link solutions manual for additional calculus topics 11th edition by raymond barnett karl byleen michael ziegler

Karl Byleen 1.f(x) = = x -1 x -2 f’(x) = -x f”(x) = 2x (3) f (Using Power Rule) -3 -4 (x) = -6x = -x f(x) = ln(1 + x) -1 f'(x) =  x = (1 + x) f"(x) = (-1)(1 + x) (Using Power Rule) (3) f (x) = (-1)(-2)(1 + x) = (1  x) -x -3 (Using Power Rule) f(x) = e f’(x) = -e f”(x) = e -x (3) f (x) = -e f(x) = ln(1 + 3x) f’(x) = = 3(1 + 3x)  3x -2 f”(x) = 3(-1)(3)(1 + 3x) Rule) f (4) f (3) (x) = (-9)(-2)(3)(1 + 3x) (x) = (54)(-3)(3)(1 + 3x) -4 5x f(x) = e f’(x) = 5e 5x 5x f”(x) = 5(5)e (3) f (Using Power -3 5x (x) = (5)e 5x 5x = e 5x = e 5x = 625e TAYLOR POLYNOMIALS AND INFINITE SERIES -3 = 54(1 + 3x) = -486(1 + 3x) 5x = e (x) = (5)e (4) f 96 -2 = -9(1 + 3x) -4 486 = - (1  3x) -1 f(x) =  x = (2 + x) f’(x) = (-1)(2 + x) f”(x) = (-1)(-2)(2 + x) f (x) = (2)(-3)(2 + x) (4) f -5 -5 -0 (x) = (-6)(-4)(2 + x) = 24(2 + x) -x f(x) = e f'(x) = -e f"(x) = f = 2(2 + x) = -6(2 + x) (3) f'(0) = -e -x e -x (x) = -e (4) f(0) = e -x f (x) = e Using 2, (3) (4) -0 f (0) = -e f (0) = e p (x) = f(0) + f'(0)x + f"(0) x 2! Thus, f(x) = e f'(x) = 4e f"(x) = 16e (3) f p (x) = - x + x 2! - x 3! 4x (x) = 64e = -1 -0 e = -0 f"(0) = -x = -0 = -1 + + 1x 4! = f (3) (0) x + f 3! = - x + (4) (0) x 4! x -1 x + 1x 24 f(0) = f'(0) = f"(0) = 16 4x (3) f (0) = 64 (3) f"(0) f (0) + Thus, p (x) = f(0) + f'(0)x + 2! x 3! x = + 4x + 16 x2 + 64 x3 = + 4x + 8x + 3! 2! f(x) = (x + 1) , f(0) = 32 x f'(x) = 3(x + 1) , f'(0) = f"(x) = 6(x + 1), f"(0) = f (x) = 6, f (0) = f (x) = f (0) = 2 p (x) = + 3x + x + x3 = + 3x + 3x + x 2! 3! EXERCISE 2-1 97 f(x) = (1 - x) , f'(x) = -4(1 - x) , f"(x) = 12(1 - x) , 10 f(0) = f'(0) = -4 f"(0) = 12 (3) f Thus, p (x) = - 4x + 11 (3) (x) = -24(1 - x), f(x) = ln(1 + 2x) 12 2! x2 f (0) = -24 24 - 3! x3 2 f'(x) =  2x (2) =1  2x 4 -2 f"(x) = -2(1 + 2x) (2) = (1  2x) (3) -3 16 f (x) = 8(1 + 2x) (2) = (1  2x) 12 f(x) = 2! 3x  = (x + 1) 1/3 + (3) f (0) x = f"(x) = 9(x  1) 10 (x) = 27(x  1) x f(0) = 31 f'(0) = 2 f (3) (0) = 16 3! (x) = + 2x - x + 16 x = 2x - 2x + x3 3! 2! f'(x) = 3(x  1) (3) f 3   = f"(0) = -4 f'(0) = f"(0) Using 2, p (x) = f(0) + f'(0)x + Thus, p = - 4x + 6x - 4x f(0) = ln(1) = 8/3 f"(0) = -9 f (3) 10 (0) = 27 (3) f"(0) f (0) p 3(x) = f(0) + f'(0)x + x + 2! 3! x3 1 Thus, p (x) = + x x + x 10 x = + x  2! 27  3! 