leq 4 \) . We write\n\n\[ \bar{x} = \frac{{\int }_{0}^{4}x \times {x}^{2}{dx}}{{\int }_{0}^{4}{x}^{2}{dx}} = \frac{{\int }_{0}^{4}{x}^{3}{dx}}{{\int }_{0}^{4}{x}^{2}{dx}} = \frac{{4}^{4}/4}{{4}^{3}/3} = 3 \]\n\nFor the \( y \) -coordinate of the centroid, the cross sectional area of the region in a plane perpendicular to the \( Y \) -axis at a position \( y \) is \( \left( {1 - \sqrt{y}}\right) \times 1 = 1 - \sqrt{y} \) for \( 0 \leq y \leq {16} \) . We write\n\n\[ \bar{y} = \frac{{\int }_{0}^{16}y \times \left( {1 - \sqrt{y}}\right) {dy}}{{\int }_{0}^{4}1 - \sqrt{y}{dy}} = \frac{{\int }_{0}^{16}y - {y}^{3/2}{dx}}{{\int }_{0}^{16}1 - {y}^{1/2}{dx}} = \frac{{\left\lbrack {y}^{2}/2 - {y}^{5/2}/\left( 5/2\right) \right\rbrack }_{0}^{16}}{{\left\lbrack y - {y}^{3/2}/\left( 3/2\right) \right\rbrack }_{0}^{16}} = \frac{24}{5}\;\blacksquare \]
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Yes
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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.
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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\[ \text{Center of mass} = \frac{{\int }_{a}^{b}{x\delta }\left( x\right) A\left( x\right) {dx}}{{\int }_{a}^{b}\delta \left( x\right) A\left( x\right) {dx}} = \frac{{\int }_{0}^{1}x \times x \times \pi \left( {1 - {x}^{2}}\right) {dx}}{{\int }_{0}^{1}x \times \pi \left( {1 - {x}^{2}}\right) {dx}} \]\n\n\[ = \frac{{\int }_{0}^{1}{x}^{2} - {x}^{4}{dx}}{{\int }_{0}^{1}x - {x}^{3}{dx}} = \frac{{\left\lbrack {x}^{3}/3 - {x}^{5}/5\right\rbrack }_{0}^{1}}{{\left\lbrack {x}^{2}/2 - {x}^{4}/4\right\rbrack }_{0}^{1}} = \frac{8}{15}\;\blacksquare \]
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Yes
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Find the length of the graph of \( y = \ln \cos x \) on \( 0 \leq x \leq \pi /4 \) .
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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\[ {\left\lbrack \ln \left( \sec x + \tan x\right) \right\rbrack }^{\prime } = \sec x \]\n\nWe can write\n\n\[ \text{Length}\; = {\int }_{0}^{\pi /4}\sqrt{1 + {\left( {f}^{\prime }\left( x\right) \right) }^{2}}{dx}\; = {\int }_{0}^{\pi /4}\sqrt{1 + {\tan }^{2}x}{dx}\; = {\int }_{0}^{\pi /4}\sec {dx} \]\n\n\[ = {\left\lbrack \ln \left( \sec x + \tan x\right) \right\rbrack }_{0}^{\pi /4} \]\n\n\[ = \ln \left( {\sec \left( {\pi /4}\right) + \tan \left( {\pi /4}\right) }\right) - \ln \left( {\sec \left( 0\right) + \tan \left( 0\right) }\right) \doteq {0.881}\;\blacksquare \]
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No
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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?
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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\nwhere \( R = {6370} \) meters is the radius of Earth. This means a satellite of mass \( m \) and altitude \( x \) experiences a gravitational force of\n\n\[ F\left( x\right) = m \times {9.8} \times \frac{{R}^{2}}{{\left( R + x\right) }^{2}}\;\text{ Newtons } \]\n\nThe work required to lift the satellite to an altitude \( A \) is\n\n\[ W\left( A\right) = {\int }_{0}^{A}m \times {9.8} \times \frac{{R}^{2}}{{\left( R + x\right) }^{2}}{dx} = {9.8} \times m \times {R}^{2}{\int }_{0}^{A}\frac{1}{{\left( R + x\right) }^{2}}{dx} \]\n\nThe integral is readily evaluated.\n\nHow much work must be done to send the satellite 'out of Earth's gravity field', so that it does not fall back to Earth? This is\n\n\[ \mathop{\lim }\limits_{{A \rightarrow \infty }}W\left( A\right) = \mathop{\lim }\limits_{{A \rightarrow \infty }}{9.8} \times m \times {R}^{2}\left\lbrack {\frac{1}{R} - \frac{1}{\left( R + A\right) }}\right\rbrack = {9.8} \times m \times R \]\n\nA finite amount of work is sufficient to send the satellite out of the Earth's gravity field.\n\nNow suppose we neglect the friction of air and ask at what velocity should the satellite be launched in order to escape the Earth? We borrow from physics the formula for kinetic energy of a body of mass \( m \) moving at a velocity \( v \) .\n\n\[ \text{Kinetic energy} = \frac{1}{2}m{v}^{2}\text{.} \]\n\nThen we equate the kinetic energy with the work required\n\n\[ \frac{1}{2}m{v}^{2} = {9.8} \times m \times R \]\n\nand solve for \( v \) . The solution is\n\n\[ v = \sqrt{{2R} \times {9.8}} = {11},{174}\text{ meters per second } \]
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Yes
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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 \( \left\lbrack {0,1}\right\rbrack \) and\n\n\[ {f}^{\prime }\left( x\right) = {\left\lbrack \sqrt{x}\right\rbrack }^{\prime } = {\left\lbrack {x}^{\frac{1}{2}}\right\rbrack }^{\prime } = \frac{1}{2}{x}^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}\;0 < x \leq 1 \]\n\nThe parabola has a vertical tangent at \( \left( {0,0}\right) \) and \( {f}^{\prime }\left( 0\right) \) is not defined. Even so, the hypothesis of the Mean Value Theorem is satisfied; the requirement is that \( {f}^{\prime }\left( x\right) \) exist for \( 0 < x < 1 \) - only required for points between 0 and 1 .\n\nWe seek a tangent parallel to the secant through \( \left( {0,0}\right) \) and \( \left( {1,1}\right) \) ; the secant has slope \( \frac{f\left( 1\right) - f\left( 0\right) }{1 - 0} = 1 \), so we look for a tangent of slope 1 .\n\nThe Mean Value Theorem asserts that there is a number \( c \) such that \( {f}^{\prime }\left( c\right) = 1 \) .
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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 \( f \) at \( \left( {1/4,1/2}\right) \) has slope 1 and is parallel to the secant from \( \left( {0,0}\right) \) to \( \left( {1,1}\right) \) (Example Figure 12.1.0.2B.) -
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Yes
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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 \( c, a < c < b \).\n\nthen there is a number \( c \) between \( a \) and \( b \) for which \( {f}^{\prime }\left( c\right) = 0 \).
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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, then \( \left( {u, f\left( u\right) }\right) \) is an interior local maximum and the tangent at \( \left( {u, f\left( u\right) }\right) \) is horizontal. Choose \( c = u \) . Then \( {f}^{\prime }\left( c\right) = 0 \).\n\nSimilarly, if \( a < v < b,\left( {v, f\left( v\right) }\right) \) is an interior local minimum, the tangent at \( \left( {v, f\left( v\right) }\right) \) is horizontal. Choose \( c = v \) . Then \( {f}^{\prime }\left( c\right) = 0 \).\n\nIf \( u \) and \( v \) are both end points of \( \left\lbrack {a, b}\right\rbrack \) (meaning \( u \) is \( a \) or \( b \) and \( v \) is \( a \) or \( b \) ) then for \( x \) in \( \left\lbrack {a, b}\right\rbrack \) , because \( f\left( a\right) = f\left( b\right) \), the largest value of \( f\left( x\right) \) is \( f\left( a\right) = f\left( b\right) \) and the least value of \( f\left( x\right) \) is \( f\left( a\right) = f\left( b\right) \) so the only value of \( f\left( x\right) \) is \( f\left( a\right) = f\left( b\right) \). See Figure 12.3. Thus the graph of \( f \) is a horizontal interval and every tangent is horizontal. For the conclusion of the theorem we can choose \( c \) any point between \( a \) and \( b \) and \( {f}^{\prime }\left( c\right) = 0 \).
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Yes
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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.
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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 \( {F}^{\prime } = 0 \). This is impossible because \( {F}^{\prime }\left( x\right) = 1 + 3{x}^{2} > 0 \) for all \( x \). Thus \( F \) is an invertible function.
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Yes
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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.
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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 \( {S}_{1} \) or \( {S}_{2} \) and every number of \( {S}_{1} \) is less than every number in \( {S}_{2} \) and \( {S}_{1} \) does not have a largest number and \( {S}_{2} \) does not have a least number.\n\nLet \( a \) be a number in \( {S}_{1} \) and \( b \) be a number in \( {S}_{2} \) . Consider the function \( f \) defined on \( \left\lbrack {a, b}\right\rbrack \) by\n\n\[ f\left( x\right) = \left\{ \begin{array}{llllll} - 1 & \text{ for } & a \leq x & \text{ and } & x & \text{ in }{S}_{1} \\ 1 & \text{ for } & x \leq b & \text{ and } & x & \text{ in }{S}_{2} \end{array}\right. \]
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Yes
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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 \( \left\lbrack {A, B}\right\rbrack \) .
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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\[ \frac{P\left( b\right) - P\left( a\right) }{b - a} = {P}^{\prime }\left( c\right) \;\text{ so that }\;P\left( b\right) - P\left( a\right) = {P}^{\prime }\left( c\right) \left( {b - a}\right) \]\n\nBy hypothesis, \( {P}^{\prime }\left( c\right) \geq 0 \) . Also \( a < b \) so that \( b - a > 0 \) . It follows that \( P\left( b\right) - P\left( a\right) \geq 0 \), so that \( P\left( a\right) \leq P\left( b\right) \) . Therefore \( P \) is nondecreasing on \( \left\lbrack {A, B}\right\rbrack \) . End of proof.
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Yes
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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 \( \left\lbrack {A, B}\right\rbrack \) .
