Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 9

Answer

a. $8a - 6a^{2}$ b. Tangent lines: $y = 2x + 3$ and $y = -8x + 19$ c. Graph
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Work Step by Step

$y = 3 + 4x^{2} - 2x^{3}$ $\lim\limits_{h \to 0} \frac{3+4(a+h)^{2}-2(a+h)^{3} - (3 + 4a^{2} - 2a^{3}}{h}$ Expand: $\lim\limits_{h \to 0} \frac{3+4a^{2} + 8ah + 4h^{2} -2a^{3} - 6a^{2}h -6ah^{2} - 2h^{3} - 3 -4a^{2} + 2a^{3}}{h}$ Simplify: $\lim\limits_{h \to 0} \frac{8ah + 4h^{2} - 6a^{2}h -6ah^{2} - 2h^{3}}{h}$ Divide by $h$. $\lim\limits_{h \to 0} \frac{h(8a + 4h - 6a^{2} -6ah - 2h^{2})}{h}$ Cancel $h$. $\lim\limits_{h \to 0} 8a + 4h - 6a^{2} - 6ah - 2h^{2}$ Replace $h$ to $0$ $\lim\limits_{h \to 0} 8a + 4(0) - 6a^{2} - 6a(0) - 2(0)^{2} = 8a - 6a^{2}$ b. Tangent line Slope from part a: $8a - 6a^{2}$ The problem gives you the intercepts of $(1,5)$ and $(2,3)$ so we use the value of $x$ to find the tangent lines. With $(1,5)$: $m = 8a - 6a^{2}$ $m = 8(1) - 6(1)^{2}$ $m = 8 -6 = 2$ Now replace it in the formula: $y - y_{1} = m(x-x_{1})$ $y = 2(x-1) + 5$ $y = 2x + 3$ With $(2,3)$: $m = 8a - 6a^{2}$ $m = 8(2) - 6(2)^{2}$ $m = 16 - 24 = -8$ Now replace it in the formula: $y - y_{1} = m(x-x_{1})$ $y = -8(x-2) + 3$ $y = -8x + 19$
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