# State and proove Rolle’s Theorem.

(Rolle’s Theorem) Suppose f (x) is continuous on *a, b+ and dif- ferentiable on (a, b). If f (a) = 0 = f (b), then there exists a point c ∈ (a, b) such that f 0 (c) = 0.

Proof. Suppose f (x) is continuous on [a, b], differentiable on (a, b) and f (a) =

0 = f (b). We will prove the theorem using two cases. First, suppose that f (x) > 0 for some x ∈ (a, b). Since f (x) is continuous on [a, b], there exists a point c ∈ [a, b] for which f (c) is the maximum value of f on [a, b]. Furthermore, f (c) > 0 implies c = a and c = b, so c ∈ (a, b) and so f 0 (c) = 0 because f (x) is differentiable on (a, b).

Now suppose f (x) ≤ 0 for all x ∈ (a, b). Then either f (x) = 0 for all x ∈ (a, b) in which case f 0 (x) = 0 for all x ∈ (a, b), or else f (x) < 0 for some x ∈ (a, b). Since f (x) is continuous on [a, b], we know that there is a point c ∈ [a, b] for which f (c) is the minimum value of f (x) on [a, b]. Since f (c) is the minimum on [a, b] and f (x) < 0 for some x ∈ (a, b), f (c) < 0. Consequently, c = a and c = b, so c ∈ (a, b) and therefore f 0 (c) = 0 because f (x) is differentiable on (a, b). This proves the theorem.

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