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Proof: The argument is the similar to that for Theorem 8.1.1 and is left as Exercise 12.2.3
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No
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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 \( {f}^{\prime \prime }\left( c\right) > 0 \) then \( c \) is a local minimum for \( f \) .\n\n2. If \( {f}^{\prime \prime }\left( c\right) < 0 \) then \( c \) is a local maximum for \( f \) .
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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 \( {f}^{\prime \prime } \) is positive throughout \( {\left( u, v\right) }^{1} \) . See the top number line in Figure 12.5. Suppose \( x \) is in \( \left( {u, v}\right) \) (the second number line in Figure 12.5). We must show that \( f\left( c\right) \leq f\left( x\right) \) .\n\nWe consider the case \( x < c \) .\n\n\n\nFigure 12.5: Number lines for the argument to Theorem 12.2.3.\n\nBy the Mean Value Theorem, there is a number, \( {c}_{1} \) between \( x \) and \( c \) for which\n\n\[ \n\frac{f\left( c\right) - f\left( x\right) }{c - x} = {f}^{\prime }\left( {c}_{1}\right) \;\text{ so that }\;f\left( c\right) - f\left( x\right) = {f}^{\prime }\left( {c}_{1}\right) \left( {c - x}\right) \n\]\n\n(12.4)\n\nSee the third number line in Figure 12.5.\n\nWatch this! By the Mean Value Theorem applied to the function \( {f}^{\prime } \) and its derivative \( {f}^{\prime \prime } \), there is a number \( {c}_{2} \) between \( {c}_{1} \) and \( c \) so that\n\n\[ \n\frac{{f}^{\prime }\left( c\right) - {f}^{\prime }\left( {c}_{1}\right) }{c - {c}_{1}} = {f}^{\prime \prime }\left( {c}_{2}\right) \;\text{ and }\;{f}^{\prime }\left( c\right) - {f}^{\prime }\left( {c}_{1}\right) = {f}^{\prime \prime }\left( {c}_{2}\right) \left( {c - {c}_{1}}\right) \n\]\n\n(12.5)\n\nSee the bottom number line in Figure 12.5.\n\nRemember that \( {f}^{\prime }\left( c\right) = 0 \) so from the last equation (Equation 12.5)\n\n\[ \n{f}^{\prime }\left( {c}_{1}\right) = - {f}^{\prime \prime }\left( {c}_{2}\right) \left( {c - {c}_{1}}\right) \n\]\n\nand from this and Equation 12.4 we get\n\n\[ \nf\left( c\right) - f\left( x\right) = - {f}^{\prime \prime }\left( {c}_{2}\right) \left( {c - {c}_{1}}\right) \left( {c - x}\right) \n\]\n\nNow \( c > x \) and \( c > {c}_{1} \) so that \( c - x \) and \( c - {c}_{1} \) are positive. Furthermore, \( {f}^{\prime \prime }\left( {c}_{2}\right) > 0 \) because \( {f}^{\prime \prime } \) is positive throughout \( \left( {u, v}\right) \) . It follows that \( f\left( c\right) - f\left( x\right) \) is negative, and that \( f\left( c\right) < f\left( x\right) \) .\n\nThe event \( c < x \) is similar.\n\nThe argument for Case 2. of Theorem 12.2.3 is similar. End of proof.
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Yes
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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\[ \n{y}^{\prime }\left( t\right) = y\left( t\right) ,\;y\left( 0\right) = 1.\n\]
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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\[ \np\left( 0\right) = y\left( 0\right) ,\;{p}^{\prime }\left( 0\right) = {y}^{\prime }\left( 0\right) ,\;{p}^{\left( 2\right) }\left( 0\right) = {y}^{\left( 2\right) }\left( 0\right) ,\;{p}^{\left( 3\right) }\left( 0\right) = {y}^{\left( 3\right) }\left( 0\right) \text{ and }{p}^{\left( 4\right) }\left( 0\right) = {y}^{\left( 4\right) }\left( 0\right) ,\n\]\n\nBecause\n\n\[ \np\left( 0\right) = {p}_{0},\;{p}^{\prime }\left( 0\right) = {p}_{1},\;{p}^{\prime \prime }\left( 0\right) = 2{p}_{2},\;{p}^{\left( 3\right) }\left( 0\right) = 3 \cdot 2{p}_{3},\text{ and }{p}^{\left( 4\right) }\left( 0\right) = 4 \cdot 3 \cdot 2 \cdot {p}_{4},\n\]\n\nit is sufficient to find values for\n\n\[ \ny\left( 0\right) ,\;{y}^{\prime }\left( 0\right) ,\;{y}^{\left( 2\right) }\left( 0\right) ,\;{y}^{\left( 3\right) }\left( 0\right) ,\;\text{ and }\;{y}^{\left( 4\right) }\left( 0\right) .\n\]\n\nWe are given \( y\left( 0\right) = 1 \) .\n\n\[ \n\text{Because}{y}^{\prime }\left( t\right) = y\left( t\right) ,\;{y}^{\prime }\left( 0\right) = y\left( 0\right) = 1\text{}.\n\]\n\nBecause \( \;{y}^{\prime }\left( t\right) = y\left( t\right) ,\;{y}^{\left( 2\right) }\left( t\right) = {y}^{\prime }\left( t\right) ,\; \) and \( \;{y}^{\left( 2\right) }\left( 0\right) = {y}^{\prime }\left( 0\right) = 1 \) .\n\nContinuing we get\n\n\[ \n{y}^{\left( 2\right) }\left( t\right) = {y}^{\prime }\left( t\right) \;{y}^{\left( 3\right) }\left( t\right) = {y}^{\left( 2\right) }\left( t\right) \;{y}^{\left( 3\right) }\left( 0\right) = {y}^{\left( 2\right) }\left( 0\right) = 1\n\]\n\n\[ \n{y}^{\left( 3\right) }\left( t\right) = {y}^{\left( 2\right) }\left( t\right) \;{y}^{\left( 4\right) }\left( t\right) = {y}^{\left( 3\right) }\left( t\right) \;{y}^{\left( 4\right) }\left( 0\right) = {y}^{\left( 3\right) }\left( 0\right) = 1\n\]\n\nBy insisting that\n\n\[ \np\left( 0\right) = y\left( 0\right) ,\;{p}^{\prime }\left( 0\right) = {y}^{\prime }\left( 0\right) ,\;{p}^{\left( 2\right) }\left( 0\right) = {y}^{\left( 2\right) }\left( 0\right) ,\;{p}^{\left( 3\right) }\left( 0\right) = {y}^{\left( 3\right) }\left( 0\right) \text{ and }\;{p}^{\left( 4\right) }\left( 0\right) = {y}^{\left( 4\right) }\left( 0\right) ,\n\]\n\nwe conclude that\n\[ \n{p}_{0} = 1,{p}_{1} = 1,{p}_{2} = \frac{1}{2},{p}_{3} = \frac{1}{3!}\text{, and}{p}_{4} = \frac{1}{4!}\text{},\n\]\n\nand that\n\[ \np\left( t\right) = 1 + t + \frac{1}{2}{t}^{2} + \frac{1}{3!}{t}^{3} + \frac{1}{4!}{t}^{4}.\n\]
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Yes
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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\[ \n{y}^{\prime \prime }\left( t\right) + y\left( t\right) = 0,\;y\left( 0\right) = 1,\;{y}^{\prime }\left( 0\right) = 0.\n\]
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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\[ \np\left( 0\right) = y\left( 0\right) ,\;{p}^{\prime }\left( 0\right) = {y}^{\prime }\left( 0\right) ,\;{p}^{\left( 2\right) }\left( 0\right) = {y}^{\left( 2\right) }\left( 0\right) ,\;{p}^{\left( 3\right) }\left( 0\right) = {y}^{\left( 3\right) }\left( 0\right) \text{ and }{p}^{\left( 4\right) }\left( 0\right) = {y}^{\left( 4\right) }\left( 0\right) ,\n\]\n\nBecause\n\n\[ \np\left( 0\right) = {p}_{0},\;{p}^{\prime }\left( 0\right) = {p}_{1},\;{p}^{\prime \prime }\left( 0\right) = 2{p}_{2},\;{p}^{\left( 3\right) }\left( 0\right) = 3 \cdot 2{p}_{3},\text{ and }{p}^{\left( 4\right) }\left( 0\right) = 4 \cdot 3 \cdot 2{p}_{4},\n\]\n\nit is sufficient to find values for\n\n\[ \ny\left( 0\right) ,\;{y}^{\prime }\left( 0\right) ,\;{y}^{\left( 2\right) }\left( 0\right) ,\;{y}^{\left( 3\right) }\left( 0\right) ,\;\text{ and }\;{y}^{\left( 4\right) }\left( 0\right) .\n\]\n\nWe are given \( y\left( 0\right) = 1 \) and \( {y}^{\prime }\left( 0\right) = 0 \) .\n\nBecause \( {y}^{\left( 2\right) }\left( t\right) + y\left( t\right) = 0,\;{y}^{\left( 2\right) }\left( t\right) = - y\left( t\right) ,\; \) and \( \;{y}^{\left( 2\right) }\left( 0\right) = - y\left( 0\right) = - 1 \) .\n\nFurthermore,\n\n\[ \n{y}^{\left( 3\right) }\left( t\right) = - {y}^{\prime }\left( t\right) \;{y}^{\left( 3\right) }\left( 0\right) = - {y}^{\prime }\left( 0\right) = 0\n\]\n\n\[ \n{y}^{\left( 4\right) }\left( t\right) = - {y}^{\left( 2\right) }\left( t\right) \;{y}^{\left( 4\right) }\left( 0\right) = - {y}^{\left( 2\right) }\left( 0\right) = 1\n\]\n\nBy matching \( p\left( t\right) \) to \( y\left( t\right) \) at \( t = 0 \) we get\n\n\[ \n{p}_{0} = 1,\;{p}_{1} = 0,\;{p}_{2} = - 1/2,\;{p}_{3} = 0,\;\text{ and }\;{p}_{4} = - 1/4!\n\]\n\nso that\n\[ \np\left( t\right) = 1 - \frac{{t}^{2}}{2} + \frac{{t}^{4}}{4!}\n\]
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Yes
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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\[ \n{y}^{\prime }\left( t\right) = y\left( t\right) \left( {1 - y\left( t\right) }\right) ,\;y\left( 0\right) = 1/2.\n\]
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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\[ \n{y}^{\left( 2\right) }\left( t\right) = y\prime \left( t\right) - {2y}\left( t\right) {y}^{\prime }\left( t\right) \;{y}^{\left( 2\right) }\left( 0\right) = 0 \n\]\n\n\[ \n{y}^{\left( 3\right) } = {y}^{\left( 2\right) } - {2y}{y}^{\left( 2\right) } - 2{\left( {y}^{\prime }\right) }^{2}\;{y}^{\left( 3\right) }\left( 0\right) = - \frac{1}{8} \n\]\n\n\[ \n{y}^{\left( 4\right) } = {y}^{\left( 3\right) } - {2y}{y}^{\left( 3\right) } - 2{y}^{\prime }{y}^{\left( 2\right) } - 4{y}^{\prime }{y}^{\left( 2\right) } \n\]\n\n\[ \n= \;{y}^{\left( 3\right) } - 6\;{y}^{\prime }\;{y}^{\left( 2\right) } - 2\;y\;{y}^{\left( 3\right) }\;{y}^{\left( 4\right) }\left( 0\right) = 0 \n\]\n\n\[ \n{y}^{\left( 5\right) } = {y}^{\left( 4\right) } - 6{y}^{\prime }{y}^{\left( 3\right) } - 6{\left( {y}^{\left( 2\right) }\right) }^{2} - {2y}{y}^{\left( 4\right) } - 2{y}^{\prime }{y}^{\left( 3\right) } \n\]\n\nBy insisting that \n\n\[ \n{a}_{k} = \frac{{y}^{\left( k\right) }\left( 0\right) }{k!},\;k = 0,1,2,\cdots ,5 \n\]\n\nwe get \n\n\[ \np\left( t\right) = \frac{1}{2} + \frac{1}{4}t - \frac{1}{48}{t}^{3} + \frac{1}{480}{t}^{5} \n\]
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Yes
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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?
