M408C: Rolle’s Theorem, Mean Value Theorem, &ReviewOctober 16, 2008Theorem 1 (Rolle’s Theorem). Suppose f is continuous on [a, b], differentiable on (a, b), andf(a) = f(b). ThenTheorem 2 (Mean Value Theorem). Suppose f is continuous on [a, b] and differentiable on (a, b).ThenCorrolary 1. If f0(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).1.2.3.14. (Practice Exam 2, 5) Which of the following statements about f(x) = x4− 3x2, −∞ < x < ∞is false?(A) it has loc mins @ x = ±q32(B) it has two infl pts (C) it has loc max @ x = 0(D) it has three 1st-order critical pts (E) it is concave up for x ∈−1√2,1√2Solution: f0(x) = 4x3− 6x = 2x(2x2− 3), so f0(x) = 0 if and only if x = 0, ±p3/2, which arethree first order critical points , so we can eliminate (D). Let’s write f0(x) = 2x(x+p3/2)(x−p3/2)and consider the sign of f0between our critical points:x < −r32⇒ f0< 0, −r32< x < 0 ⇒ f0> 0, 0 < x <r32⇒ f0< 0, x >r32⇒ f0> 0.Thus ±p3/2 are local mins while x = 0 is a local max. This eliminate (A) and (C). We nextconsider f00(x) = 12x2− 6 = 6(2x2− 1) which is zero at x = ±1/√2. Let’s next consider how thesign of f00changes be tween these points:x < −1√2=⇒ f00> 0, −1√2< x <1√2=⇒ f00< 0, x >1√2=⇒ f00> 0.Thus f has two inflection points, so we can eliminate (B). This only leaves (C), which we see isfalse since f00< 0 on (−1/√2, 1/√2), which means f is concave down there.5. (Practice Exam 2, 13) Let f(x) be differentiable and satisfy f0(x) ≥ 2 for all x, and supposef(0) = 0. Which of the following statements is false?(A)f(x) > 0 for all x > 0 (B) f (x) 6= 0 for all x 6= 0 (C) f(5) ≥ 10(D) f (3) ≤ 3 (E) f(b) 6= f(a) for any numbers b 6= aSolution: Let’s do process of elimination.(A) is true since f(0) = 0 and f is always increasing so it must be greater than 0 for all x > 0.(B) is true since if f(x) = 0 for some x 6= 0, we c ould apply Rolle’s theorem: Suppose first thatx > 0. then on (0, x), f is differentiable and continuous and f(0) = f (x) = 0. By Rolle’s theorem,there is a c ∈ (0, x) such that f0(c) = 0. But this cannot happen since f0(x) ≥ 2 for all x. On theother hand, supp os e that x < 0. Then we apply the same argument to (x, 0) to conclude that thereis a c ∈ (x, 0) so that f0(c) = 0, which cannot happen. Thus B must be true.(C) is maybe true or not. Let’s come back to it later.(D) is maybe true or not. Let’s come back to it later(E) is true since if f(b) = f (a) for some b 6= a, then by Rolle’s theorem, there would be ac ∈ (a, b) (suppose that a > b – if b < a, just consider the interval (b, a) and apply the sameargument) such that f0(c) = 0, which cannot happen since f0(x) ≥ 2. Thus (E) is true.This just leave (C) and (D). We know that f is increasing at a rate faster than 2 at each pointsince f0(x) ≥ 2. So by the time x is 3, f(x) must be greater than or equal to 6. Thus (D) is
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