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18.014 Problem Set Solutions Sam Elder November 2, 2015 Problem 1. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). Prove that there exists c ∈ (a, b) such that f (b) − f (a) a + b + . f 0 (c) + c = b−a 2 Solution. Let g(x) = f (x)+ 21 x2 . This is also continuous on [a, b] and differentiable on (a, b) since polynomials are continuous and differentiable everywhere. Then g 0 (x) = f 0 (x) + x, so by the Mean Value Theorem for Derivatives, there is some c ∈ (a, b) such that g 0 (c)(b − a) = g(b) − g(a) b2 − a2 2 f (b) − f (a) b + a f 0 (c) + c = + , b−a 2 (f 0 (c) + c)(b − a) = f (b) − f (a) + as desired. Problem 2. Find the maximum and minimum values of the function √ √ f (x) = 2 x + 1 − x on the interval [0, 1]. Answer. 1 ≤ f (x) ≤ √ 5 on [0, 1]. Solution. We know that all√extrema occur at critical points (where f 0 (x) = 0), endpoints, or anywhere where f 0 (x) is not defined. Since x is differentiable for x > 0, f 0 (x) is defined on (0, 1), so we restrict our attention 1 to the endpoints and critical points. For critical points, if f 0 (x) = √1x − 2√1−x = 0, we have 1 1 4 √ = √ =⇒ x = 4(1 − x) =⇒ x = . 5 x 2 1−x It suffices to compute f (0), f ( 45 ) and f (1). We have √ √ f (0) = 2 0 + 1 − 0 = 1 r r √ 4 4 1 5 f =2 + =√ = 5 5 5 5 5 √ √ f (1) = 2 1 + 1 − 1 = 2. Therefore, √ the minimum value of f (x) on [0, 1] is 1, which occurs at x = 0, and the maximum value of f (x) on [0, 1] is 5, which occurs at x = 45 . 1 Problem 3. Find the maximum value of the function f (x) = x + 2 1 + x2 x+1 on the interval [1, 2]. Answer. f (x) ≤ 3 on [1, 2]. Solution. Notice that f (x) is a sum of rational functions whose denominators are never zero on [1, 2]. Therefore, f (x) is differentiable everywhere, and its maximum must occur either at an endpoint or a critical point. Moreover, we compute f 0 (x) = 1 − 2 2 − ; 3 x (x + 1)2 f 00 (x) = 4 6 + . 4 x (x + 1)3 Therefore, for 1 ≤ x ≤ 2, f 00 (x) > 0 and f is convex. Hence any critical points will be minima and not 2 maxima, so the maximum must occur at an endpoint. Since f (1) = 1 + 112 + 1+1 = 3 and f (2) = 2 + 212 + 32 = 2 + 11 12 < 3, the maximum value of the function on [1, 2] is 3. Problem 4. Define f : R → R by the integral x2 Z f (x) = x 1 dt. 1 + t2 Compute the derivative f 0 (x). Solution. Define g : R → R by the integral Z x g(x) = 0 1 dt. 1 + t2 By the first part of the Fundamental Theorem of Calculus, g 0 (x) = the Chain Rule, f 0 (x) = 2xg 0 (x2 ) − g 0 (x) = 1 . Since f (x) = g(x2 ) − g(x), by 1 + x2 1 2x − . 1 + x4 1 + x2 Problem 5. Compute the integral Z 0 1 √ x3 dx. 1 + x2 √ Answer. The integral evaluates to Solution. Let f (x) = √ 2− 2 . 3 1 + x2 , and g(x) = (1 + x2 )3/2 . We compute by the Chain Rule that 1 x (2x)(1 + x2 )−1/2 = √ 2 1 + x2 p 3 3x(1 + x2 ) g 0 (x) = (2x)(1 + x2 )1/2 = 3x 1 + x2 = √ . 2 1 + x2 f 0 (x) = 2 x3 x3 = 13 g 0 (x) − f 0 (x), so 13 g(x) − f (x) is a primitive for √ . By the second part of the 1 + x2 1 + x2 Fundamental Theorem of Calculus, therefore, Therefore, √ Z 1 0 x3 √ dx = 1 + x2 √ √ 1 1 1 1 √ 2− 2 . g(1) − f (1) − g(0) − f (0) = 2 2− 2 − ·1−1 = 3 3 3 3 3 Problem 6. Let f : R → R be continuous and thrice-differentiable (so f 0 , f 00 , and f 000 all exist everywhere in R). Let c ∈ R be a point such that f 0 (c) = f 00 (c) = 0 and f 000 (c) > 0. Part 6.1. Prove that there exists δ > 0 such that f 00 (x) > 0 for c < x < c+δ and f 00 (x) < 0 for c−δ < x < c. f 00 (x) − f 00 (c) , Q(c) = f 000 (c). Because f 000 (c) exists and is positive, x−c Q(x) is continuous at c, with Q(c) > 0. By the sign-preserving property of continuous functions, there is some δ > 0 such that Q(x) > 0 for |x − c| < δ. When x > c, this implies that on c < x < c + δ, Solution. 1 Define the function Q(x) = f 00 (x) − f 00 (c) > 0 =⇒ f 00 (x) > f 00 (c) = 0, x−c as desired. When x < c, this implies that on c − δ < x < c, f 00 (x) − f 00 (c) > 0 =⇒ f 00 (x) < f 00 (c) = 0, x−c as desired. Part 6.2. Prove that f 0 (x) > 0 for 0 < |x − c| < δ, where δ is as in part (a). Solution. By Theorem 4.7a, since f 00 (x) > 0 on (c, c + δ), f 0 (x) is strictly increasing on [c, c + δ], so f 0 (x) > f 0 (c) = 0 for c < x < c + δ. By Theorem 4.7b, since f 00 (x) < 0 on (c − δ, c), f 0 (x) is strictly decreasing on [c − δ, c], so f 0 (x) > f 0 (c) = 0 for c − δ < x < c. Putting these together, if 0 < |x − c| < δ, f 0 (x) > 0. Part 6.3. Conclude that f does not have a local maximum or minimum at c. Solution. By Theorem 4.7 again, since f 0 (x) > 0 on (c, c + δ) and (c − δ, c), f is strictly increasing on both [c − δ, c] and [c, c + δ], so f is strictly increasing on [c − δ, c + δ]. In other words, for c < x < c + δ, f (x) > f (c), and for c − δ < x < c, f (x) < f (c). Therefore, f does not have a local minimum or maximum at c. 1 This solution is motivated by the proof of Theorem 4.3. 3