3 + x 81 1/4 f(x) = x  16 = (x + 16) f(0) = 1 f'(x) = 4(x + 16) f'(0) = 32 f"(x) = - (x + 16) f"(0) = - 16 2048 x p (x) = + x 32 4096 14 (A) f(x) = x - f(0) = -1 f'(x) = 4x f'(0) = f"(x) = 12x f"(0) = 13 -3/4 (3) f (3) (x) = 24x f (0) = Using 2, p (x) = f(0) + f'(0)x + f"(0) 2! Thus, p3(x) = -1 |p 3(x) - f(x)| = |-1 - (x Now 4 + (3) f (0) 3! x - 1)| = |x | = |x| < 0.1 implies |x| < (0.1) and |x| Therefore, -0.562 < x < 0.562 (B) From part (A), (4) 1/4 x ≈ 0.562 (4) (4) f (x) = 24 f (0) = 24 (3) f (0) Using 2, p (x) = f(0) + f'(0)x + f"(0) x + f (0) x + x4 2! 3! 4! 24 4 Thus, p 4(x) = -1 + 4! x = -1 + x = x - = f(x) |p4(x) - f(x)| = < 0.1 for f(0) = f(x) = x f'(x) = 5x f'(0) = f"(x) = 20x f"(0) = f (x) = 60x f (0) = all x 0 0 and 15 (A) (4) f (4) (x) = 120x f p4(x) = (0) = 5 |p4(x) - f(x)| = |0 - x | = |x| 1/5 < 0.01 or |x| < (0.01) = 0.398 Therefore, -0.398 < x < 0.398 (3) (B) f(0) = f'(0) = f"(0) = f = f (5) 5! x (5) |p5(x) - f(x)| = |x (4) (0) = f (x) = 120 and hence f 120 p5(x) = = (0) (0) = = x - x | = < 0.01 Thus for all x, |p5(x) - f(x)| < 0.01 EXERCISE 2-1 99 f(x) = x 16 f f'(x) = 3x f"(x) = 6x (3) (x) = f(1) = f'(1) = f"(1) = (3) f (1) = p (x) = + 3(x - 1) + 2! (x - 1) = + 3(x - 1) + 3(x - 1) 17 + (x - 1) 3! f(x) = x - 6x + 10 f'(x) = 2x - f"(x) = p (x) = + 2 (x - 3) 2! + (x - 1) f(3) = f'(3) = f"(4) = = + (x - 3) 18 f(x) f'(x) f"(x) f (x) (4) f = = = = ln(2 - x) -(2 - x) -(2 - x) -2(2 - x) f(1) = f'(1) = -1 f"(1) = -1 -4 (x) = -6(2 - x) p (x) = -(x - 1) - (x - 1) 19 f(x) = (x - 1) -2x e -2x -2x f"(x) = 4e f - 2x (x) = -8e f Thus, p (x) = - 2x + Now, e = e (x - 1) 4! - (x - 1) - 4(x - 1) (3) Using 2, p (x) = f(0) + f'(0)x + - 0.5 (1) = -6 f(0) = f'(0) = -2 f"(0) = f'(x) = -2e (3) f 3! = -(x - 1) - (1) = -2 (4) - (x - 1) - 2! f -2(0.25) x - 2! (0) = -8 (3) f"(0) x + f (0) x 2! 3! - x = - 2x + 2x 3! 3 x = f(0.25) ≈ p (0.25) = - 2(0.25) + 2(0.25) (0.25) = 0.60416667 -3 20 f(x) = f'(x) = = x1/2 x x-1/2 x-3/2 f"(x) = - (3) f (4) f(1) = 1 f'(1) = (x) = x-5/2 15 f (x) = - 16 x Thus, -7/2 f"(1) = - (3) f (4) f (1) = 15 (1) = - 16 (3) (4) f"(1) f (1) f (1) p4(x) = f(1) + f'(1)(x - 1) + + + 2! (x - 1) 3! (x - 1) 4! (x - 1) = + (x - 1) (x - 1) + 15 (x - 1) (x - 1) 16  4!  3! 2! 1 = + (x - 1) - (x - 1) Now, 1.2 = f(1.2) ≈ p4(1.2) + 16 (x - 1) - 128 (x - 1) (1.2 - 1) + (1.2 - 1) = + (1.2 - 1) - 16 = + (0.2) - (0.2) + (0.2) (0.2) 16 128 = 1.0954375 - (1.2 - 1) 128 -1 21 f(x) = f = (4 - x)  x -2 -2 (x) = 2(-3)(4 - x) (4) f (n) f (-1) = 1·2·3(1 - x) (x) = 2·3(-4)(4 - x) -5 -4 (-1) = 1·2·3·4(1 - x) -5 -(n+1) (x) = n!