## Calculus 8th Edition

Published by Cengage

# Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 59

#### Answer

$(-3,0)$

#### Work Step by Step

Given $$y = [\ln(x+4)]^2$$ Since \begin{align*} y'&= 2[\ln(x+4)](x+4)'\\ &=\frac{2\ln(x+4)}{x+4} \end{align*} The tangent is horizontal when $y'=0$ \begin{align*} y'&=0\\ \frac{2\ln(x+4)}{x+4}&=0\\ \ln(x+4)&=0\\ x+4&=1\\ x&=-3 \end{align*} and $y=0$, then at $(-3,0)$ the tangent is horizontal

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