96 4.2 The Mean Value Theorem Figure 4.1: Illustration of Fermat’s Rule. The function f has a local minimum at a and a local maximum at b. Taking into account the differentiability of f at c yields f (x) − f (c) f (x) − f (c) f ′ (c)= lim = lim ≥ 0. x→c x − c x→c+ x − c Similarly, f (x) − f (c) ≤ 0 for all x ∈ (c − δ ,c). x − c It follows that f (x) − f (c) f (x) − f (c) f ′ (c)= lim = lim ≤ 0. x→c x − c x→c− x − c Therefore, f ′ (c)= 0. The proof is similar for the case where f has a local maximum at c. □ Theorem 4.2.2 — Rolle’s Theorem. Let a,b ∈ R with a < b and f : [a,b] → R. Suppose f is continuous on [a,b] and differentiable on (a,b) with f (a)= f (b). Then there exists c ∈ (a,b) such that f ′ (c)= 0. (4.1) Proof: Since f is continuous on [a,b], by the extreme value theorem (Theorem 3.4.2) there exist xm ∈ [a,b] and xM ∈ [a,b] such that f (xm)= min{ f (x) : x ∈ [a,b]} and f (xM)= max{ f (x) : x ∈ [a,b]}. Then f (xm) ≤ f (x) ≤ f (xM) for all x ∈ [a,b]. (4.2) If xm ∈ (a,b) or xM ∈ (a,b), then f has a local minimum at xm or f has a local maximum at xM. By Theorem 4.2.1, f ′ (xm)= 0 or f ′ (xM)= 0, and (4.1) holds with c = xm or c = xM. If both xm and xM are the endpoints of [a,b], then f (xm)= f (xM) because f (a)= f (b). By (4.2), f is a constant function, so f ′ (c)= 0 for any c ∈ (a,b). □ We are now ready to use Rolle’s Theorem to prove the Mean Value Theorem presented below.
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