Mean Value Theorem MATH 161 Calculus I J. Robert Buchanan Department of Mathematics Summer 2018
Background: Corollary to the Intermediate Value Theorem Corollary Suppose f is continuous on the closed interval [a, b] and f (a) and f (b) have opposite signs (in other words f (a)f (b) < 0). Then there is at least one number c (a, b) for which f (c) = 0. f(b) y y=f(x) a f(c)=0 b x f(a)
Preview The Mean Value Theorem (MVT) is of fundamental importance in calculus. From the MVT many important other concepts will be derived. We begin with an elementary version of the MVT called Rolle s Theorem.
Rolle s Theorem Theorem (Rolle s Theorem) Suppose f is continuous on interval [a, b] and differentiable on interval (a, b) and f (a) = f (b). Then there is a number c (a, b) such that f (c) = 0. y f'(c)=0 f(a)=f(b) y=f(x) a c b x
Example Suppose f (x) = 1 + sin x on the interval [ π, π]. Find the number c referred to in Rolle s Theorem. We note that f ( π) = 1 = f (π) and that f (x) is continuous on [ π, π] and differentiable on ( π, π). Thus according to Rolle s Theorem there exists π < c < π for which f (c) = 0 cos c = 0 c = π 2. The number c = π/2 would also solve the equation.
Consequences of Rolle s Theorem (1 of 2) Theorem If f is continuous on [a, b], differentiable on (a, b) and f (x) = 0 has two solutions in [a, b], then f (x) = 0 has at least one solution in (a, b). Proof. Let the two solutions of f (x) = 0 be x = s and x = t where s < t. Since f (s) = f (t) = 0 then by Rolle s Theorem f (c) = 0 for some s < c < t.
Consequences of Rolle s Theorem (2 of 2) Theorem For any integer n > 0, if f is continuous on [a, b], differentiable on (a, b) and f (x) = 0 has n solutions in [a, b], then f (x) = 0 has at least (n 1) solutions in (a, b). Proof. Let the n solutions of f (x) = 0 be {x 1, x 2,..., x n } with a x 1 < x 2 < < x n b. There are n 1 pairs of adjacent solutions x i and x i+1. Since f (x i ) = f (x i+1 ) = 0 then by Rolle s Theorem f (c i ) = 0 for some x i < c i < x i+1.
Example Prove that x 3 + 4x 1 = 0 has just one real number solution. Hint: show that it has at least one solution, then suppose the equation has two solutions and reach a contradiction.
Solution Let f (x) = x 3 + 4x 1. Since f is a polynomial, it is continuous and differentiable at all real numbers. Since f (0) = 1 < 0 < 4 = f (1) then according to the Intermediate Value Theorem (IVT) there is at least one solution to the equation f (x) = 0 in the interval (0, 1). Suppose there are two solutions, in other words there exist numbers x 1 < x 2 such that f (x 1 ) = 0 = f (x 2 ). By Rolle s Theorem there exists a number x 1 < c < x 2 for which f (c) = 0 3c 2 + 4 = 0 Since there are no real number solutions to this equation, then this contradicts the assumption that f (x) = 0 has two different solutions. Therefore it can have only one solution.
Mean Value Theorem Theorem (Mean Value Theorem) Suppose f is continuous on [a, b] and differentiable on (a, b). There exists a number c (a, b) such that f (c) = f (b) f (a). b a Remark: the MVT says the slope of the tangent line somewhere in (a, b) equals the slope of the secant line across [a, b].
