Find the intervals in which the function f given by

Solution:

$f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$

$\therefore f^{\prime}(x)=\frac{(2+\cos x)(4 \cos x-2-\cos x+x \sin x)-(4 \sin x-2 x-x \cos x)(-\sin x)}{(2+\cos x)^{2}}$

$=\frac{(2+\cos x)(3 \cos x-2+x \sin x)+\sin x(4 \sin x-2 x-x \cos x)}{(2+\cos x)^{2}}$

$=\frac{6 \cos x-4+2 x \sin x+3 \cos ^{2} x-2 \cos x+x \sin x \cos x+4 \sin ^{2} x-2 x \sin x-x \sin x \cos x}{(2+\cos x)^{2}}$ Now, $f^{\prime}(x)=0$

$=\frac{4 \cos x-4+3 \cos ^{2} x+4 \sin ^{2} x}{(2+\cos x)^{2}}$

$=\frac{4 \cos x-4+3 \cos ^{2} x+4-4 \cos ^{2} x}{(2+\cos x)^{2}}$

$=\frac{4 \cos x-\cos ^{2} x}{(2+\cos x)^{2}}=\frac{\cos x(4-\cos x)}{(2+\cos x)^{2}}$

cos x = 0 or cos x = 4

But, cos x ≠ 4

∴cos x = 0

Now, $x=\frac{\pi}{2}$ and $x=\frac{3 \pi}{2}$ divides $(0,2 \pi)$ into three disjoint intervals i.e.,

$\left(0, \frac{\pi}{2}\right),\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$, and $\left(\frac{3 \pi}{2}, 2 \pi\right)$

Thus, $f(x)$ is increasing for $0

In the interval $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right), f^{\prime}(x)<0$.

 

Thus, $f(x)$ is decreasing for $\frac{\pi}{2}