If f : R → R is defined by f(x) = |x|^3 show that f′′(x) exists for all real x and find it.

Question

If \(f : R \to R\) is defined by \(f(x) = |x|^3\), show that \(f'' (x) \) exists for all real x and find it.

Answer 1

We have, 

\(f(x)=|x|^{3},\left\{\begin{array}{l}x^{3}, \text { if } x \geq 0 \\ (-x)^{3}=-x^{3}, \text { if } x<0\end{array}\right.\)

Now, (LHD at x = 0) 

\(=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)

\(=\lim\limits _{x \rightarrow 0^{-}}\left(\frac{-x^{3}-0}{x}\right)\)

\(=\lim\limits _{x \rightarrow 0^{-}}\left(-x^{2}\right)=0\)

(RHD at x = 0) 

\(=\lim\limits_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)

\(=\lim\limits _{x \rightarrow 0^{+}}\left(\frac{x^{3}-0}{x}\right)\)

\(=\lim \limits_{x \rightarrow 0}\left(-x^{2}\right)=0\)

\(\therefore\) (LHD of f(x) at x = 0) = (RHD of f(x) at x = 0)

So, f(x) is differentiable at x = 0 and the derivative of f(x) is given by

\(f^{\prime}(x)=\left\{\begin{array}{l}3 x^{2}, \text { if } x \geq 0 \\ -3 x^{2}, \text { if } x<0\end{array}\right.\)

Now,

(LHD of \(f'(x)\) at x = 0) \(=\lim\limits _{x \rightarrow 0^{-}} \frac{f^{\prime}(x)-f^{\prime}(0)}{x-0}=\lim\limits _{x \rightarrow 0^{-}}\left(\frac{-3 x^{2}-0}{x}\right)=\lim\limits _{x \rightarrow 0^{-}}(-3 x)=0\)

(RHD of \(f'(x)\) at x = 0) \(=\lim\limits _{x \rightarrow 0^{+}} \frac{f^{\prime}(x)-f^{\prime}(0)}{x-0}=\lim\limits _{x \rightarrow 0^{+}}\left(\frac{3 x^{2}-0}{x-0}\right)=\lim\limits_{x \rightarrow 0^{+}}(3 x)=0\)

\(\therefore\) (LHD of \(f'(x)\) at x = 0) = (RHD of \(f'(x)\) at x = 0)

So, \(f'(x)\) is differentiable at x = 0.

Hence, \(f^{\prime \prime}(x)=\left\{\begin{array}{l}6 x, \text { if } x \geq 0 \\ -6 x, \text { if } x<0 .\end{array}\right.\)