Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Section 13.3 - Tangent Lines and Derivatives - 13.3 Exercises - Page 922: 31

Answer

(a) $f'(a)=3a^2-2$ (b) $y=-2x+4$. $y=x+2$. $y=10x-12$ (c) See graph.

Work Step by Step

(a) Use the formula given in this section, we have: $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to 0}\frac{(a+h)^3-2(a+h)+4-a^3+2a-4}{h}=\lim_{h\to 0}\frac{ 3a^2h+3ah^2+h^3-2h}{h}=\lim_{h\to 0}(3a^2+3ah+h^2-2)=3a^2-2$$ (b) At $x=0$, we have $f(0)=4$ and $f'(0)=-2$ which is the slope of the tangent line. Knowing the slope and a point $(0,4)$ on the line, we can get the equation as $y=-2x+4$. At $x=1$, we have $f(1)=3$ and $f'(1)=1$ which is the slope of the tangent line. Knowing the slope and a point $(1,3)$ on the line, we can get the equation as $y=(x-1)+3$ or $y=x+2$. At $x=2$, we have $f(2)=2^3-4+4=8$ and $f'(2)=3\times2^2-2=10$ which is the slope of the tangent line. Knowing the slope and a point $(2,8)$ on the line, we can get the equation as $y=10(x-2)+8$ or $y=10x-12$ (c) See graph.
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