Question

The set of points where the function f (x) = x |x| is differentiable is
(a) $\left(-\infty ,\infty \right)$
(b) $\left(-\infty ,0\right)\cup \left(0,\infty \right)$
(c) $\left(0,\infty \right)$
(d) $\left[0,\infty \right]$

Answer 1

(a) $\left(-\infty ,\infty \right)$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=x\left|x\right| \\ \Rightarrow f\left(x\right)=\left\{\begin{array}{l}\begin{array}{l}-x^2,x<0\\ 0,x=0\end{array}\\ x^2,x>0\end{array}\right. \\ \mathrm{When},x<0,\mathrm{we}\mathrm{have} \\ f\left(x\right)=-x^2\mathrm{which}\mathrm{being}\mathrm{a}\mathrm{polynomial}\mathrm{function}\mathrm{is}\mathrm{continuous}\mathrm{and}\mathrm{differentiable}\mathrm{in}\left(-\infty ,0\right) \\ \mathrm{When},x>0,\mathrm{we}\mathrm{have} \\ f\left(x\right)=x^2\mathrm{which}\mathrm{being}\mathrm{a}\mathrm{polynomial}\mathrm{function}\mathrm{is}\mathrm{continuous}\mathrm{and}\mathrm{differentiable}\mathrm{in}\left(0,\infty \right) \\ \mathrm{Thus}\mathrm{possible}\mathrm{point}\mathrm{of}\mathrm{non}-\mathrm{differentiability}\mathrm{of}f\left(x\right)\mathrm{is}x=0 \\ \mathrm{Now},\mathrm{LHD}\left(\mathrm{at}x=0\right)=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{x\to 0^{-}}{\lim }\frac{-x^2-0}{x} \\ =\underset{h\to 0}{\lim }\frac{-{\left(-h\right)}^2}{-h} \\ =\underset{h\to 0}{\lim }h \\ =0 \\ \mathrm{And}\mathrm{RHD}\left(\mathrm{at}x=0\right)=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{x\to 0^{+}}{\lim }\frac{x^2-0}{x} \\ =\underset{h\to 0}{\lim }\frac{h^2}{h} \\ =\underset{h\to 0}{\lim }h \\ =0 \\ \therefore \mathrm{LHD}\left(\mathrm{at}x=0\right)=\mathrm{RHD}\left(\mathrm{at}x=0\right) \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{also}\mathrm{differentiable}\mathrm{at}x=0 \\ \mathrm{i}.\mathrm{e}.f\left(x\right)\mathrm{is}\mathrm{differentiable}\mathrm{in}\left(-\infty ,\infty \right) \end{gathered}$$