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Answer:  Not applicable

Hint: Using Lagrange’s mean theorem and limits.

Given: $f(x)=|x| \text { on }[-1,1]$

Solution:

Lagrange’s mean value theorem states if a function$f(x)$ is continuous on a closed interval    [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that

Differentiability at x = 0:

LHD =

\begin{aligned} &\lim _{x \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{-h} \\ &\lim _{x \rightarrow 0^{-}} \frac{-(0-h)-0}{-h} \end{aligned}

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

\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{h}{-h} \\ =&-1 \end{aligned}

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

\begin{aligned} &=\lim _{x \rightarrow 0^{-}} \frac{(0-h)-0}{-h} \\ &=\lim _{x \rightarrow 0^{-}} \frac{-h-0}{-h} \\ &=\lim _{x \rightarrow 0^{-}} \frac{-h}{-h} \\ &=1 \end{aligned}

$\mathrm{LHS} \neq \mathrm{RHS}$

$\therefore f(x)$ is not differentiable at x = 0

$\therefore$Lagrange’s mean value theorem is not applicable for function f (x) = |x| on [-1, 1]

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