Question

find the intervals in which f(x)=sin3x-cos3x is strictly increasing or decreasing. xgreater than 0 but less than pie​

Answer 1

Answer:

f(x) is increasing in $(0,\frac{\pi}{2})$

and decreasing in $(\frac{\pi}{2},\pi)$

Step-by-step explanation:

Given : $f(x)=\sin 3x-\cos 3x$

To find : The intervals in which  $f(x)=\sin 3x-\cos 3x$ is strictly increasing or decreasing.

Solution :

$f(x)=\sin 3x-\cos 3x$

Differentiate w.r.t x

$f'(x)=3\cos 3x+3\sin 3x$

Interval $0<x<\pi$

Cosine function is positive in the 1st quadrant and negative in 2nd quadrant

and sin function is positive in both quadrant.

When $0<x<\frac{\pi}{2}$

$f'(x)>0$

Hence, it is increasing

When $\frac{\pi}{2}<x<\pi$

$f'(x)<0$

Hence, it is decreasing

Therefore, f(x) is increasing in $(0,\frac{\pi}{2})$

and decreasing in $(\frac{\pi}{2},\pi)$