# Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 19

$x=\frac{\pi}{4}$ is the critical point of $f$. The maximum value is $\sqrt 2$ and the minimum value is 1.

#### Work Step by Step

$f'(x)=\cos x-\sin x$ $f'(x)$ exists for all $x$. $f'(x)=\cos x-\sin x=0\implies$ $\cos x=\sin x\implies x=\frac{\pi}{4}$ $x=\frac{\pi}{4}$ is the critical point of $f$. We compare the values of the function at the critical points and at the end points to find the maximum and minimum values. $f(\frac{\pi}{4})=\sin\frac{\pi}{4}+\cos \frac{\pi}{4}=\sqrt 2$ $f(\frac{\pi}{2})=\sin\frac{\pi}{2}+\cos\frac{\pi}{2}=1$ $f(0)=\sin 0+\cos0=1$ The maximum value is $\sqrt 2$ and the minimum value is 1.

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