Math 512A. Homework 9. Solutions

Math 512A. Homework 9. Solutions Problem 1. (i) Suppose that g(x) = f (x + c) for all x. Prove, starting from the definition of the derivative, that g 0 (x) = f 0 (x + c) for all x. (ii) Prove that if g(x) = f (cx), then g 0 (x) = c · f 0 (cx). (iii) Suppose that f is differentiable and periodic, with period a, i.e., f (x + a) = f (x) for all x. Prove that f 0 is also periodic with period a. (iv) (Not required) Prove that if f is even, i.e., f (x) = f (−x), then f 0 (x) = −f 0 (−x). (v) (Not required) Prove that if f is odd, i.e., f (−x) = −f (x), then f 0 (x) = f 0 (−x). Solution. (i) According to the definition, g 0 (x) g(x + h) − g(x) h→0 h f (x + h + c) − f (x + c) = lim h→0 h = f 0 (x + c) = lim because g(y) = f (y + h) (ii) By the definition g 0 (x) = = = = = g(x + h) − g(x) h f (cx + ch) − f (cx) lim h→0 h f (cx + ch) − f (cx) lim c h→0 ch f (cx + k) − f (cx) c lim k→0 k cf 0 (cx) lim h→0 make k = ch (iii) Follows directly from (i). Problem 2. (i) Let f (x) = x2 if x is rational, and f (x) = 0 if x is irrational. Prove that f is differentiable at 0. (ii) Let f be a function such that |f (x)| ≤ x2 for all x. Prove that f is differentiable at 0. (iii) (Not required) Let α > 1. Prove that if f satisfies |f (x)| ≤ |x|α , then f is differentiable at 0. Solution. (i) We compute f (0 + h) − f (0) = h It follows that lim h→0  h, h is rational 0, h is irrational f (0 + h) − f (0) = 0, so that f is differentiable at 0 and f 0 (0) = 0. h 1 (ii) It follows from |f (x)| ≤ x2 that −x2 ≤ f (x) ≤ x2 and f (0) = 0, so that −h ≤ f (h) − f (0) ≤h h and hence that f is differeniable at 0 and f 0 (0) = 0. Problem 3. Suppose that a and b are two consecutive roots of the polynomial function f , but that a and b are not double roots, so that we can write f (x) = (x − a)(x − b)g(x) where g(a) 6= 0 and g(b) 6= 0. (i) Prove that g(a) and g(b) have the same sign. (ii) Prove that there is some number x with a < x < b and f 0 (x) = 0. (iii) (Not required) Prove that (ii) holds true even if a and b are multiple roots. Hint: If f (x) = (x−a)n (x−b)m g(x) where g(a) 6= 0 and g(b) 6= 0, consider the polynomial function h(x) = f 0 (x)/(x − a)n−1 (x − b)m−1 . Solution. (i) If g(a) and g(b) have opposite sign, then, by the Intermediate Value Theorem, there is c in (a, b) such that g(c) = 0. Then f (c) = 0 also, contradicting that a and b are consecutive roots of f . (ii) Because a and b are roots of f , f (a) = f (b) = 0. Moreover, f is continuous on [a, b] and differentiable on (a, b) because it is a polynomial. Thus Rolle’s Theorem implies that f 0 (x) = 0 for some x in (a, b). Note that the derivative f 0 (x) = (x − a)g(x) + (x − b)g(x) + (x − a)(x − b)g 0 (x), so that f 0 (a) = (a − b)g(a) 6= 0 and f 0 (b) = (b − a)g(b) 6= 0. Problem 4. (i) If a1 < a2 < · · · < an , find the minimum value of f (x) = n X (x − ai )2 . i=1 (ii) Find the minimum value of f (x) = n X |x − ai |. i=1 (iii) (Not required) Let a > 0. Prove that the maximum value of f (x) = 1 1 + 1 + |x| 1 + |x − a| is (2 + a)/(1 + a). Solution. Done in class. Problem 5. (i) Suppose that |f (x) − f (y)| ≤ |x − y|α for some α > 1. Prove that f is constant. (ii) Find a function f other than a constant function such that |f (x) − f (y)| ≤ |x − y| Solution. (i) Becasue α > 1, this inequality implies that f 0 (x) = 0 for all x: |f 0 (x)| = lim y→0 (ii) f (x) = x/2. |f (x) − f (y)| ≤ lim |x − y|α−1 = 0. y→0 |x − y|