Question

Find the intervals in which the function $f$ given by
$f\left(x\right)=\sin x+\cos x,0\le x\le 2\pi$ is strictly increasing or strictly decreasing.

Answer 1

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

$f^{\prime }\left(x\right)=\cos x-\sin x$

$f^{\prime }\left(x\right)=0$

$\cos x-\sin x=0$

$\tan x=1$

$x=\frac{\pi }{4},\frac{5\pi }{4},a$ & $0\le x\le 2\pi$

The point $x=\frac{\pi }{4}$ and $\frac{5\pi }{4}$ divides, the interval $[0,2\pi ]$ into $3$ disjoint intervals.

i.e., $[(0,\frac{\pi }{4}),(\frac{\pi }{4},\frac{5\pi }{4}),(\frac{5\pi }{4},2\pi )]$

$f^{\prime }\left(x\right)>0$ if $x\in [(0,\frac{\pi }{4})\cup (\frac{5\pi }{4},2\pi )]$

or f is in the intervals.

$[(0,\frac{\pi }{4}),(\frac{5\pi }{4},2\pi )]$

Also, $f^{\prime }\left(x\right)<0$, if $x\in (\frac{\pi }{4},\frac{5\pi }{4})$

So fxn is strictly decreasing in $(\frac{\pi }{4},\frac{5\pi }{4})$.