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

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