Proof f (b) f (a) Slope of secant line: m sec = b a Equation of secant line: y = m sec (x a) + f (a) Define s(x) = f (x) (m sec (x a) + f (a)). Note that s(x) is continuous on [a, b] and differentiable on (a, b) and that s(a) = 0 = s(b). Hence by Rolle s Theorem s (c) = 0 for some a < c < b. s (x) = f (x) m sec 0 = f f (b) f (a) (c) b a f (c) = f (b) f (a) b a
Visualization of the Mean Value Theorem y y=f'(c)(x-c)+f(c) f(a) f(b) y=m sec (x-a)+f(a) y=f(x) a c b x
Example Consider the function f (x) = x 3 + x 2 + 1 on the interval [ 1, 1]. Find the value of c guaranteed by the Mean Value Theorem such that f (c) = f (b) f (a). b a 3c 2 + 2c = f (1) f ( 1) 1 ( 1) 3c 2 + 2c = 3 1 2 3c 2 + 2c = 1 3c 2 + 2c 1 = 0 (3c 1)(c + 1) = 0 c = 1 3
Illustration f (x) = x 3 + x 2 + 1 3.0 y 2.5 2.0 1.5 1.0 0.5-1.0-0.5 0.5 1.0 x
Constants and the MVT Recall: if c is a constant then d dx [c] = 0. Question: are there any non-constant, differentiable functions such that f (x) = 0 on an interval? Theorem Suppose f (x) = 0 for all x in an open interval (a, b), then f (x) is constant on (a, b). Proof. Let s and t be any two numbers in the interval (a, b) with s < t. According to the MVT, f (s) f (t) s t = f (c) (for some s < c < t) f (s) f (t) = 0 f (s) = f (t)
Derivatives that Agree Note: If f (x) = x 3 + 2x + 1 and g(x) = x 3 + 2x then f (x) = g (x) for all x even though f (x) g(x) for all x. Question: suppose two functions have the same derivative, what is the relationship between the functions? Corollary Suppose f (x) and g(x) are functions such that f (x) = g (x) for all x in some open interval I, then there exists a constant C such that f (x) = g(x) + C for all x in I.
Example Find all the functions f (x) such that f (x) = 9x 4 + 3x + 1. Remark: this process is known as antidifferentiation. We need to raise each power of x by 1 and divide the coefficient by the new power. f (x) = 9 4 + 1 x 4+1 + 3 1 + 1 x 1+1 + x + C = 9 5 x 5 + 3 2 x 2 + x + C where C is an arbitrary constant.
Inequalities and the MVT Use the MVT to prove the inequality cos u cos v u v. Let f (x) = cos x, then by the MVT f (u) f (v) u v = f (c) (c between u and v) cos u cos v = sin c u v cos u cos v u v = sin c 1 cos u cos v u v 1 cos u cos v u v.
Increasing and Decreasing Functions Definition Function f (x) is increasing on the interval I if for all a and b in I with a < b then f (a) < f (b). Function f (x) is decreasing on the interval I if for all a and b in I with a < b then f (a) > f (b). Theorem If f (x) > 0 for all x in interval open I, then f (x) is increasing on interval I. If f (x) < 0 for all x in interval open I, then f (x) is decreasing on interval I.
Proof Suppose f (x) > 0 for all x in interval I. Let a and b be any two numbers in interval I with a < b. By the MVT, f (b) f (a) b a = f (c) (for some a < c < b) f (b) f (a) b a > 0 (since f (x) > 0 for all x in interval I) f (b) f (a) > 0 (since b a > 0) f (b) > f (a). Since a and b were arbitrary, then f (x) is increasing on interval I.
Examples Determine whether the following functions are increasing, decreasing, or neither on the interval (, ). f (x) = x 5 + 3x 3 + x 1 Increasing since f (x) = 5x 4 + 9x 2 + 1 > 0 for all x. f (x) = x 4 + 2x 2 + 1 Neither since f (x) = 4x 3 + 4x = 4x(x 2 + 1). However, f (x) is increasing for x > 0 since f (x) > 0 for x > 0 and f (x) is decreasing for x < 0 since f (x) < 0 for x < 0. f (x) = e x Decreasing since f (x) = e x < 0 for all x. f (x) = cos x Neither since f (x) = sin x which takes on both positive and negative values.
Homework Read Section 2.10 Exercises: 1 45 odd