Math 414 Professor Lieberman April 7, 2003 HOMEWORK #10 SOLUTIONS Section 4.4 2. (e) FALSE. The function f (x) = x is uniformly continuous on D = [0, ∞), but it is not bounded on D. 7. (e) For n ∈ N, set xn = 2 , (2n + 1)π yn = 2 . (2n − 1)π Then |xn − yn | = 4 , π(4n2 − 1) and 2 2 8n = |f (xn ) − f (yn )| = + , 2 (2n + 1)π (2n − 1)π (4n − 1)π so |f (xn ) − f (yn )| = 2n|xn − yn |. Therefore f is NOT Lipschitzian. 10. For all examples, we take D = R. (a) The function f (x) = x + 1 is Lipschitzian with Lipschitz constant L = 1 and has no fixed points in D. (b) The function f (x) = x is Lipschitzian with Lipschitz constant L = 1 and has infinitely many fixed points in D. (c) The function f (x) = −x is Lipschitzian with Lipschitz constant L = 1 and has exactly one fixed point in D. Section 4.6 11. From Theorem 4.6.6, E is closed and bounded so Theorem 4.3.1 implies that f (E) is bounded. To show that f (E) is closed, let a be an accumulation point of f (E). Then there is a sequence (yn ) with yn ∈ f (E) \ {a} and yn → a. Because yn ∈ f (E), there is a number xn ∈ E such that f (xn ) = yn . Now Theorem 2.6.4 implies that there is a convergent subsequence (xnk ) (because E is bounded and non-empty). Let us set x0 = limk→∞ xnk . Because E is closed, x0 ∈ E. Now the continuity of f implies that f (xnk ) → f (x0 ). In addition, Theorem 2.6.5 implies that f (xnk ) → a, so a = f (x0 ), which means that a ∈ f (E), and therefore f (E) is closed. Section 5.1 5. (b) It is NOT true. The function f (x) = x3 + x has derivative f 0 (x) = 3x2 + 1, which is never zero. 1 2 12. First, let’s do some algebra: f (x + h) − f (x) f (x)f (h) − f (x) f (h) − 1 = = f (x) , h h h and f (x) = f (x + 0) = f (x)f (0), so f (0) = 1. Now lim f (x) h→0 f (h) − 1 f (h) − 1 = f (x) lim = f (x)f 0 (0), h→0 h h so f (x + h) − f (x) = f 0 (0)f (x). h→0 h lim Section 5.2 15. (a) From exercise 10a, the function f is increasing on (a, c) and decreasing on (c, b). It follows that f (x) ≤ f (y) for all x and y with a ≤ x ≤ y < c. Because f (c) = limy→c f (y), it follows that f (x) ≤ f (c) for all x ∈ [a, c]. Similarly, f (x) ≤ f (c) for all x ∈ [c, b] and therefore f (x) ≤ f (c) for all x ∈ [a, b], so f has a relative maximum at x = c. 20. (h) Consider t ∈ (0, 1) and b as fixed and define f (a) = at b1−t − ta − (1 − t)b. Then a t−1 f 0 (a) = tat−1 b1−t − t = t −1 . b Because t ∈ (0, 1), it follows that f 0 (a) > 0 if a < b and f 0 (a) < 0 if a > b. Exercise 15(a) then implies that f has a maximum at a = b. (The argument given here shows that it is a strict maximum.) Because f (b) = 0, it follows that f (a) < f (b) = 0 if a 6= b. If t = 0 or t = 1, then the equality follows from simple algebra. 29. Given ε > 0, there is a positive constant M such that |f 0 (y)| < ε if y ≥ M . In addition, the mean value theorem says that, for any x, there is a y ∈ (x, x + 1) such that g(x) = f (x + 1) − f (x) = f 0 (y). If x ≥ M , then y ≥ M , so |g(x)| < ε. This says that lim g(x) = 0. x→∞ 40. Suppose first that f+0 (a) > f−0 (b). Because k is between f+0 (a) and f−0 (b), there is a constant ε1 such that f+0 (a) + ε1 < k < f−0 (b) − ε1 . By the definition of the derivative, there are δ1 > 0 and δ2 > 0 such that 0 f (a + h) − f (a) < ε1 f+ (a) − h for x ∈ (a, a + δ1 ) and 0 f (b − h) − f (b) f− (b) − < ε1 −h for y ∈ (b − δ2 , b). Now set δ = 21 min{δ1 , δ2 , b − a}. Then 0 f+ (a) − f (a + δ) − f (a) < ε1 , δ and 0 f (b) − f (b − δ) f− (b) − < ε1 . δ 3 Now set g(x) = (f (x+δ)−f (x))/δ. It follows that g(a) < f+0 (a)+ε1 and g(b−δ) > f−0 (b)−ε1 , so k is between g(a) and g(b − δ). In addition, g is continuous on [a, b], so the intermediate value theorem says that there is a point γ ∈ (a, b − δ) such that g(γ) = k. We finish the proof by applying the mean value theorem to f on the interval [γ, γ + δ]: there is a point c ∈ (γ, γ + δ) such that f (γ + δ) − f (γ) = g(γ) = k. δ If f+0 (a) > f−0 (b), a similar argument can be used. f 0 (c) = Section 5.3 3. f (x) = x2 |x|. 