## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Review - Exercises: 36

#### Answer

The equation of the tangent line to $f$ at the point with - $x=0$: $$(l_1):y=6x+2$$ - $x=-1$: $$(l_2):y=\frac{3}{8}x+\frac{7}{8}$$

#### Work Step by Step

$$y=f(x)=\frac{2}{1-3x}$$ First, we look for the slope of the tangent line to the curve $f$ at a number a $(m_a)$: $$m_a=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$ $$m_a=\lim\limits_{x\to a}\frac{\frac{2}{1-3x}-\frac{2}{1-3a}}{x-a}$$ $$m_a=\lim\limits_{x\to a}\frac{2(1-3a)-2(1-3x)}{(1-3x)(1-3a)(x-a)}$$ $$m_a=\lim\limits_{x\to a}\frac{2-6a-2+6x}{(1-3x)(1-3a)(x-a)}$$ $$m_a=\lim\limits_{x\to a}\frac{6x-6a}{(1-3x)(1-3a)(x-a)}$$ $$m_a=\lim\limits_{x\to a}\frac{6}{(1-3x)(1-3a)}$$ $$m_a=\frac{6}{(1-3a)(1-3a)}$$ $$m_a=\frac{6}{(1-3a)^2}$$ The equation of the tangent line to the curve $f$ at a number a would have the following form: $$(l): y=m_ax+b$$ 1) At $x=a=0$ Then $$m_a=m(0)=\frac{6}{(1-3\times0)^2}=6$$ And $$f(a)=f(0)=\frac{2}{1-3\times0}=2$$ Now we find $b$ by applying the above found variables to the equation of $l$: $$2=6\times0+b$$$$b=2$$ Therefore, the equation of the tangent line to $f$ at the point with $x=0$ is $$(l_1):y=6x+2$$ 2) At $x=a=-1$ Then $$m_a=m(-1)=\frac{6}{[1-3\times(-1)]^2}=\frac{3}{8}$$ And $$f(a)=f(-1)=\frac{2}{1-3\times(-1)}=\frac{1}{2}$$ Now we find $b$ by applying the above found variables to the equation of $l$: $$\frac{1}{2}=\frac{3}{8}\times(-1)+b$$$$b=\frac{7}{8}$$ Therefore, the equation of the tangent line to $f$ at the point with $x=-1$ is $$(l_2):y=\frac{3}{8}x+\frac{7}{8}$$

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