Week 5. The Mean Value Theorem

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In this lecture, we introduce Rolle’s Theorem and the Mean Value Theorem. We then look at the relation between the sign of the derivative and the monotonicity of a function. 1. Rolle’s Theorem Rolle’s Theorem is a simple but important theorem about real functions. Theorem 9.1 (Rolle’s Theorem). Suppose that the function f is continuous on [a, b] and differentiable on (a, b), and that f (a) = f (b). Then there is a point c ∈ (a, b) such that f ′ ( c) = 0 . Proof. Since f is continuous on [a, b], by the Maximum-Minimum Theorem, f is bounded and attains its bounds. We write M for the maximum and m for the minimum. Since m ≤ f (a) = f (b) ≤ M by definition, then it must be true that f (a) = f (b) < M , or m < f (a) = f (b), or m = f (a) = f (b) = M . Consider first the possibility that M > f (a). In this case, there is a point c ∈ (a, b) such that f (c) = M , and this is a local maximum. Now f is differentiable in (a, b), so f ′ (c) = 0, by Theorem 8.11. Consider next the possibility that m < f (a). In this case, a similar argument shows that there is a point c ∈ (a, b) such that f (c) = m, and f ′ (c) = 0. The last possibility is that m = M . In this case, the function f is constant, and the derivative of a constant function is 0, so f ′ (c) = 0 for all c ∈ (a, b). Thus in each case, there is a point c with the required property. 2 Corollary 9.2. If f : (a, b) → R is differentiable, then there is a zero of f ′ between any two different zeros of f . Proof. If x1 and x2 are two distinct zeros of f , and x1 < x2 , then f (x1 ) = f (x2 ), so, by Rolle’s Theorem, there exists c in the interval (x1 , x2 ) such that f ′ (c) = 0, that is, c is a root of f ′ lying between x1 and x2 . 2 Corollary 9.3. If f : (a, b) → R is differentiable and has n distinct zeros, then f ′ has at least n − 1 distinct zeros. Equivalently, if f ′ has m distinct zeros, then f has at most m + 1 distinct zeros.
2. The Mean Value Theorem The Mean Value Theorem is an important consequence and generalization of Rolle’s Theorem. Theorem 9.5 (Mean Value Theorem). Suppose that the real function f is continuous on [a, b] and differentiable on (a, b). Then there exists c ∈ (a, b) such that f ′ ( c) = f (b) − f (a) . b−a f (b) − f (a) . b−a
Proof. Let L be the linear function such that L(a) = f (a) and L(b) = f (b). Then L′ (x) = Define ϕ : [a, b] → R by
ϕ(x) = f (x) − L(x).
Now ϕ is continuous on [a, b] and differentiable on (a, b), and ϕ(a) = ϕ(b) = 0 by definition. By Rolle’s Theorem, there exists c ∈ (a, b) such that ϕ′ (c) = 0. Hence 0 = f ′ (c) − L′ (c) = f ′ (c) − which implies the desired conclusion. f (b) − f (a) , b−a 2
p′ has exactly one real root, which must be negative. It follows that p has at most 2 real roots. Since p(−1) = −1 and limx→+∞ p(x) = limx→−∞ p(x) = +∞, p has at least two real △ roots. Thus p has exactly 2 real roots.
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9. THE MEAN VALUE THEOREM
In particular, if f is a nonzero real polynomial of degree n, then f has at most n zeros. Exercise 9.4. Define the real polynomial p by p(x) = x4 − x2 + 2x + 1. How many real roots does p have? First of all, p′ (x) = 4x3 − 2x + 2 and p′′ (x) = 12x2 − 2. So p′′ (x) = 0 when Answer. √ x = ±1/ 6. Thus p′ is a cubic with two turning points. Since √ √ 4 −2 ) 6 + 2 ≈ 1.456 p′ (1/ 6) = ( 36 6 and √ √ 4 −2 ) 6 + 2 ≈ 2.544, p′ (−1/ 6) = −( 36 6
3. APPLICATIONS OF THE MEAN VALUE THEOREM
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3. Applications of the Mean Value Theorem While the Mean Value Theorem guarantees the existence of a point c ∈ (a, b) such that f (b) − f (a) = f ′ ( c) , b−a it does not tell us anything about how to find c. For specific functions, we may compute c, but in general we cannot. Nevertheless, the Mean Value Theorem is useful. Exercise 9.6. Suppose that a, b ∈ [−π/4, π/4] and a = b. Show that |tan b − tan a| < 2 |b − a| . Answer. Suppose that a < b. The function x → tan x is continuous and differentiable on (−π/2, π/2), and hence continuous on [a, b] and differentiable on (a, b), because [a, b] and (a, b) are subsets of (−π/2, π/2). Hence by the Mean Value Theorem, there exists c ∈ (a, b) such that tan b − tan a = sec2 c, b−a and this implies that |tan b − tan a| = (sec2 c) |b − a| . Now c ∈ (a, b) ⊆ (−π/4, π/4), so cos2 c > 1/2 and sec2 c < 2; we deduce that The same argument holds if b < a, except that [a, b] and (a, b) must be replaced by [b, a] and (b, a). △ Exercise 9.7. Show that Answer. Define f (x) = ex − 1 − x. We need to show that f (x) ≥ 0 for all x ∈ R. Note that f ′ (x) = ex − 1, so f ′ (x) > 0 when x > 0 and f ′ (x) < 0 when x < 0. If x > 0, then we consider f in the interval [0, x]. By the Mean Value Theorem, there exists c ∈ (0, x) such that f (x) − f (0) = f ′ ( c) > 0 . x−0 It follows that f (x) > 0. If x < 0, then we consider f in the interval [x, 0]. By the Mean Value Theorem, there exists c ∈ (x, 0) such that f (x) − f (0) = f ′ ( c) < 0 . x−0 It follows that f (x) > 0. △ Thus f (x) ≥ 0 for all x; equality holds only when x = 0. The Mean Value Theorem is very useful for proving results of this type. 1 + x ≤ ex ∀x ∈ R. |tan b − tan a| < 2 |b − a| .