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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 \( A\left( 1\right) = {1050} \) . If we knew \( {A}^{\prime }\left( 1\right) \) and \( {A}^{\prime \prime }\left( 1\right) \) we could compute the degree-2 polynomial, \( P\left( t\right) \), that matches \( A\left( t\right) \) at the anchor point \( a = 1 \) and use \( P\left( 3\right) \) as our estimate of \( A\left( 3\right) \) . We use the following estimates:\n\n\[ \n{A}^{\prime }\left( a\right) \doteq \frac{A\left( {a + h}\right) - A\left( {a - h}\right) }{2h}\;{A}^{\prime \prime }\left( a\right) \doteq \frac{A\left( {a + h}\right) - {2A}\left( a\right) + A\left( {a - h}\right) }{{h}^{2}} \n\] \n\nwhere for our problem, \( a = 1 \) and \( h = 1 \) . The first estimate is what we have called the centered difference quotient estimate of the derivative; we call the second estimate the centered difference quotient estimate of the second derivative. A rationale for these two estimates appears in the discussion of Solution 2. With these estimates, we have \n\n\[ \n{A}^{\prime }\left( 1\right) \doteq \frac{A\left( 2\right) - A\left( 0\right) }{2} \n\] \n\n\[ \n= \frac{{1080} - {1000}}{2} = {40} \n\] \n\n\[ \n{A}^{\prime \prime }\left( 1\right) \doteq \frac{A\left( 2\right) - {2A}\left( 1\right) + A\left( 0\right) }{{1}^{2}} \n\] \n\n\[ \n= \frac{{1080} - 2 \cdot {1050} + {1000}}{1} = - {20} \n\] \nTherefore, \n\n\[ \nP\left( t\right) = A\left( 1\right) + {A}^{\prime }\left( 1\right) \left( {t - 1}\right) + \frac{1}{2}{A}^{\prime \prime }\left( 1\right) {\left( t - 1\right) }^{2} \n\] \n\n\[ \n= {1050} + {40}\left( {t - 1}\right) + \frac{1}{2}\left( {-{20}}\right) {\left( t - 1\right) }^{2} \n\] \n\n\[ \nP\left( 3\right) = {1050} + {40} \cdot 2 + \frac{1}{2}\left( {-{20}}\right) {\left( 2\right) }^{2} = {1090} \n\] \n\nOur estimate of the January 1, 2003 adult population is 1090 .
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Yes
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What is the error in\n\\[ \sin \frac{\pi }{4} \doteq \frac{\pi }{4} - \frac{1}{3!}{\\left( \frac{\pi }{4}\\right) }^{3}? \\]
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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} \\) . We do a worst case analysis. The largest value of \\( \\sin {c}_{3} \\) for \\( {c}_{3} \\) in \\( \\left\\lbrack {0,\\frac{\\pi }{4}}\\right\\rbrack \\) is for \\( {c}_{3} = \\frac{\\pi }{4} \\) and \\( \\sin \\frac{\\pi }{4} = \\frac{\\sqrt{2}}{2} \\doteq {0.7071} \\) . Therefore, we say that the error in\n\\[ \sin \\frac{\\pi }{4} \\doteq \\frac{\\pi }{4} - \\frac{1}{3!}{\\left( \\frac{\\pi }{4}\\right) }^{3}\\;\\text{is no bigger than}\\;\\frac{\\frac{\\sqrt{2}}{2}}{4!}{\\left( \\frac{\\pi }{4}\\right) }^{4} \\doteq {0.0112} \\]
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Yes
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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 \( f \) .
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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 in \( \left( {p, q}\right) \) . Then because \( {f}^{\prime }\left( a\right) = 0 \)\n\n\[ f\left( b\right) = f\left( a\right) + \frac{{f}^{\left( 2\right) }\left( {c}_{1}\right) }{2!}{\left( b - a\right) }^{2}. \]\n\n\( {c}_{1} \) is between \( b \) and \( a \) and therefore is in \( \left( {p, q}\right) \) so that\n\n\[ f\left( b\right) = f\left( a\right) + \;\text{ a negative number. } \]\n\nThus, \( \left( {a, f\left( a\right) }\right) \) is a local maximum for \( f \) .
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Yes
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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
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\[ f\left( {a + h}\right) = f\left( a\right) + {f}^{\prime }\left( a\right) h + \frac{{f}^{\left( 2\right) }\left( {c}_{1}\right) }{2!}{h}^{2} \] we can solve for \( {f}^{\prime }\left( a\right) \) and get \[ {f}^{\prime }\left( a\right) = \frac{f\left( {a + h}\right) - f\left( a\right) }{h} - \frac{{f}^{\left( 2\right) }\left( {c}_{1}\right) }{2!}h \] \( \left( {12.22}\right) \)
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Yes
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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 to the graph of \( f \) at \( \left( {a, f\left( a\right) }\right) \) .
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\[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{\left| f\left( x\right) - L\left( x\right) \right| }{\left| x - a\right| } = \mathop{\lim }\limits_{{x \rightarrow a}}\left| \frac{f\left( x\right) - f\left( a\right) - {f}^{\prime }\left( a\right) \left( {x - a}\right) }{x - a}\right| \] \[ = \left| {\mathop{\lim }\limits_{{x \rightarrow a}}\frac{f\left( x\right) - f\left( a\right) }{x - a} - {f}^{\prime }\left( a\right) }\right| \] (13.3) \[ = 0\text{.} \]
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Yes
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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 \( z = 0 \) plane is the tangent plane to each of the graphs at \( \left( {0,0}\right) \).
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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 \( F \), the linear approximation to \( F \) at \( \left( {0,0}\right) \) is \( L\left( {x, y}\right) = 0 \), and the tangent plane is \( z = 0 \), or the \( x, y \) plane. The same is true for \( G \) and \( H \) ; the tangent plane at \( \left( {0,0,0}\right) \) is \( z = 0 \) for all three examples.
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Yes
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Fit a line to the data in Example Figure 13.2.3.3.
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\[ {S}_{x} = 1 + 2 + 4 + 6 + 8 = {21}, \] \[ {S}_{y} = {0.5} + {0.8} + {1.0} + {1.7} + {1.8} = {5.8} \] \[ {S}_{xx} = {1}^{2} + {2}^{2} + {4}^{2} + {6}^{2} + {8}^{2} = {121}, \] \[ {S}_{xy} = 1 \cdot {0.5} + 2 \cdot {0.8} + 4 \cdot {1.0} + 6 \cdot {1.7} + 8 \cdot {1.8} = {30.7} \] \[ \Delta = n{S}_{xx} - {\left( {S}_{x}\right) }^{2} = 5 \cdot {121} - {\left( {21}\right) }^{2} = {164} \] \[ a = \left( {{S}_{xx}{S}_{y} - {S}_{x}{S}_{xy}}\right) /\Delta = \left( {{121} \cdot {5.8} - {21} \cdot {30.7}}\right) /{164} = {0.348} \] \[ b = \left( {n{S}_{xy} - {S}_{x}{S}_{y}}\right) /\Delta = \left( {5 \cdot {30.7} - {21} \cdot {5.8}}\right) /{164} = {0.193} \] The line \( y = {0.348} + {0.193x} \) is the closest to the data in the sense of least squares and is drawn in Example Figure 13.2.3.3.
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Yes
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Find the dimensions of the largest box (rectangular solid) that will fit in a hemisphere of radius \( R \) .
