Solve this

(a) f is continuous
(c) f' is continuous

We have,

$f(x)=(x+|x|)|x|$

$=x|x|+(|x|)^{2}$

$=x|x|+x^{2}$

$f(x)= \begin{cases}2 x^{2} & x \geq 0 \\ 0 & x<0\end{cases}$

To check continuity of $f(x)$ at $x=0$

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)$

$=\lim _{x \rightarrow 0^{+}} 2 x^{2}$

$=0$

And $f(0)=0$

Here, $\mathrm{LHL}=\mathrm{RHL}=f(0)$

Therefore, $f(x)$ is continuous at $x=0$

Hence, $f(x)$ is continuous everywhere.

To check the differentiability of $f(x)$ at $x=0$

$(\mathrm{LHD}$ at $x=0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{x \rightarrow 0^{-}} \frac{0-0}{x}=0$

$(\mathrm{RHD}$ at $x=0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{x \rightarrow 0^{-}} \frac{2 x^{2}-0}{x}$

$=\lim _{x \rightarrow 0^{-}} \frac{2 x^{2}-0}{x}$

$=\lim _{x \rightarrow 0^{-}} 2 x=0$

$\mathrm{LHD}=\mathrm{RHD}$

Therefore, $f(x)$ is derivative at $x=0$

Hence, $f(x)$ is differentiable everywhere.

$f^{\prime}(x)=\left\{\begin{array}{lc}4 x & x \geq 0 \\ 0 & x<0\end{array}\right.$

To check continuity of $f^{\prime}(x)$ at $x=0$

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f^{\prime}(x)$

$=\lim _{x \rightarrow 0^{-}} 0$

$=0$

$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f^{\prime}(x)$

$=\lim _{x \rightarrow 0^{+}} 4 x$

$=0$

And $f^{\prime}(0)=0$

Here, $\mathrm{LHL}=\mathrm{RHL}=f^{\prime}(0)$

Therefore, $f^{\prime}(x)$ is continuous at $x=0$

Hence, $f^{\prime}(x)$ is continuous everywhere.

$f^{\prime \prime}(x)= \begin{cases}4 & x \geq 0 \\ 0 & x<0\end{cases}$

To check continuity of $f^{\prime \prime}(x)$ at $x=0$

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f^{\prime \prime}(x)$

$=\lim _{x \rightarrow 0^{-}} 0$

$=0$

$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f^{\prime \prime}(x)$

$=\lim _{x \rightarrow 0^{+}} 4$

$=4$

Therefore, LHL $\neq$ RHL

Therefore, $f^{\prime \prime}(x)$ is not continuous at $x=0$

Hence, $f^{\prime \prime}(x)$ is not continuous everywhere.