Question

Let $f\left(x\right)=\left(x+\left|x\right|\right)\left|x\right|$. Then, for all x
(a) f is continuous
(b) f is differentiable for some x
(c) f' is continuous
(d) f'' is continuous

Answer 1

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

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=\left(x+\left|x\right|\right)\left|x\right| \\ =x\left|x\right|+{\left(\left|x\right|\right)}^2 \\ =x\left|x\right|+x^2 \\ f\left(x\right)=\left\{\begin{array}{l}2x^{2}x\ge 0\\ 0x<0\end{array}\right. \\ \mathrm{To}\mathrm{check}\mathrm{continuity}\mathrm{of}f\left(x\right)\mathrm{at}x=0 \\ \left(\mathrm{LHL}\mathrm{at}x=0\right)=\underset{x\to 0^{-}}{\lim }f\left(x\right) \\ =\underset{x\to 0^{-}}{\lim }0 \\ =0 \\ \left(\mathrm{RHL}\mathrm{at}x=0\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right) \\ =\underset{x\to 0^{+}}{\lim }2x^2 \\ =0 \\ \mathrm{And}f\left(0\right)=0 \\ \mathrm{Here},\mathrm{LHL}=\mathrm{RHL}=f\left(0\right) \\ \mathrm{Therefore},f\left(x\right)\mathrm{is}\mathrm{continuous}\mathrm{at}x=0 \\ \mathrm{Hence},f\left(x\right)\mathrm{is}\mathrm{continuous}\mathrm{everywhere}. \end{gathered}$$

$$\begin{gathered} \mathrm{To}\mathrm{check}\mathrm{the}\mathrm{differentiability}\mathrm{of}f\left(x\right)\mathrm{at}x=0 \\ \left(\mathrm{LHD}\mathrm{at}x=0\right)=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{x\to 0^{-}}{\lim }\frac{0-0}{x}=0 \\ \left(\mathrm{RHD}\mathrm{at}x=0\right)=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{x\to 0^{-}}{\lim }\frac{2x^2-0}{x} \\ =\underset{x\to 0^{-}}{\lim }\frac{2x^2-0}{x} \\ =\underset{x\to 0^{-}}{\lim }2x=0 \\ \mathrm{LHD}=\mathrm{RHD} \\ \mathrm{Therefore},f\left(x\right)\mathrm{is}\mathrm{derivative}\mathrm{at}x=0 \\ \mathrm{Hence},\mathit{}f\left(x\right)\mathrm{is}\mathrm{differentiable}\mathrm{everywhere}. \end{gathered}$$

$$\begin{gathered} f\text{'}\left(x\right)=\left\{\begin{array}{l}4xx\ge 0\\ 0x<0\end{array}\right. \\ \mathrm{To}\mathrm{check}\mathrm{continuity}\mathrm{of}f\text{'}\left(x\right)\mathrm{at}x=0 \\ \left(\mathrm{LHL}\mathrm{at}x=0\right)=\underset{x\to 0^{-}}{\lim }f\text{'}\left(x\right) \\ =\underset{x\to 0^{-}}{\lim }0 \\ =0 \\ \left(\mathrm{RHL}\mathrm{at}x=0\right)=\underset{x\to 0^{+}}{\lim }f\text{'}\left(x\right) \\ =\underset{x\to 0^{+}}{\lim }4x \\ =0 \\ \mathrm{And}f\text{'}\left(0\right)=0 \\ \mathrm{Here},\mathrm{LHL}=\mathrm{RHL}=f\text{'}\left(0\right) \\ \mathrm{Therefore},f\text{'}\left(x\right)\mathrm{is}\mathrm{continuous}\mathrm{at}x=0 \\ \mathrm{Hence},f\text{'}\left(x\right)\mathrm{is}\mathrm{continuous}\mathrm{everywhere}. \end{gathered}$$
$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\left\{\begin{array}{l}4x\ge 0\\ 0x<0\end{array}\right. \\ \mathrm{To}\mathrm{check}\mathrm{continuity}\mathrm{of}f\text{'}\text{'}\left(x\right)\mathrm{at}x=0 \\ \left(\mathrm{LHL}\mathrm{at}x=0\right)=\underset{x\to 0^{-}}{\lim }f\text{'}\text{'}\left(x\right) \\ =\underset{x\to 0^{-}}{\lim }0 \\ =0 \\ \left(\mathrm{RHL}\mathrm{at}x=0\right)=\underset{x\to 0^{+}}{\lim }f\text{'}\text{'}\left(x\right) \\ =\underset{x\to 0^{+}}{\lim }4 \\ =4 \\ \mathrm{Therefore},\mathrm{LHL}\ne \mathrm{RHL} \\ \mathrm{Therefore},f\text{'}\text{'}\left(x\right)\mathrm{is}\mathrm{not}\mathrm{continuous}\mathrm{at}x=0 \\ \mathrm{Hence},f\text{'}\text{'}\left(x\right)\mathrm{is}\mathrm{not}\mathrm{continuous}\mathrm{everywhere}. \end{gathered}$$