`f(x) = 1 - |x - 1|, [0,2]` Explain why Rolle’s Theorem does not apply to the function even though there exist `a` and `b` such that `f(a) = f(b)`.
This function satisfies some conditions of Rolle's Theorem: it is continuous on [0, 2], differentiable almost everywhere on (0, 2) and f(0) = f(2) (=0).
But there is one point where f isn't differentiable: x=1. From the left (x->1-0)
`lim[(f(x) - f(1))/(x-1)] = lim[(x-1)/(x-1)] = 1,`
while from the right (x->1+0)
`lim[(f(x) - f(1))/(x-1)] = lim[(1-x)/(x-1)] = -1.`
So there is no `f'(1) = lim_(x-gt1)[(f(x) - f(1))/(x-1)]` .