Question
Discuss the applicability of Lagrange's mean value theorem for the function:
f(x) = |x| on [−1, 1]

Answer

Here,
f(x) = |x| on [−1, 1]
$\text{f}(\text{x})=\begin{cases}-\text{x},&\text{x}<0\\\text{x},&\text{x}\geq0\end{cases}$
For differentiability at x = 0
$\text{LHD}=\lim_\limits{\text{x}\rightarrow0^-}\frac{\text{f}(0-\text{h})-\text{f}(0)}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-(0-\text{h})-0}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{-\text{h}}$
$\text{LHD}=-1$
$\text{RHD}=\lim_\limits{\text{x}\rightarrow0^+}\frac{\text{f}(0+\text{h})-\text{f}(0)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(0+\text{h})-0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}}$
$=1$
$\text{LHD}\neq\text{RHD}$
⇒ f(x) is not differentiable at $\text{x}=0\in(-1,1)$
Hence, Lagrange's mean value theorem is verified.

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