In the xc xc following we use theorem 222 a f c g c

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Unformatted text preview: no, x = 2 (c) no, x = 1 (d) (c) yes (b) y (d) yes (d) yes y 1 3 x 1 3 x yes 55 Chapter 2 (c) y (d) y 1 x 1 2 -1 0 8. f (x) = 1/x, g (x) = 9. (a) x if x = 0 1 sin x 3 if x = 0 (b) C One second could cost you one dollar. $4 t 1 10. 2 (a) no; disasters (war, flood, famine, pestilence, for example) can cause discontinuities (b) continuous (c) not usually continuous; see Exercise 9 (d) continuous 11. none 14. f is not defined at x = ±1 15. f is not defined at x = ±4 √ −7 ± 57 f is not defined at x = 2 17. f is not defined at x = ±3 f is not defined at x = 0, −4 19. none 16. 18. 12. none 13. none 20. f is not defined at x = 0, −3 16 is continuous on 4 < x; x lim− f (x) = lim+ f (x) = f (4) = 11 so f is continuous at x = 4 21. none; f (x) = 2x + 3 is continuous on x < 4 and f (x) = 7 + x→4 22. 23. x→4 lim f (x) does not exist so f is discontinuous at x = 1 x→1 (a) f is continuous for x < 1, and for x > 1; lim− f (x) = 5, lim+ f (x) = k , so if k = 5 then f is x→1 x→1 continuous for all x (b) f is continuous for x < 2, and for x > 2; lim− f (x) = 4k , lim+ f (x) = 4 + k , so if 4k = 4 + k , k = 4/3 then f is continuous for all x 24. x→2 x→2 no, f is not defined for x ≤ 2 (a) no, f is not defined at x = 2 (b) (c) yes (d) no, f is not defined for x ≤ 2 Exercise Set 2.4 25. (a) 56 (b) y x c 26. (a) x c f (c) = lim f (x) (b) y lim f (x) = 2 x→c lim g (x) = 1 x→1 x→1 y y 1 1 x -1 1 x (c) (a) x = 0, lim− f (x) = −1 = +1 = lim+ f (x) so the discontinuity is not removable (b) x = −3; define f (−3) = −3 = lim f (x), then the discontinuity is removable (c) 27. Define f (1) = 2 and redefine g (1) = 1. f is undefined at x = ±2; at x = 2, lim f (x) = 1, so define f (2) = 1 and f becomes continuous x→0 x→0 x→−3 x→2 there; at x = −2, lim does not exist, so the discontinuity is not removable x→−2 1 1 x+2 = , so define f (2) = and f becomes x2 + 2x + 4 3 3 (a) f is not defined at x = 2; lim f (x) = lim (b) 28. continuous there lim− f (x) = 1 = 4 = lim+ f (x), so f has a nonremovable discontinuity at x = 2 (c) 29. (a) x→2 x→2 x→2 x→2 lim f (x) = 8 = f (1), so f has a removable discontinuity at x = 1 x→1 discontinuity at x = 1/2, not removable; at x = −3, removable y 5 5 -5 x (b) 2x2 + 5x − 3 = (2x − 1)(x + 3) 57 30. Chapter 2 (a) there appears to be one discontinuity near x = −1.52 (b) one discontinuity at x = −1.52 4 -3 3 –4 31. For x > 0, f (x) = x3/5 = (x3 )1/5 is the composition (Theorem 2.4.6) of the two continuous functions g (x) = x3 and h(x) = x1/5 and is thus continuous. For x < 0, f (x) = f (−x) which is the composition of the continuous functions f (x) (for positive x) and the continuous function y = −x. Hence f (−x) is continuous for all x > 0. At x = 0, f (0) = lim f (x) = 0. x→0 32. x4 + 7x2 + 1 ≥ 1 > 0, thus f (x) is the composition of the polynomial x4 + 7x2 + 1, the square root √ x, and the function 1/x and is therefore continuous by Theorem 2.4.6. 33. (a) Let f (x) = k for x = c and f (c) = 0; g (x) = l for x = c and g (c) = 0. If k = −l then f + g is continuous; otherwise it’s not. (b) f (x) = k for x = c, f (c) = 1; g (x) = l = 0 for x = c, g (c) = 1. If kl = 1, then f g is continuous; otherwise it is not. 34. A rational function is the quotient f (x)/g (x) of two polynomials f (x) and g (x). By Theorem 2.4.2 f and g are continuous everywhere; by Theorem 2.4.3 f /g is continuous except when g (x) = 0. 35. Since f and g are continuous at x = c we know that lim f (x) = f (c) and lim g (x) = g (c). In the x→c x→c following we use Theorem 2.2.2. (a) f (c) + g (c) = lim f (x) + lim g (x) = lim(f (x) + g (x)) so f + g is continuous at x = c. x→c x→c x→c (b) same as (a) except the + sign becomes a − sign (c) lim f (x) f f (c) f (x) = x→c = lim so is continuous at x = c x→c g (x) g (c) lim g (x) g x→c 36. h(x) = f (x) − g (x) satisfies h(a) > 0, h(b) < 0. Use the Intermediate Value Theorem or Theorem 2.4.9. 37. Of course such a function must be discontinuous. Let f (x) = 1 on 0 ≤ x < 1, and f (x) = −1 on 1 ≤ x ≤ 2. 38. A square whose diagonal has length r has area f (r) = r2 /2. Note that f (r) = r2 /2 < πr2 /2 < 2r2 = f (2r). By the Intermediate Value Theorem there must be a value c between r and 2r such that f (c) = πr2 /2, i.e. a square of diagonal c whose area is πr2 /2. 39. The cone has volume πr2 h/3. The function V (r) = πr2 h (for variable r and fixed h) gives the volume of a right circular cylinder of height h and radius r, and satisfies V (0) < πr2 h/3 < V (r). By the Intermediate Value Theorem there is a value c between 0 and r such that V (c) = πr2 h/3, so the cylinder of radius c (and height h) has volume equal to that of the cone. 40. If f (x) = x3 − 4x + 1 then f (0) = 1, f (1) = −2. Use Theorem 2.4.9. 41. If f (x) = x3 + x2 − 2x then f (−1) = 2, f (1) = 0. Use the Intermediate Value Theorem. Exercise Set 2.4 42....
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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