## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 118: 13

#### Answer

The equation of the tangent line is $y=-2x+9$.

#### Work Step by Step

The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$g(x)=\frac{x}{x-2}\hspace{1cm}A(3,3)$$ 1) Find the slope $m$ of the tangent: $$m=\lim_{h\to0}\frac{g(a+h)-g(a)}{h}$$ Here $a=3$ and $g(a)=b=3$ $$m=\lim_{h\to0}\frac{g(3+h)-3}{h}=\lim_{h\to0}\frac{\frac{3+h}{3+h-2}-3}{h}=\lim_{h\to0}\frac{\frac{h+3}{h+1}-3}{h}$$ $$m=\lim_{h\to0}\frac{\frac{h+3-3h-3}{h+1}}{h}=\lim_{h\to0}\frac{-2h}{h(h+1)}=\lim_{h\to0}\frac{-2}{h+1}$$ $$m=\frac{-2}{0+1}=-2$$ 2) Find the equation of the tangent line at $A(3,3)$: The tangent line would have this form: $$y=-2x+m$$ Substitute $A(3,3)$ here to find $m$: $$-2\times3+m=3$$ $$-6+m=3$$ $$m=9$$ So the equation of the tangent line is $y=-2x+9$.

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