If the solve the problem


$f(x)=\sin x-\cos x, 0


Given : $f(x)=\sin x-\cos x$

$\Rightarrow f^{\prime}(x)=\cos x+\sin x$

For a local maximum or a local minimum, we must have


$\Rightarrow \cos x+\sin x=0$

$\Rightarrow \cos x=-\sin x$

$\Rightarrow \tan x=-1$

$\Rightarrow x=\frac{3 \pi}{4}$ or $\frac{7 \pi}{4}$

Since $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $\frac{3 \pi}{4}, x=\frac{3 \pi}{4}$ is the point of local maxima.

The local maximum value of $f(x)$ at $x=\frac{3 \pi}{4}$ is given by

$\sin \left(\frac{3 \pi}{4}\right)-\cos \left(\frac{3 \pi}{4}\right)=\sqrt{2}$

Since $f^{\prime}(x)$ changes from negative to positive when $x$ increases through $\frac{7 \pi}{4}, x=\frac{7 \pi}{4}$ is the point of local minima.

The local minimum value of $f(x)$ at $x=\frac{7 \pi}{4}$ is given by

$\sin \left(\frac{7 \pi}{4}\right)-\cos \left(\frac{7 \pi}{4}\right)=-\sqrt{2}$

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