Question

Find the local maxima and local minima, of the given functions. Also find the local maximum and local minimum values:

$h\left(x\right)=\sin x+\cos x,0<x<\frac{\pi }{2}$

Answer 1

$h\left(x\right)=\sin \left(x\right)+\cos \left(x\right)$
Dividing and multiplying by $\sqrt{2}$
$=\sqrt{2}(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)$
$=\sqrt{2}(\cos 45\sin x+\sin 45\cos x)$
$=\sqrt{2}\left(\sin \right(x+45\left)\right)$
$\Rightarrow h\left(x\right)=\sqrt{2}\left(\sin \right(x+45\left)\right)$
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 $h\left(x\right)$ is $\sqrt{2}$ and $-\sqrt{2}$
But since domain has been constrained to $0<x<\frac{\pi }{2}$, therefor $h\left(x\right)$ will attain a minimum value of $1$ at $x=0$ and $x=\frac{\pi }{2}$