Question

Find the local maxima and local minima for the given function and also find the local maximum and local minimum values $f\left(x\right)=\sin x-\cos x,0<x<2\pi$

Answer 1

$f\left(x\right)=\sin \left(x\right)-\cos \left(x\right)$

Dividing and multiplying by $\sqrt{2}$

$=\sqrt{2}(\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x)$

$=\sqrt{2}(\cos 45\sin x-\sin 45\cos x)$

$=\sqrt{2}\left(\sin \right(x-45\left)\right)$

$\Rightarrow f\left(x\right)=\sqrt{2}\left(\sin \right(x-45\left)\right)$

We know that

$-1\le \sin \left(x\right)\le 1$

$\Rightarrow -\sqrt{2}\le \sqrt{2}\sin \left(x\right)\le \sqrt{2}$

Here $x$ can take any value

$\Rightarrow -\sqrt{2}\le \sqrt{2}\sin (x-45)\le \sqrt{2}$

So maximum and minimum value of the function $f\left(x\right)$ is $\sqrt{2}$ and $-\sqrt{2}$