Given as f(x) = |x| on [– 1, 1]
Therefore, f(x) can be defined as \(=
\begin{cases}
-x, & \quad \text{ } x < 0 \text{ }\\
x, & \quad \text{} x ≥ 0 \text{}
\end{cases}\)
For the differentiability at x = 0,
For RHD
Here, f(x) is not diffferentiable at x = 0
Thus, lagrange's mean value theorem is not applicable for the function f(x) = |x| on [-1,1].