5. (a) We need to have the derivative equal to 6, and f 0 (x) = x2 + x, so f 0 (x) = 6 for x = 2 and x = −3. The points on the curve are (2, 5/3) and (−3, −11/2). 6. By induction, for each k ∈ N, there is a polynomial pk such that ( pk (1/x) exp(−1/x2 ) if x > 0, (k) f (x) = 0 if x < 0, so f is infinitely differentiable except possibly at x = 0. In addition, pk (1/x) exp(−1/x2 ) = 0, x→0+ x so f is also infinitely differentiable at 0. Therefore every Taylor polynomial is identically zero. The same argument works for Example 5.3.3. f (k+1) (0) = lim 25. Set g(x) = f (x + a) − f (x) for x > 0. Then g 0 (x) = f 0 (x + a) − f 0 (x) ≤ 0 because f 0 ≤ 0 implies that f 0 is decreasing. It follows that g 0 is decreasing, so g(x) ≤ g(0) = f (a). Therefore f (x + a) − f (x) ≤ f (a), or f (x + a) ≤ f (x) + f (a). To complete the proof substitute x = b. Section 5.5 1. FALSE. We can rewrite f (x) = exp(x ln(4x − 1)), so 0 f (x) = exp(x ln(4x − 1)) ln(4x − 1) + x 4 . 4x − 1 8. TRUE. By differentiation, 2 2 2 x5 x x 8 x 8 8 0 3 f (x) = 4x exp − sin 3 − exp − sin 3 − 24 exp − cos 3 4 x 2 4 x 4 x if x 6= 0 and 2 h 8 0 3 f (0) = lim h exp − sin 3 = 0. h→0 4 h Now comes the big calculation. We use the identity p A sin t + B cos t = A2 + B 2 sin(t + δ), 4 √ √ where cos δ = A/ A2 + B 2 and sin δ = B/ A2 + B 2 to see that 2r x x5 0 |f (x)| ≤ exp − (4x3 − )2 + 242 . 4 2 for x ∈ [−1, 1] \ {0}. Now, we set 2r x x5 g(x) = exp − (4x3 − )2 + 242 , 4 2 and we compute # r 2" 3 − x5 /2)(12x2 − 5x4 /2) 5 x x x 2(4x 0 p − (4x3 − )2 + 242 + g (x) = exp − 4 2 2 (4x3 − x5 /2)2 + 242 2 x exp − x4 5 4 3 4 2 2 2 4 2 =q −(4x − x ) − 24 + 2(4x − x )(12x − x ) . 5 2 (4x3 − x2 )2 + 242 Now we estimate: 5 2(4x2 − x4 )(12x2 − x4 ) ≤ 2(4)(12) < 242 2 0 for 0 < x ≤ 1, so g is negative and therefore g is strictly decreasing on [0, 1]. Because g is even, it follows that g(x) ≤ g(0) = 24 for x ∈ [−1, 1] and g(x) < 24 if x 6= 0. Hence |f 0 (x)| < 24 if x ∈ [−1, 1], so f 0 is bounded. Now, let’s look at xn = (4/(nπ))1/3 (which makes 8/x3n = 2nπ) and yn = (8/((2n + 1)π))1/3 (which makes 8/yn3 = (2n + 1)π). Then lim f 0 (xn ) = −24, n→∞ and lim f 0 (yn ) = 24. n→∞ Therefore sup f 0 = 24 and inf f 0 = −24 and f 0 doesn’t attain either value, so it does not attain its absolute extrema on [−1, 1]. Section 5.6 1. (a) We have 3f (x + h) − f (x) − 2f (x − h) = 3[f (x) + f 0 (x)h + O(h2 )] − f (x) − 2[f (x) + f 0 (x)(−h) + O(h2 )] = 5f 0 (x)h + O(h2 ), so 5 3f (x + h) − f (x) − 2f (x − h) = f 0 (x) + O(h). ch c If we take c = 5, we conclude that 3f (x + h) − f (x) − 2f (x − h) + O(h2 ). ch (b) Taylor’s theorem gives f 0 (x) = 1 f (x + h) = f (x) + f 0 (x)h + f 00 (x)h2 + O(h3 ), 2 and f (x + 2h) = f (x) + 2f 0 (x)h + 2f 00 (x)h2 + O(h3 ), 5 so f (x + 2h) − 2f (x + h) + f (x) 1 = [f (x) + 2f 0 (x)h + 2f 00 (x)h2 ] − 2[f (x) + f 0 (x)h + f 00 (x)h2 ] + f (x) + O(h3 ) 2 = f 00 (x)h3 + O(h3 ). Therefore f 00 (x) = f (x + 2h) − 2f (x + h) + f (x) + O(h). h2 Section 6.1 4. FALSE. Take f (x) = 0, h(x) = 4, and ( 0 if x is rational g(x) = 4 if x is irrational. Then, clearly, f (x) ≤ g(x) ≤ h(x) on the interval [−2, 3], and f and g are Riemann integrable on [−2, 3] by Exercise 6.1.3, but g is not Riemann integrable [−2, 3] by Example 6.1.8.