## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises - Page 149: 27

#### Answer

$f'(1)=3$ The equation of the tangent line is $$(l):y=3x-1$$

#### Work Step by Step

$$f(x)=3x^2-x^3$$ 1) The derivative of $f(x)$ at a number $a$ is calculated as follows: $$f'(a)=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{(3x^2-x^3)-(3a^2-a^3)}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{3(x^2-a^2)-(x^3-a^3)}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{3(x-a)(x+a)-(x-a)(x^2+a^2+ax)}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{(x-a)[3(x+a)-x^2-a^2-ax]}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}[3(x+a)-x^2-a^2-ax]$$ $$f'(a)=3(a+a)-a^2-a^2-a\times a$$ $$f'(a)=6a-3a^2$$ So $f'(1)=6\times1-3\times1^2=3$ 2) The equation of the tangent line $l$ of the curve $y$ at point $(1,2)$ would have the following form $$(l):y=f'(1)x+b$$$$(l):y=3x+b$$ Since point $(1,2)$ lies in the tangent line $l$, we can find $b$ by applying the point into the equation of the tangent line $l$: $$3\times1+b=2$$$$3+b=2$$$$b=-1$$ Overall, the equation of tangent line $l$ is $$(l):y=3x-1$$

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