Find the intervals in which $f(x)$ is increasing or decreasing:

Question:

Find the intervals in which $f(x)$ is increasing or decreasing:

i. $f(x)=x|x|, x \in R$

ii. $f(x)=\sin x+|\sin x|, 0 iii.$f(x)=\sin x(1+\cos x), 0

Solution:

(i): Consider the given function,

$f(x)=x|x|, x \in R$

$\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}-\mathrm{x}^{2}, \mathrm{x}<0 \\ \mathrm{x}^{2}, \mathrm{x}>0\end{array}\right.$

$\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}-2 \mathrm{x}, \mathrm{x}<0 \\ 2 \mathrm{x}, \mathrm{x}>0\end{array}\right.$

$\Rightarrow f^{\prime}(x)>0$

Therefore, $f(x)$ is an increasing function for all real values.

(ii): Consider the given function,

$f(x)=\sin x+|\sin x|, 0$\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}2 \sin \mathrm{x}, 0<\mathrm{x} \leq \pi \\ 0, \pi<\mathrm{x} \leq 2 \pi\end{array}\right.\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}2 \cos \mathrm{x}, 0<\mathrm{x} \leq \pi \\ 0, \pi<\mathrm{x} \leq 2 \pi\end{array}\right.$The function$2 \cos x$will be positive between$\left(0, \frac{\pi}{2}\right)$Hence the function$f(x)$is increasing in the interval$\left(0, \frac{\pi}{2}\right)$The function$2 \cos x$will be negative between$\left(\frac{\pi}{2}, \pi\right)$Hence the function$f(x)$is decreasing in the interval$\left(\frac{\pi}{2}, \pi\right)$The value of$f^{\prime}(x)=0$, when,$\pi

Therefore, the function $f(x)$ is neither increasing nor decreasing in the interval $(\pi, 2 \pi)$

(iii): consider the function,

$f(x)=\sin x(1+\cos x), 0$\Rightarrow f^{\prime}(x)=\cos x+\sin x(-\sin x)+\cos x(\cos x)\Rightarrow f^{\prime}(x)=\cos x-\sin ^{2} x+\cos ^{2} x\Rightarrow f^{\prime}(x)=\cos x+\left(\cos ^{2} x-1\right)+\cos ^{2} x\Rightarrow f^{\prime}(x)=\cos x+2 \cos ^{2} x-1\Rightarrow f^{\prime}(x)=(2 \cos x-1)(\cos x+1)$for$f(x)$to be increasing, we must have,$f^{\prime}(x)>0\left.\Rightarrow f^{\prime}(x)\right)=(2 \cos x-1)(\cos x+1)\Rightarrow 0

So, $f(x)$ to be decreasing, we must have,

$f^{\prime}(x)<0$

$\left.\Rightarrow f^{\prime}(x)\right)=(2 \cos x-1)(\cos x+1)$

$\Rightarrow \frac{\pi}{3}$\Rightarrow x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$So,$f(x)$is decreasing in$\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\$