## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 6

#### Answer

The equation of the tangent line $l$ at point $A(2,3)$ is $$(l):y=9x-15$$

#### Work Step by Step

$$y=f(x)=x^3-3x+1$$ Given point $A(2,3)$ According to definition, the slope of the tangent line $l$ at point $A(2, 3)$ is $$m_l=\lim\limits_{x\to2}\frac{f(x)-f(2)}{x-2}$$$$m_l=\lim\limits_{x\to2}\frac{(x^3-3x+1)-3}{x-2}$$$$m_l=\lim\limits_{x\to2}\frac{x^3-3x-2}{x-2}$$$$m_l=\lim\limits_{x\to2}\frac{(x-2)(x+1)^2}{x-2}$$$$m_l=\lim\limits_{x\to2}(x+1)^2$$$$m_l=(2+1)^2=9$$ Therefore, the equation of the tangent line $l$ would have the following form: $$(l): y=9x+b$$ Since $l$ passes through point $A(2,3)$, we have $$9\times2+b=3$$$$18+b=3$$$$b=-15$$ In conclusion, the equation of the tangent line $l$ at point $A(2,3)$ is $$(l):y=9x-15$$

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