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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\[ V\left( {x, y}\right) = {2x} \times {2y} \times z = {4xy}\sqrt{{R}^{2} - {x}^{2} - {y}^{2}} \]\n\nBefore launching into partial differentiation, it is perhaps clever, and certainly useful, to observe that the values of \( x \) and \( y \) for which \( V \) is a maximum are also the values for which \( {V}^{2}/{16} \) is a maximum.\n\n\[ \frac{{V}^{2}\left( {x, y}\right) }{16} = W\left( {x, y}\right) = \frac{{16}{x}^{2}{y}^{2}\left( {{R}^{2} - {x}^{2} - {y}^{2}}\right) }{16} = {R}^{2}{x}^{2}{y}^{2} - {x}^{4}{y}^{2} - {x}^{2}{y}^{4}. \]\n\nIt is easier to analyze \( W\left( {x, y}\right) \) than it is to analyze \( V\left( {x, y}\right) \) .\n\n\[ {W}_{1}\left( {x, y}\right) = 2{R}^{2}x{y}^{2} - 4{x}^{3}{y}^{2} - {2x}{y}^{4} \]\n\n\[ = {2x}{y}^{2}\left( {{R}^{2} - 4{x}^{2} - 2{y}^{2}}\right) \]\n\n\[ {W}_{2}\left( {x, y}\right) = 2{R}^{2}{x}^{2}y - 2{x}^{4}{y}^{2} - 4{x}^{2}{y}^{3} \]\n\n\[ = 2{x}^{2}y\left( {{R}^{2} - 2{x}^{2} - 4{y}^{2}}\right) \]\n\nSolving for \( {W}_{1}\left( {x, y}\right) = 0 \) and \( {W}_{2}\left( {x, y}\right) = 0 \) yields \( x = R/\sqrt{3} \) and \( y = R/\sqrt{3} \), for which \( z = R/\sqrt{3} \). The dimensions of the box are \( {2R}/\sqrt{3},{2R}/\sqrt{3} \), and \( R/\sqrt{3} \).
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Yes
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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 region below the graph of \( F \) is
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\[ {\int }_{0}^{1}{\int }_{0}^{1}\left( {\sin {\pi x}}\right) \left( {\cos \frac{\pi }{2}y}\right) {dydx} = {\int }_{0}^{1}{\left\lbrack \left( \sin \pi x\right) \left( \sin \frac{\pi }{2}y\right) \times \frac{2}{\pi }\right\rbrack }_{y = 0}^{y = 1}{dx} \] \[ = \frac{2}{\pi }{\int }_{0}^{1}\sin {\pi xdx} \] \[ = \frac{2}{\pi } \cdot \frac{2}{\pi } \doteq {0.405285} \]
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Yes
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Show that\n\n\[ \n{u}_{t}\left( {x, t}\right) = \frac{1}{{\pi }^{2}}{u}_{xx}\left( {x, t}\right) ,\;u\left( {x,0}\right) = {30} * \left( {1 - \cos {\pi x}}\right) ,\;\text{ and }\;\begin{array}{l} {u}_{x}\left( {0, t}\right) = 0 \\ {u}_{x}\left( {2, t}\right) = 0. \end{array} \n\]
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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\[ {u}_{x}\left( {x, t}\right) = {30}\left( {0 - {e}^{-t}\left( {-\sin {\pi x}}\right) \left( \pi \right) = {30\pi }{e}^{-t}\sin {\pi x}}\right. \]\n\n\[ {u}_{xx}\left( {x, t}\right) = {30}{\pi }^{2}{e}^{-t}\cos {\pi x} \]\n\nFrom Equations 13.30 and 13.32\n\n\[ {u}_{t}\left( {x, t}\right) = {30}{e}^{-t}\cos \left( {\pi x}\right) = \frac{1}{{\pi }^{2}}{30}{\pi }^{2}{e}^{-t}\cos \left( {\pi x}\right) = \frac{1}{{\pi }^{2}}{u}_{xx}. \]\n\nFrom Equation 13.27\n\n\[ u\left( {x,0}\right) = {30} * {\left. \left( 1 - {e}^{-t}\cos \pi * x\right) \right| }_{t = 0} = {30}\left( {1 - \cos {\pi x}}\right) . \]\n\nFrom Equation 13.31,\n\n\[ {u}_{x}\left( {0, t}\right) = {\left. {30}\pi {e}^{-t}\sin \pi x\right| }_{x = 0} = 0\;\text{ and }\;{u}_{x}\left( {2, t}\right) = {\left. {30}\pi {e}^{-t}\sin \pi x\right| }_{x = 2} = 0 \]\n\nThus all of Equations 13.28 are satisfied.
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Yes
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Express the following sets of numbers using interval notation.
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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 describe the real numbers \( x \) in one of these or the other, we have \( \{ x \mid x \leq - 2 \) or \( x \geq 2\} = \left( {-\infty , - 2\rbrack \cup \lbrack 2,\infty }\right) \) .
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Yes
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Example 1.1.2. Plot the following points: \( A\left( {5,8}\right), B\left( {-\frac{5}{2},3}\right), C\left( {-{5.8}, - 3}\right), D\left( {{4.5}, - 1}\right), E\left( {5,0}\right) \) , \( F\left( {0,5}\right), G\left( {-7,0}\right), H\left( {0, - 9}\right), O\left( {0,0}\right) .{}^{10} \)
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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.
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Yes
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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.
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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 its opposite to get \( \left( {-2, - 3}\right) \).\n\n2. To find the point symmetric about the \( y \) -axis, we replace the \( x \) -coordinate with its opposite to get \( \left( {2,3}\right) \).\n\n3. To find the point symmetric about the origin, we replace the \( x \) - and \( y \) -coordinates with their opposites to get \( \left( {2, - 3}\right) \).
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Yes
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Find and simplify the distance between \( P\left( {-2,3}\right) \) and \( Q\left( {1, - 3}\right) \) .
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\[ d = \sqrt{{\left( {x}_{1} - {x}_{0}\right) }^{2} + {\left( {y}_{1} - {y}_{0}\right) }^{2}} \] \[ = \sqrt{{\left( 1 - \left( -2\right) \right) }^{2} + {\left( -3 - 3\right) }^{2}} \] \[ = \sqrt{9 + {36}} \] \[ = 3\sqrt{5} \] So the distance is \( 3\sqrt{5} \) .
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Yes
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Find all of the points with \( x \) -coordinate 1 which are 4 units from the point \( \\left( {3,2}\\right) \) .
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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 distance from \( \\left( {3,2}\\right) \) to \( \\left( {1, y}\\right) \) be 4 . The Distance Formula, Equation 1.1, yields\n\n\\[ d = \\sqrt{{\\left( {x}_{1} - {x}_{0}\\right) }^{2} + {\\left( {y}_{1} - {y}_{0}\\right) }^{2}} \\]\n\n\\[ 4 = \\sqrt{{\\left( 1 - 3\\right) }^{2} + {\\left( y - 2\\right) }^{2}} \\]\n\n\\[ 4 = \\sqrt{4 + {\\left( y - 2\\right) }^{2}} \\]\n\n\\[ {4}^{2} = {\\left( \\sqrt{4 + {\\left( y - 2\\right) }^{2}}\\right) }^{2} \\] squaring both sides\n\n\\[ {16} = 4 + {\\left( y - 2\\right) }^{2} \\]\n\n\\[ {12} = {\\left( y - 2\\right) }^{2} \\]\n\n\\[ {\\left( y - 2\\right) }^{2} = {12} \\]\n\n\\[ y - 2 = \\pm \\sqrt{12} \\] extracting the square root\n\n\\[ y - 2 = \\pm 2\\sqrt{3} \\]\n\n\\[ y = 2 \\pm 2\\sqrt{3} \\]\n\nWe obtain two answers: \( \\left( {1,2 + 2\\sqrt{3}}\\right) \) and \( \\left( {1,2 - 2\\sqrt{3}}\\right) \) . The reader is encouraged to think about why there are two answers.
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Yes
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Find the midpoint of the line segment connecting \( P\left( {-2,3}\right) \) and \( Q\left( {1, - 3}\right) \) .
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\[ M = \left( {\frac{{x}_{0} + {x}_{1}}{2},\frac{{y}_{0} + {y}_{1}}{2}}\right) \] \[ = \left( {\frac{\left( {-2}\right) + 1}{2},\frac{3 + \left( {-3}\right) }{2}}\right) = \left( {-\frac{1}{2},\frac{0}{2}}\right) \] \[ = \left( {-\frac{1}{2},0}\right) \] The midpoint is \( \left( {-\frac{1}{2},0}\right) \).
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Yes
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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 \( \left( {b, a}\right) \) .
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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 \( y = x \), as required.
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Yes
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1. \( A = \{ \left( {0,0}\right) ,\left( {-3,1}\right) ,\left( {4,2}\right) ,\left( {-3,2}\right) \} \)
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1. To graph \( A \), we simply plot all of the points which belong to \( A \), as shown below on the left.
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No
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Determine whether or not \( \left( {2, - 1}\right) \) is on the graph of \( {x}^{2} + {y}^{3} = 1 \) .
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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 \( {x}^{2} + {y}^{3} = 1 \) .
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Yes
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Example 1.2.3. Graph \( {x}^{2} + {y}^{3} = 1 \) .
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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 the corresponding value \( y \), and plot the resulting point \( \left( {x, y}\right) \) . For example, substituting \( x = - 3 \) into the equation yields\n\n\[ \ny = \sqrt[3]{1 - {x}^{2}} = \sqrt[3]{1 - {\left( -3\right) }^{2}} = \sqrt[3]{-8} = - 2, \n\]\n\nso the point \( \left( {-3, - 2}\right) \) 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.\n\n<table><thead><tr><th>\( x \)</th><th>\( y \)</th><th>\( \left( {x, y}\right) \)</th></tr></thead><tr><td>\( - 3 \)</td><td>\( - 2 \)</td><td>\( \left( {-3, - 2}\right) \)</td></tr><tr><td>\( - 2 \)</td><td>\( - \sqrt[3]{3} \)</td><td>\( \left( {-2, - \sqrt[3]{3}}\right) \)</td></tr><tr><td>\( - 1 \)</td><td>0</td><td>\( \left( {-1,0}\right) \)</td></tr><tr><td>0</td><td>1</td><td>\( \left( {0,1}\right) \)</td></tr><tr><td>1</td><td>0</td><td>\( \left( {1,0}\right) \)</td></tr><tr><td>2</td><td>\( - \sqrt[3]{3} \)</td><td>\( \left( {2, - \sqrt[3]{3}}\right) \)</td></tr><tr><td>3</td><td>\( - 2 \)</td><td>\( \left( {3, - 2}\right) \)</td></tr></table>
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Yes
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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.
