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 maximumat b. Taking into account the differentiability of f at c yields f ′(c)=lim x→c f (x)−f (c) x−c = lim x→c+ f (x)−f (c) x−c ≥ 0. Similarly, f (x)−f (c) x−c ≤ 0 for all x ∈(c−δ,c). It follows that f ′(c)=lim x→c f (x)−f (c) x−c = lim x→c− f (x)−f (c) x−c ≤ 0. 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 ∈Rwith 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] andxM∈[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 withc =xm or c =xM. If bothxm andxM 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 anyc ∈(a,b). □ We are now ready to use Rolle’s Theorem to prove the Mean Value Theorem presented below.
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