Question

Find the intervals in which two function f given by f (x)= sinx+cosx , $o<x<2$ is
strictly increasing,
strictly deacresing.

Answer 1

We have,

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

Differentiation this with respect to x and we get,

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

When,

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

$\cos x-\sin x=0$

$\cos x=\sin x$

$\frac{\sin x}{\cos x}=1$

$\tan x=1$

$\tan x=\tan \frac{\pi }{4}=\tan \frac{5\pi }{4}$

$x=\frac{\pi }{4},\frac{5\pi }{4}as0\le x\le 2\pi$

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

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

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

Or $f\left(x\right)$ is strictly increasing in the intervals $[(0,\frac{\pi }{4})\text{and}\frac{5\pi }{4},2\pi ]$

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

Then, $f\left(x\right)$ is strictly decreasing in $(\frac{\pi }{4},\frac{5\pi }{4})$

Hence, this is the answer.