## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 9

#### Answer

$x=±1$

#### Work Step by Step

Using the quotient rule, we get $f'(x)=\frac{\frac{d}{dx}(x)\times(x^{2}+1)-x\frac{d}{dx}(x^{2}+1)}{(x^{2}+1)^{2}}$ $=\frac{(x^{2}+1)-2x^{2}}{(x^{2}+1)^{2}}$ $=\frac{-x^{2}+1}{(x^{2}+1)^{2}}$ If $f'(x)$ does not exist or $f'(x)=0$, this implies $x$ is a critical point of the function. But, $f'(x)$ exists for all $x$ as the denominator does not vanish ($x^{2}+1$ is never equal to 0) If $f'(x)=\frac{-x^{2}+1}{(x^{2}+1)^{2}}=0$ $\implies -x^{2}+1=0$ $\implies x^{2}=1$ Or $x= ±1$ Thus, we know that: $x=±1$ are the critical points.

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