The value of $c$ in Lagrange's theorem for the function $|x|$ in the interval $[-1,1]$ is:

Lagrange's Mean Value Theorem (LMVT) states that if a function $f(x)$ is:Continuous on $[a,b]$ Differentiable on $(a,b)$ Then there exists at least one $c\in (a,b)$ such that $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$.For $f(x)=|x|$ in the interval $[-1,1]$:The function is continuous on $[-1,1]$.However, $f(x)=|x|$ is not differentiable at $x=0$ because it has a sharp corner there (the left-hand derivative is $-1$ and the right-hand derivative is $1$).Since $0\in (-1,1)$, the condition of differentiability on the open interval is not satisfied.Thus, LMVT is not applicable, and the value of $c$ is non-existent.