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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\[x = 2 \pm 1\]\n\n\[x = 3,1\]\n\nWe get two answers for \( x \) which correspond to two \( x \) -intercepts: \( \left( {1,0}\right) \) and \( \left( {3,0}\right) \) . Turning our attention to \( y \) -intercepts, we set \( x = 0 \) and solve\n\n\[{\left( x - 2\right) }^{2} + {y}^{2} = 1\]\n\n\[{\left( 0 - 2\right) }^{2} + {y}^{2} = 1\]\n\n\[4 + {y}^{2} = 1\]\n\n\[{y}^{2} = - 3\]\n\nSince 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.\n\nPlotting the data we have so far, we get\n\n\n\nMoving 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 \( \left( {1,0}\right) \) is on the graph would mean \( \left( {-1,0}\right) \) would have to be on the graph. (Why?) Since \( \left( {-1,0}\right) \) 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 \( \left( {x, - y}\right) \) into the equation and simplify\n\n\[{\left( x - 2\right) }^{2} + {y}^{2} = 1\]\n\n\[{\left( x - 2\right) }^{2} + {\left( -y\right) }^{2}\overset{?}{ = }1\]\n\n\[{\left( x - 2\right) }^{2} + {y}^{2} \triangleq 1\]\n\nSince 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\n\n
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Yes
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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. \( {R}_{2} = \{ \left( {-2,1}\right) ,\left( {1,3}\right) ,\left( {2,3}\right) ,\left( {3, - 1}\right) \} \)
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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 \( x \) -coordinate has only one corresponding \( y \) -coordinate. So, \( {R}_{2} \) does represent \( y \) as a function of \( x \) .
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Yes
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Use the Vertical Line Test to determine which of the following relations describes \( y \) as a function of \( x \) .
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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 \( x \) .
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Yes
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Use the Vertical Line Test to determine which of the following relations describes \( y \) as a function of \( x \) .
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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 \( {S}_{1} \), there is a point on the curve with \( x \) -coordinate 1 just below \( \left( {1,2}\right) \), which means that both \( \left( {1,2}\right) \) and this point on the curve lie on the vertical line \( x = 1 \) . (See the picture below and the left.) Hence, the graph of \( {S}_{1} \) fails the Vertical Line Test, so \( y \) is not a function of \( x \) here. However, in \( {S}_{2} \) 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 \( {S}_{2} \) only at the point \( \left( {1,2}\right) \) . Indeed, any vertical line will cross the graph at most once, so we have that the graph of \( {S}_{2} \) passes the Vertical Line Test. Thus it describes \( y \) as a function of \( x \) .
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Yes
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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.
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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 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 \( \left( {1,3}\right) \) indicates that the point \( \left( {1,3}\right) \) isn’t on the graph, but all points on the curve leading up to that point are. 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 \( \left( {-\infty ,1}\right) \) . To determine the range of \( G \), we project the curve to the \( y \) -axis as follows: The graph of \( G \) Note that even though there is an open circle at \( \left( {1,3}\right) \), we still include the \( y \) value of 3 in our range, since the point \( \left( {-1,3}\right) \) 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, \( ( - \infty ,4\rbrack \) .
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Yes
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Example 1.3.5. Determine which equations represent \( y \) as a function of \( x \) .
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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\n\[ \ny = \pm \sqrt{1 - {x}^{3}} \n\] \n\nIf we substitute \( x = 0 \) into our equation for \( y \), we get \( y = \pm \sqrt{1 - {0}^{3}} = \pm 1 \), so that \( \left( {0,1}\right) \) and \( \left( {0, - 1}\right) \) are on the graph of this equation. Hence, this equation does not represent \( y \) as a function of \( x \) . 2.\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\nFor every choice of \( x \), the equation \( y = \sqrt[3]{1 - {x}^{2}} \) returns only one value of \( y \) . Hence, this equation describes \( y \) as a function of \( x \) .\n\n3.\n\n\[ \n{x}^{2}y = 1 - {3y} \n\] \n\n\[ \n{x}^{2}y + {3y} = 1 \n\] \n\n\[ \ny\left( {{x}^{2} + 3}\right) = 1\;\text{factor} \n\] \n\n\[ \ny = \frac{1}{{x}^{2} + 3} \n\] \n\nFor each choice of \( x \), there is only one value for \( y \), so this equation describes \( y \) as a function of \( x \) .
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Yes
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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 \( g\\left( x\\right) \) .
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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 parentheses: \( 3\\left( {x + 4}\\right) = {3x} + {12} \) . Hence, \( g\\left( x\\right) = {3x} + {12} \) . We can check our formula by replacing \( x \) with 5 to get \( g\\left( 5\\right) = 3\\left( 5\\right) + {12} = {15} + {12} = {27}\\checkmark \) .
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Yes
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1. (a) \( f\left( {-1}\right), f\left( 0\right), f\left( 2\right) \)
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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, \( f\left( 0\right) = - {\left( 0\right) }^{2} + 3\left( 0\right) + 4 = 4 \), and \( f\left( 2\right) = - {\left( 2\right) }^{2} + 3\left( 2\right) + 4 = - 4 + 6 + 4 = 6 \) .
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Yes
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Find the domain \( {}^{3} \) of the following functions.\n\n1. \( g\left( x\right) = \sqrt{4 - {3x}} \) 2. \( h\left( x\right) = \sqrt[5]{4 - {3x}} \)
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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 \( g\left( x\right) \) returns a real number. Our domain is \( \left( {-\infty ,\frac{4}{3}}\right\rbrack \) .\n\n2. The formula for \( h\left( x\right) \) is hauntingly close to that of \( g\left( x\right) \) with one key difference - whereas the expression for \( g\left( x\right) \) includes an even indexed root (namely a square root), the formula for \( h\left( x\right) \) 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 \( \left( {-\infty ,\infty }\right) \) .
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Yes
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1. Find and interpret \( h\left( {10}\right) \) and \( h\left( {60}\right) \) .
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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 \( 0 \leq t \leq {20} \), we use the first formula listed, \( h\left( t\right) = - 5{t}^{2} + {100t} \), to find \( h\left( {10}\right) \) . We get \( h\left( {10}\right) = - 5{\left( {10}\right) }^{2} + {100}\left( {10}\right) = {500} \) . Since \( t \) represents the number of seconds since lift-off and \( h\left( t\right) \) is the height above the ground in feet, the equation \( h\left( {10}\right) = {500} \) means that 10 seconds after lift-off, the model rocket is 500 feet above the ground. To find \( h\left( {60}\right) \), we note that \( t = {60} \) satisfies \( t > {20} \), so we use the rule \( h\left( t\right) = 0 \) . This function returns a value of 0 regardless of what value is substituted in for \( t \), so \( h\left( {60}\right) = 0 \) . 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.
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Yes
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1. Find \( \left( {f + g}\right) \left( {-1}\right) \)
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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 \( \left( {f + g}\right) \left( {-1}\right) = f\left( {-1}\right) + g\left( {-1}\right) = 8 + 4 = {12}. \)
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Yes
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Find and simplify the difference quotients for the following functions 1. \( f\left( x\right) = {x}^{2} - x - 2 \)
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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 \[ f\left( {x + h}\right) = {\left( x + h\right) }^{2} - \left( {x + h}\right) - 2 \] \[ = {x}^{2} + {2xh} + {h}^{2} - x - h - 2\text{.} \] So the difference quotient is \[ \frac{f\left( {x + h}\right) - f\left( x\right) }{h} = \frac{\left( {{x}^{2} + {2xh} + {h}^{2} - x - h - 2}\right) - \left( {{x}^{2} - x - 2}\right) }{h} \] \[ = \frac{{x}^{2} + {2xh} + {h}^{2} - x - h - 2 - {x}^{2} + x + 2}{h} \] \[ = \frac{{2xh} + {h}^{2} - h}{h} \] \[ = \frac{h\left( {{2x} + h - 1}\right) }{h} \] factor \[ = \frac{h\left( {{2x} + h - 1}\right) }{h} \] cancel \[ = {2x} + h - 1\text{.} \]
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Yes
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1. Find and interpret \( C\left( 0\right) \) .
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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 might be.
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Yes
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Example 1.6.1. Graph \( f\left( x\right) = {x}^{2} - x - 6 \) .
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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 \( y = f\left( x\right) \), this means \( f\left( x\right) = 0 \) .\n\n\[ f\left( x\right) = {x}^{2} - x - 6 \]\n\n\[ 0 = {x}^{2} - x - 6 \]\n\n\[ 0 = \left( {x - 3}\right) \left( {x + 2}\right) \text{factor} \]\n\n\[ x - 3 = 0\text{ or }x + 2 = 0 \]\n\n\[ x = - 2,3 \]\n\nSo we get \( \left( {-2,0}\right) \) and \( \left( {3,0}\right) \) as \( x \) -intercepts. To find the \( y \) -intercept, we set \( x = 0 \) . Using function notation, this is the same as finding \( f\left( 0\right) \) and \( f\left( 0\right) = {0}^{2} - 0 - 6 = - 6 \) . Thus the \( y \) -intercept is \( \left( {0, - 6}\right) \) . 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.\n\n<table><thead><tr><th>\( x \)</th><th>\( f\left( x\right) \)</th><th>\( \left( {x, f\left( x\right) }\right) \)</th></tr></thead><tr><td>\( - 3 \)</td><td>6</td><td>\( \left( {-3,6}\right) \)</td></tr><tr><td>\( - 2 \)</td><td>0</td><td>\( \left( {-2,0}\right) \)</td></tr><tr><td>\( - 1 \)</td><td>\( - 4 \)</td><td>\( \left( {-1, - 4}\right) \)</td></tr><tr><td>0</td><td>\( - 6 \)</td><td>\( \left( {0, - 6}\right) \)</td></tr><tr><td>1</td><td>\( - 6 \)</td><td>\( \left( {1, - 6}\right) \)</td></tr><tr><td>2</td><td>\( - 4 \)</td><td>\( \left( {2, - 4}\right) \)</td></tr><tr><td>3</td><td>0</td><td>\( \left( {3,0}\right) \)</td></tr><tr><td>4</td><td>6</td><td>\( \left( {4,6}\right) \)</td></tr></table>
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Yes
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Graph: \( f\left( x\right) = \left\{ \begin{matrix} 4 - {x}^{2} & \text{ if } & x < 1 \\ x - 3, & \text{ if } & x \geq 1 \end{matrix}\right. \)
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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 \( f\left( x\right) = 4 - {x}^{2} \) . Setting \( f\left( x\right) = 0 \) gives \( 0 = 4 - {x}^{2} \), so that \( x = \pm 2 \) . 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 \( \left( {-2,0}\right) \) . For \( x \geq 1, f\left( x\right) = x - 3 \) . Setting \( f\left( x\right) = 0 \) gives \( 0 = x - 3 \), or \( x = 3 \) . Since \( x = 3 \) satisfies the inequality \( x \geq 1 \), we get \( \left( {3,0}\right) \) as another \( x \) -intercept. Next, we seek the \( y \) -intercept. Notice that \( x = 0 \) falls in the domain \( x < 1 \) . Thus \( f\left( 0\right) = 4 - {0}^{2} = 4 \) yields the \( y \) -intercept \( \left( {0,4}\right) \) . As far as symmetry is concerned, you can check that the equation \( y = 4 - {x}^{2} \) 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 \( f\left( x\right) = x - 3 \), so the point on the graph \( \left( {1, f\left( 1\right) }\right) \) is \( \left( {1, - 2}\right) \) . However, for all values less than 1, we use the formula \( f\left( x\right) = 4 - {x}^{2} \) . 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 \( f\left( x\right) = 4 - {x}^{2} \) . Making a table as before, we see that as the \( x \) values sneak up to \( x = 1 \) in this fashion, the \( f\left( x\right) \) values inch closer and closer \( {}^{1} \) to \( 4 - {1}^{2} = 3 \) . To indicate this graphically, we use an open circle at the point \( \left( {1,3}\right) \) . Putting all of this information together and plotting additional points, we get
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Yes
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Determine analytically if the following functions are even, odd, or neither even nor odd. Verify your result with a graphing calculator.
