`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)`.

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Chapter 3, 3.2 - Problem 3 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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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)]` .

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