Rolle's Theorem states as follows:Verify rolle's theorem for |x| in [-1 1]
I will assume that the exercise actually told you to work with something like "\(\displaystyle f(x)\, =\, |\,x\,|\)" on the given interval.Suppose \(\displaystyle f(x)\) is a function that satisfies all of the following:
Then there is a number \(\displaystyle \,c\,\) such that \(\displaystyle \,a\, <\, c\, <\, b\,\) and \(\displaystyle \,f'(c)\, =\, 0.\) Or, in other words \(\displaystyle \,f(x)\,\) has a critical point in \(\displaystyle \,(a,\, b)\).
- \(\displaystyle f(x)\) is continuous on the closed interval \(\displaystyle \,[a,\, b] \)
- \(\displaystyle f(x)\) is differentiable on the open interval \(\displaystyle \,(a,\, b)\)
- \(\displaystyle f(a)\, =\, f(b)\)
.The graph of mod(x) has a sharp turn at x=0 which means we can not draw a unique tangent at that point. This makes it non- differentiable at that point.
Aliter : the differential of mod(x) = mod (x) / x , wherever it is defined. This can be proved from first principles. This differential is undefined at x=0.
Hence, Rolle's theorem can not be applied.
But \(\displaystyle \dfrac{x}{|x|} \equiv \dfrac{|x|}{x}\)If f(x) = mod(x) is supposed to be the absolute value function, then f'(x) does not equal (mod(x))/x. It's the "flip" of it.
If f(x) = |x|, \(\displaystyle \ \ \) then f'(x) = x/|x|