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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\nHence, \( f \) is even. The graphing calculator furnishes the following.\n\n \n\nThis suggests that the graph of \( f \) is symmetric about the \( y \)-axis, as expected.
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Yes
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1. Find the domain of \( f \).
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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 \( \left\lbrack {-4,4}\right\rbrack \) .
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Yes
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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.
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Solution. Entering this function into the calculator gives\n\n \n\nUsing the Minimum and Maximum features, we get\n\n\n\nTo two decimal places, \( f \) appears to have its only local minimum at \( \left( {-{1.73}, - {4.33}}\right) \) and its only local maximum at \( \left( {{1.73},{4.33}}\right) \) . Given the symmetry about the origin suggested by the graph, the relation between these points shouldn't be too surprising. The function appears to be increasing on \( \left\lbrack {-{1.73},{1.73}}\right\rbrack \) and decreasing on \( \left( {-\infty , - {1.73}\rbrack \cup \lbrack {1.73},\infty }\right) \) . This makes -4.33 the (absolute) minimum and 4.33 the (absolute) maximum.
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Yes
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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.
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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\[ = \sqrt{{x}^{2} + {\left\lbrack {\left( x - 3\right) }^{2}\right\rbrack }^{2}}\;\text{Since}y = {\left( x - 3\right) }^{2} \]\n\n\[ = \sqrt{{x}^{2} + {\left( x - 3\right) }^{4}} \]\n\nGiven a value for \( x \), the formula \( d = \sqrt{{x}^{2} + {\left( x - 3\right) }^{4}} \) is the distance from \( \left( {0,0}\right) \) to the point \( \left( {x, y}\right) \) on the curve \( y = {\left( x - 3\right) }^{2} \) . What we have defined, then, is a function \( d\left( x\right) \) which we wish to minimize over all values of \( x \) . To accomplish this task analytically would require Calculus so as we've mentioned before, we can use a graphing calculator to find an approximate solution. Using the calculator, we enter the function \( d\left( x\right) \) as shown below and graph. \n\nUsing the Minimum feature, we see above on the right that the (absolute) minimum occurs near \( x = 2 \) . Rounding to two decimal places, we get that the minimum distance occurs when \( x = {2.00} \) . To find the \( y \) value on the parabola associated with \( x = {2.00} \), we substitute 2.00 into the equation to get \( y = {\left( x - 3\right) }^{2} = {\left( {2.00} - 3\right) }^{2} = {1.00} \) . So, our final answer is \( \left( {{2.00},{1.00}}\right) {.}^{16} \)
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Yes
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Theorem 1.2. Vertical Shifts. Suppose \( f \) is a function and \( k \) is a positive number.
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- 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 subtracting \( k \) from the \( y \)-coordinates of the points on the graph of \( f \).
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Yes
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Theorem 1.3. Horizontal Shifts. Suppose \( f \) is a function and \( h \) is a positive number.
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- 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 \( h \) units by adding \( h \) to the \( x \) -coordinates of the points on the graph of \( f \) .
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Yes
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1. Graph \( f\left( x\right) = \sqrt{x} \) . Plot at least three points.
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## 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<table><thead><tr><th>\( x \)</th><th>\( f\left( x\right) \)</th><th>\( \left( {x, f\left( x\right) }\right) \)</th></tr></thead><tr><td>0</td><td>0</td><td>\( \left( {0,0}\right) \)</td></tr><tr><td>1</td><td>1</td><td>\( \left( {1,1}\right) \)</td></tr><tr><td>4</td><td>2</td><td>\( \left( {4,2}\right) \)</td></tr></table>\n\n\n\n\( y = f\left( x\right) = \sqrt{x} \)
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Yes
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Theorem 1.4. Reflections. Suppose \( f \) is a function.
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- 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 multiplying the \( x \) -coordinates of the points on the graph of \( f \) by -1 .
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Yes
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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. \( m\left( x\right) = 3 - \sqrt{x} \)
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## 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, if \( x = - 4, - x = 4 \), thus \( \sqrt{-x} = \sqrt{-\left( {-4}\right) } = 2 \) is perfectly well-defined. To find the domain analytically, we set \( - x \geq 0 \) which gives \( x \leq 0 \), so that the domain of \( g \) is \( ( - \infty ,0\rbrack \) . Since \( g\left( x\right) = f\left( {-x}\right) \), Theorem 1.4 tells us that the graph of \( g \) is the reflection of the graph of \( f \) across the \( y \) -axis. We accomplish this by multiplying each \( x \) -coordinate on the graph of \( f \) by -1, so that the points \( \left( {0,0}\right) ,\left( {1,1}\right) \), and \( \left( {4,2}\right) \) move to \( \left( {0,0}\right) ,\left( {-1,1}\right) \), and \( \left( {-4,2}\right) \) , respectively. Graphically, we see that the domain of \( g \) is \( ( - \infty ,0\rbrack \) and the range of \( g \) is the same as the range of \( f \), namely \( \lbrack 0,\infty ) \) .\n\n
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Yes
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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 \( a \) .
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- 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 \( \frac{1}{a} \) .
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Yes
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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. \( g\left( x\right) = 3\sqrt{x} \)
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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 is a vertical scaling of the graph of \( f \) by a factor of 3. We find the range of \( g \) is also \( \lbrack 0,\infty ) \) .
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Yes
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Theorem 1.7. Transformations. Suppose \( f \) is a function. If \( A \neq 0 \) and \( B \neq 0 \), then to graph\n\n\[ g\left( x\right) = {Af}\left( {{Bx} + H}\right) + K \]
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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, but may also include a reflection about the \( y \) -axis if \( B < 0 \) .\n\n3. Multiply the \( y \) -coordinates of the points on the graph obtained in Step 2 by \( A \) . This results in a vertical scaling, but may also include a reflection about the \( x \) -axis if \( A < 0 \) .\n\n4. Add \( K \) to each of the \( y \) -coordinates of the points on the graph obtained in Step 3. This results in a vertical shift up if \( K > 0 \) or down if \( K < 0 \) .\n\nTheorem 1.7 can be established by generalizing the techniques developed in this section. Suppose \( \left( {a, b}\right) \) is on the graph of \( f \) . Then \( f\left( a\right) = b \), and to make good use of this fact, we set \( {Bx} + H = a \) and solve. We first subtract the \( H \) (causing the horizontal shift) and then divide by \( B \) . If \( B \)\nis a positive number, this induces only a horizontal scaling by a factor of \( \frac{1}{B} \) . If \( B < 0 \), then we have a factor of -1 in play, and dividing by it induces a reflection about the \( y \) -axis. So we have \( x = \frac{a - H}{B} \) as the input to \( g \) which corresponds to the input \( x = a \) to \( f \) . We now evaluate \( g\left( \frac{a - H}{B}\right) = {Af}\left( {B \cdot \frac{a - H}{B} + H}\right) + K = {Af}\left( a\right) + K = {Ab} + K \) . We notice that the output from \( f \) is first multiplied by \( A \) . As with the constant \( B \), if \( A > 0 \), this induces only a vertical scaling. If \( A < 0 \), then the -1 induces a reflection across the \( x \) -axis. Finally, we add \( K \) to the result, which is our vertical shift. A less precise, but more intuitive way to paraphrase Theorem 1.7 is to think of the quantity \( {Bx} + H \) is the ’inside’ of the function \( f \) . What’s happening inside \( f \) affects the inputs or \( x \) -coordinates of the points on the graph of \( f \) . To find the \( x \) -coordinates of the corresponding points on \( g \), we undo what has been done to \( x \) in the same way we would solve an equation. What’s happening to the output can be thought of as things happening ’outside’ the function, \( f \) . Things happening outside affect the outputs or \( y \) -coordinates of the points on the graph of \( f \) . Here, we follow the usual order of operations agreement: we first multiply by \( A \) then add \( K \) to find the corresponding \( y \) -coordinates on the graph of \( g \) .
