Question

Find the absolute maximum/minimum value of following function.
$f\left(x\right)=\sin x+\cos x;x$ in $[0,\pi ]$.

Answer 1

$f\left(x\right)=\sin x+\cos x[0,\pi ]$

$f^{{}^{\prime }}\left(x\right)=\cos x-\sin x{f}^{{}^{\prime }}\left(x\right)=0\Rightarrow \cos x-\sin x=0\Rightarrow \cos x=\sin x\tan x=1\Rightarrow x=\frac{\pi }{4}$

Checking the value of $f\left(x\right)$ at $x=0,\frac{\pi }{4},\pi$

$f\left(0\right)=\sin 0+\cos 0=1f\left(\pi \right)=\sin \pi +\cos \pi =-1f\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

$\therefore$ min value of $f\left(x\right)=-1$

and max value of $f\left(x\right)=\sqrt{2}$