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Yes
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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 unit\n\n4. Horizontal stretching by a factor of 2
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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 \( x \) -axis using Theorem 1.4: \( {g}_{2}\left( x\right) = - {g}_{1}\left( x\right) = - \left( {{x}^{2} + 2}\right) = \) \( - {x}^{2} - 2 \) . We shift the graph to the right 1 unit, according to Theorem 1.3, by setting \( {g}_{3}\left( x\right) = \) \( {g}_{2}\left( {x - 1}\right) = - {\left( x - 1\right) }^{2} - 2 = - {x}^{2} + {2x} - 3 \) . Finally, we induce a horizontal stretch by a factor of 2 using Theorem 1.6 to get \( g\left( x\right) = {g}_{3}\left( {\frac{1}{2}x}\right) = - {\left( \frac{1}{2}x\right) }^{2} + 2\left( {\frac{1}{2}x}\right) - 3 \) which yields \( g\left( x\right) = - \frac{1}{4}{x}^{2} + x - 3 \) . We use the calculator to graph the stages below to confirm our result.
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Yes
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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.
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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. \( m = \frac{4 - 2}{3 - \left( {-1}\right) } = \frac{2}{4} = \frac{1}{2} \)\n\n\n\n3. \( m = \frac{-3 - 3}{2 - \left( {-2}\right) } = \frac{-6}{4} = - \frac{3}{2} \)\n\n\n\n4. \( m = \frac{2 - 2}{4 - \left( {-3}\right) } = \frac{0}{7} = 0 \)\n\n\n\n\n\n5. \( m = \frac{-1 - 3}{2 - 2} = \frac{-4}{0} \), which is undefined \n\n6. \( m = \frac{-1 - 3}{{2.1} - 2} = \frac{-4}{0.1} = - {40} \)\n\n
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Yes
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1. Find the slope of the line containing the points \( \left( {6,{24}}\right) \) and \( \left( {{10},{32}}\right) \) .
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1. For the slope, we have \( m = \frac{{32} - {24}}{{10} - 6} = \frac{8}{4} = 2 \) .
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Yes
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Example 2.1.3. Write the equation of the line containing the points \( \left( {-1,3}\right) \) and \( \left( {2,1}\right) \) .
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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 \( \left( {-1,3}\right) \) and leave it to the reader to check that using \( \left( {2,1}\right) \) results in the same equation. Substituting into the point-slope form of the line, we get\n\n\[ y - {y}_{0} = m\left( {x - {x}_{0}}\right) \]\n\n\[ y - 3 = - \frac{2}{3}\left( {x - \left( {-1}\right) }\right) \]\n\n\[ y - 3 = - \frac{2}{3}\left( {x + 1}\right) \]\n\n\[ y - 3 = - \frac{2}{3}x - \frac{2}{3} \]\n\n\[ y = - \frac{2}{3}x + \frac{7}{3} \]\n\nWe can check our answer by showing that both \( \left( {-1,3}\right) \) and \( \left( {2,1}\right) \) are on the graph of \( y = - \frac{2}{3}x + \frac{7}{3} \) algebraically, as we did in Section 1.2.1.
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Yes
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Graph the following functions. Identify the slope and \( y \) -intercept.
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1. To graph \( f\left( x\right) = 3 \), we graph \( y = 3 \). This is a horizontal line \( \left( {m = 0}\right) \) through \( \left( {0,3}\right) \).
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No
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1. Find and interpret \( C\left( {10}\right) \) .
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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, \( C\left( {10}\right) = {950} \) means it costs \( \$ {950} \) to produce 10 PortaBoys for the local retailer.
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Yes
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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.
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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 \( p\left( x\right) = {mx} + b \) . To determine \( m \) and \( b \), we use the fact that 20 PortaBoys were sold during the week when the price was 220 dollars and 40 units were sold when the price was 190 dollars. Using function notation, these two facts can be translated as \( p\left( {20}\right) = {220} \) and \( p\left( {40}\right) = {190} \) . Since \( m \) represents the rate of change of \( p \) with respect to \( x \), we have\n\n\[ m = \frac{\Delta p}{\Delta x} = \frac{{190} - {220}}{{40} - {20}} = \frac{-{30}}{20} = - {1.5}. \]\n\nWe now have determined \( p\left( x\right) = - {1.5x} + b \) . To determine \( b \), we can use our given data again. Using \( p\left( {20}\right) = {220} \), we substitute \( x = {20} \) into \( p\left( x\right) = {1.5x} + b \) and set the result equal to 220 : \( - {1.5}\left( {20}\right) + b = {220} \) . Solving, we get \( b = {250} \) . Hence, we get \( p\left( x\right) = - {1.5x} + {250} \) . We can check our formula by computing \( p\left( {20}\right) \) and \( p\left( {40}\right) \) to see if we get 220 and 190, respectively. You may recall from page 82 that the function \( p\left( x\right) \) is called the price-demand (or simply demand) function for this venture.
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Yes
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1. Find and simplify an expression for the weekly revenue \( R\left( x\right) \) as a function of weekly sales \( x \) .
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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 restricted to these values of \( x \) as well.
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Yes
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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: \( \left| {ab}\right| = \left| a\right| \left| b\right| \)
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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 \( \left| {ab}\right| = \left| a\right| \left| b\right| \) . We need to show that this equation is true for all real numbers \( a \) and \( b \) . If \( a \) and \( b \) are both positive, then so is \( {ab} \) . Hence, \( \left| a\right| = a,\left| b\right| = b \) and \( \left| {ab}\right| = {ab} \) . Hence, the equation \( \left| {ab}\right| = \left| a\right| \left| b\right| \) is the same as \( {ab} = {ab} \) which is true. If both \( a \) and \( b \) are negative, then \( {ab} \) is positive. Hence, \( \left| a\right| = - a,\left| b\right| = - b \) and \( \left| {ab}\right| = {ab} \) . The equation \( \left| {ab}\right| = \left| a\right| \left| b\right| \) becomes \( {ab} = \left( {-a}\right) \left( {-b}\right) \), which is true. Suppose \( a \) is positive and \( b \) is negative. Then \( {ab} \) is negative, and we have \( \left| {ab}\right| = - {ab},\left| a\right| = a \) and \( \left| b\right| = - b \) . The equation \( \left| {ab}\right| = \left| a\right| \left| b\right| \) reduces to \( - {ab} = a\left( {-b}\right) \) which is true. A symmetric argument shows the equation \( \left| {ab}\right| = \left| a\right| \left| b\right| \) holds when \( a \) is negative and \( b \) is positive. Finally, if either \( a \) or \( b \) (or both) are zero, then both sides of \( \left| {ab}\right| = \left| a\right| \left| b\right| \) are zero, so the equation holds in this case, too. All of this rhetoric has shown that the equation \( \left| {ab}\right| = \left| a\right| \left| b\right| \) holds true in all cases.
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Yes
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1. \( \left| {{3x} - 1}\right| = 6 \)
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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 get \( x = - \frac{5}{3} \) . We may check both of these solutions by substituting them into the original equation and showing that the arithmetic works out.
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Yes
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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 of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute extrema, if they exist.
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## 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 \( \left( {0,0}\right) \) as our only \( x \) -intercept. To find the \( y \) -intercept, we set \( x = 0 \), and find \( y = f\left( 0\right) = 0 \) , so that \( \left( {0,0}\right) \) is our \( y \) -intercept as well. \( {}^{1} \) Using Definition 2.4, we get\n\n\[ f\left( x\right) = \left| x\right| = \begin{cases} - x, & \text{ if }x < 0 \\ x, & \text{ if }x \geq 0 \end{cases} \]\n\nHence, for \( x < 0 \), we are graphing the line \( y = - x \) ; for \( x \geq 0 \), we have the line \( y = x \) . Proceeding as we did in Section 1.6, we get\n\n\n\n\n\nNotice that we have an ’open circle’ at \( \left( {0,0}\right) \) in the graph when \( x < 0 \) . As we have seen before, this is due to the fact that the points on \( y = - x \) approach \( \left( {0,0}\right) \) as the \( x \) -values approach 0 . Since \( x \) is required to be strictly less than zero on this stretch, the open circle is drawn at the origin. However, notice that when \( x \geq 0 \), we get to fill in the point at \( \left( {0,0}\right) \), which effectively 'plugs' the hole indicated by the open circle. Thus we get,\n\n\n\n---\n\n\( {}^{1} \) Actually, since functions can have at most one \( y \) -intercept (Do you know why?), as soon as we found \( \left( {0,0}\right) \) as the \( x \) -intercept, we knew this was also the \( y \) -intercept.\n\n---\n\nBy projecting the graph to the \( x \) -axis, we see that the domain is \( \left( {-\infty ,\infty }\right) \) . Projecting to the \( y \) -axis gives us the range \( \lbrack 0,\infty ) \) . The function is increasing on \( \lbrack 0,\infty ) \) and decreasing on \( ( - \infty ,0\rbrack \) . The relative minimum value of \( f \) is the same as the absolute minimum, namely 0 which occurs at \( \left( {0,0}\right) \) . There is no relative maximum value of \( f \) . There is also no absolute maximum value of \( f \), since the \( y \) values on the graph extend infinitely upwards.
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Yes
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Graph the following functions starting with the graph of \( f\left( x\right) = \left| x\right| \) and using transformations.
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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\( f\left( x\right) = \left| x\right| \)\n\n1. Since \( g\left( x\right) = \left| {x - 3}\right| = f\left( {x - 3}\right) \), Theorem 1.7 tells us to add 3 to each of the \( x \) -values of the points on the graph of \( y = f\left( x\right) \) to obtain the graph of \( y = g\left( x\right) \) . This shifts the graph of \( y = f\left( x\right) \) to the right 3 units and moves the point \( \left( {-1,1}\right) \) to \( \left( {2,1}\right) ,\left( {0,0}\right) \) to \( \left( {3,0}\right) \) and \( \left( {1,1}\right) \) to \( \left( {4,1}\right) \) . Connecting these points in the classic ’ \( \vee \) ’ fashion produces the graph of \( y = g\left( x\right) \) .\n\n\n\n\( f\left( x\right) = \left| x\right| \)
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Yes
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1. \( f\left( x\right) = \frac{\left| x\right| }{x} \)
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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 \( \left| x\right| = 0 \), which, from Theorem 2.1, implies \( x = 0 \) . However, \( x = 0 \) is not in the domain of \( f \) , which means we have, in fact, no \( x \) -intercepts. We have no \( y \) -intercepts either, since \( f\left( 0\right) \) is undefined. Re-writing the absolute value in the function gives \[ f\left( x\right) = \left\{ {\begin{matrix} \frac{-x}{x}, & \text{ if }x < 0 \\ \frac{x}{x}, & \text{ if }x > 0 \end{matrix} = \left\{ \begin{array}{rr} - 1, & \text{ if }x < 0 \\ 1, & \text{ if }x > 0 \end{array}\right. }\right. \] To graph this function, we graph two horizontal lines: \( y = - 1 \) for \( x < 0 \) and \( y = 1 \) for \( x > 0 \) . We have open circles at \( \left( {0, - 1}\right) \) and \( \left( {0,1}\right) \) (Can you explain why?) so we get  \[ f\left( x\right) = \frac{\left| x\right| }{x} \] As we found earlier, the domain is \( \left( {-\infty ,0}\right) \cup \left( {0,\infty }\right) \) . The range consists of just two \( y \) -values: \( \{ - 1,1\} \) . The function \( f \) is constant on \( \left( {-\infty ,0}\right) \) and \( \left( {0,\infty }\right) \) . The local minimum value of \( f \) is the absolute minimum value of \( f \), namely -1 ; the local maximum and absolute maximum values for \( f \) also coincide - they both are 1 . Every point on the graph of \( f \) is simultaneously a relative maximum and a relative minimum. (Can you remember why in light of Definition 1.11? This was explored in the Exercises in Section 1.6.2.)
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Yes
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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. \( g\left( x\right) = {\left( x + 2\right) }^{2} - 3 \)
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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 \( \left( {-2,4}\right) \) to \( \left( {-4,4}\right) ,\left( {-1,1}\right) \) to \( \left( {-3,1}\right) ,\left( {0,0}\right) \) to \( \left( {-2,0}\right) ,\left( {1,1}\right) \) to \( \left( {-1,1}\right) \) and \( \left( {2,4}\right) \) to \( \left( {0,4}\right) \) . Next, we subtract 3 from each of the \( y \) -values of these new points. This moves the graph down 3 units and moves \( \left( {-4,4}\right) \) to \( \left( {-4,1}\right) ,\left( {-3,1}\right) \) to \( \left( {-3, - 2}\right) ,\left( {-2,0}\right) \) to \( \left( {-2,3}\right) \) , \( \left( {-1,1}\right) \) to \( \left( {-1, - 2}\right) \) and \( \left( {0,4}\right) \) to \( \left( {0,1}\right) \) . We connect the dots in parabolic fashion to get\n\n\n\nFrom the graph, we see that the vertex has moved from \( \left( {0,0}\right) \) on the graph of \( y = f\left( x\right) \) to \( \left( {-2, - 3}\right) \) on the graph of \( y = g\left( x\right) \) . This sets \( \lbrack - 3,\infty ) \) as the range of \( g \) . We see that the graph of \( y = g\left( x\right) \) crosses the \( x \) -axis twice, so we expect two \( x \) -intercepts. To find these, we set \( y = g\left( x\right) = 0 \) and solve. Doing so yields the equation \( {\left( x + 2\right) }^{2} - 3 = 0 \), or \( {\left( x + 2\right) }^{2} = 3 \) . Extracting square roots gives \( x + 2 = \pm \sqrt{3} \), or \( x = - 2 \pm \sqrt{3} \) . Our \( x \) -intercepts are \( \left( {-2 - \sqrt{3},0}\right) \approx \left( {-{3.73},0}\right) \) and \( \left( {-2 + \sqrt{3},0}\right) \approx \left( {-{0.27},0}\right) \) . The \( y \) -intercept of the graph, \( \left( {0,1}\right) \) was one of the points we originally plotted, so we are done.
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Yes
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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 \( \left( {h, k}\right) \) .
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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 positive number. If \( a > 0 \), then \( a{\left( x - h\right) }^{2} \) is positive, thus \( y = a{\left( x - h\right) }^{2} + k \) is always a number larger than \( k \) . This means that when \( a > 0,\left( {h, k}\right) \) is the lowest point on the graph and thus the parabola must open upwards, making \( \left( {h, k}\right) \) the vertex. A similar argument shows that if \( a < 0,\left( {h, k}\right) \) is the highest point on the graph, so the parabola opens downwards, and \( \left( {h, k}\right) \) is also the vertex in this case.
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Yes
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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. \( f\left( x\right) = {x}^{2} - {4x} + 3 \)
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## 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 target perfect square quantity is \( {\left( x - 2\right) }^{2} \) . To get an expression equivalent to \( {\left( x - 2\right) }^{2} \), we need to add \( {\left( -2\right) }^{2} = 4 \) to the \( {x}^{2} - {4x} \) to create a perfect square trinomial, but to keep the balance, we must also subtract it. We collect the terms which create the perfect square and gather the remaining constant terms. Putting it all together, we get\n\n\[ f\left( x\right) = {x}^{2} - {4x} + 3\;\text{(Compute}\frac{1}{2}\left( {-4}\right) = - 2\text{.)} \]\n\n\[ = \left( {{x}^{2} - {4x} + \underline{4} - \underline{4}}\right) + 3\;\text{(Add and subtract}{\left( -2\right) }^{2} = 4\text{to}\left( {{x}^{2} + {4x}}\right) \text{.)} \]\n\n\[ = \left( {{x}^{2} - {4x} + 4}\right) - 4 + 3\;\text{(Group the perfect square trinomial.)} \]\n\n\[ = {\left( x - 2\right) }^{2} - 1\;\text{(Factor the perfect square trinomial.)} \]\n\nOf course, we can always check our answer by multiplying out \( f\left( x\right) = {\left( x - 2\right) }^{2} - 1 \) to see that it simplifies to \( f\left( x\right) = {x}^{2} - {4x} - 1 \) . In the form \( f\left( x\right) = {\left( x - 2\right) }^{2} - 1 \), we readily find the vertex to be \( \left( {2, - 1}\right) \) which makes the axis of symmetry \( x = 2 \) . To find the \( x \) -intercepts, we set \( y = f\left( x\right) = 0 \) . We are spoiled for choice, since we have two formulas for \( f\left( x\right) \) . Since we recognize \( f\left( x\right) = {x}^{2} - {4x} + 3 \) to be easily factorable, \( {}^{8} \) we proceed to solve \( {x}^{2} - {4x} + 3 = 0 \) . Factoring gives \( \left( {x - 3}\right) \left( {x - 1}\right) = 0 \) so that \( x = 3 \) or \( x = 1 \) . The \( x \) -intercepts are then \( \left( {1,0}\right) \) and \( \left( {3,0}\right) \) . To find the \( y \) -intercept, we set \( x = 0 \) . Once again, the general form \( f\left( x\right) = {x}^{2} - {4x} + 3 \) is easiest to work with here, and we find \( y = f\left( 0\right) = 3 \) . Hence, the \( y \) -intercept is \( \left( {0,3}\right) \) . With the vertex, axis of symmetry and the intercepts, we get a pretty good graph without the need to plot additional points. We see that the range of \( f \) is \( \lbrack - 1,\infty ) \) and we are done.
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Yes
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1. Determine the weekly profit function \( P\left( x\right) \) .
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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 restricted to these values.
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Yes
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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 material. If he makes the pasture adjacent to a stream (so no fencing is required on that side), what are the dimensions of the pasture which maximize the area? What is the maximum area? If an average alpaca needs 25 square feet of grazing area, how many alpaca can Donnie keep in his pasture?
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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 pasture and we let \( l \) denote the length of the pasture. Since the units given to us in the statement of the problem are feet, we assume \( w \) and \( l \) are measured in feet. The area of the pasture, which we’ll call \( A \), is related to \( w \) and \( l \) by the equation \( A = {wl} \) . Since \( w \) and \( l \) are both measured in feet, \( A \) has units of feet \( {}^{2} \), or square feet. We are given the total amount of fencing available is 200 feet, which means \( w + l + w = {200} \), or, \( l + {2w} = {200} \) . We now have two equations, \( A = {wl} \) and \( l + {2w} = {200} \) . In order to use the tools given to us in this section to maximize \( A \), we need to use the information given to write \( A \) as a function of just one variable, either \( w \) or \( l \) . This is where we use the equation \( l + {2w} = {200} \) . Solving for \( l \), we find \( l = {200} - {2w} \) , and we substitute this into our equation for \( A \) . We get \( A = {wl} = w\left( {{200} - {2w}}\right) = {200w} - 2{w}^{2} \) . We now have \( A \) as a function of \( w, A\left( w\right) = {200w} - 2{w}^{2} = - 2{w}^{2} + {200w} \) .\n\nBefore we go any further, we need to find the applied domain of \( A \) so that we know what values of \( w \) make sense in this problem situation. \( {}^{9} \) Since \( w \) represents the width of the pasture, \( w > 0 \) . Likewise, \( l \) represents the length of the pasture, so \( l = {200} - {2w} > 0 \) . Solving this latter inequality, we find \( w < {100} \) . Hence, the function we wish to maximize is \( A\left( w\right) = - 2{w}^{2} + {200w} \) for \( 0 < w < {100} \) . Since \( A \) is a quadratic function (of \( w \) ), we know that the graph of \( y = A\left( w\right) \) is a parabola. Since the coefficient of \( {w}^{2} \) is -2, we know that this parabola opens downwards. This means that there is a maximum value to be found, and we know it occurs at the vertex. Using the vertex formula, we find \( w = - \frac{200}{2\left( {-2}\right) } = {50} \), and \( A\left( {50}\right) = - 2{\left( {50}\right) }^{2} + {200}\left( {50}\right) = {5000} \) . Since \( w = {50} \) lies in the applied domain, \( 0 < w < {100} \), we have that the area of the pasture is maximized when the width is 50 feet. To find the length, we use \( l = {200} - {2w} \) and find \( l = {200} - 2\left( {50}\right) = {100} \), so the length of the pasture is 100 feet. The maximum area is \( A\left( {50}\right) = {5000} \), or 5000 square feet. If an average alpaca requires 25 square feet of pasture, Donnie can raise \( \frac{5000}{25} = {200} \) average alpaca.
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Yes
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