(4 - x) = 4(1 +x) -x  x f'(x) = -4(1 + x) f"(x) = (-1) (4)(2)(1 + x) 22 -3 f'(x) = -1(4 - x)-3(-1) = (4 - x) f"(x) = -2(4 - x) (-1) = 1·2(4 - x) (3) -4 f(x) = f (3) = 4(2!)(-1) (1 + x) -4 (x) = (-1) (4)(3)(2)(1 + x) M (n) f n -(n+1) (x) = 4(n!)(-1) (1 + x) 3x 23 f(x) = e 3x 3x f'(x) = e (3) = 3e 3x 3x 3x = e f"(x) = 3e (3) = 9e (3) 3x f (x) = 9e f (x) = 27e (4) (n) f 24 (3) = 27e 3x 3x (3) = 81e 3x 3x = e 3x = e n 3x (x) = e f(x) = ln(2x + 1) -1 f'(x) = 2(2x + 1) f"(x) = -(2) (2x + 1) -2 (3) 1) f (n) f 25 (x) = (-1) (x) = (-1) = (-1) (2x + 1) -2 = (-1) -1 (2x + (2!)(2x + 1) -3 n+1 n ((n - 1)!)(2x + 1) -n f(x) = ln(6 - x) = -(6 - x) f'(x) = (-1) = - x x -2 -2 f"(x) = (6 - x) (-1) = -(6 - x) (4) f -4 (x) = 1·2·3(6 - x) -4 (n) x) f -4 = 4(3!)(-1) (1 + x) (-1) = -1·2·3(6 - -n (x) = -(n - 1)!(6 - x) f(x) 26 x/2 = e f'(x) = x/2 e  1  f"(x) = ex/2 = 2 (3) ex/2 (x) = 13 ex/2 (n ) f f (x) = n ex/2 27 From Problem 31, 1 2! (3) 3! (n) n! f(0) = , f'(0) = , f"(0) = , f (0) = , … , f (0) = 4 4 Thus, 1 x + x + p (x) = x + x + … + n 4 n 1 4 4n 1 -1 28 f(x) =  x = 4(1 + x) From problem 32, f (x) = 4(n!)(-1) (1 + x) and (n) hence f (0) = 4(n!)(-1)n n The coefficient of x (3) , f Thus, = 1, f'(0) = 3e 0 x + 3 2! n (n ) x (0) = n + … + 3! n = 4(-1) n! = , …, f p (x) = + 3x +  n! n 4(n!)(1) = 3, f"(0) = e (0) = e (0) is 29.From Problem 33, f(0) = e (n) f = n e n = n x n! 30.f(x) = ln(2x + 1) From problem 34, (n) n n+1 n -n n+1 (n ) f (0) n! n f (x) = (-1) ((n - 1)!)(2x + 1) and = (-1) n+1 2n x n Thus, p (x) = 2x - x2 + 23 x - 24 x4 + … + (-1) n n 31 From Problem 35, (3) (n) (n  1)! 1 2! f(0) = ln 6, f'(0) = - , f"(0) = - , f (0) = - , …, f (0) = n 6 6 Thus, x x3 - … x p (x) = ln - x n 2  3 6 n n 6 EXERCISE 2-1 x/2 32 f(x) = e From problem 36, f (n) (x) = e x/2 and (n) f (0) = Thus, 105 p (x) = + n x + 2!2 2x f"(x) = (-1) (3) f (x) = (-1) f (x) = (-1) (n) n 3!2 3x 2(2!)x f"(x) = - x (3) f (x) = 23 x n!2 nx n!2 n +x -4 2(n!)x -(n+1) n = 2(-1) and - 2(x - 1) n + … + 2(-1) (x - 1) Step f(1) = Step a = f(1) = f'(1) = a1 = f'(1) = 1 2 -3 pn(x) = - 2(x - 1) + 2(x - 1) f'(x) = x + … + f(1) = n! 34 Step f(x) = ln x 2(3!)x f(n) (1) Therefore + -1 f(x) = = 2x x f'(x) = -2x 33 f"(1) f"(1) = -1 (3) f (1) = 2 = = 2! = - 2! = - (1) (3) f a 3! = 3! = n (4) (x) = - x f f  (n) f (x) = n 1 (1) f (n  1)! (4) (1) (n) = -3! a (1) = n+1 (-1) (4) a f = n = f (1) n! xn 4! = -4! = - (n) (n - 1)! 3! (1)  1)! n 1 = (1) Step The nth degree Taylor polynomial is: p (x) = (x - 1) - (x - 1) + (x - 1) - (x n 34 (n n! n 1 = (1) (1)n 1 n (x - 1) 1)4 + … + n x f(x) = e 35 f (n) (n) x (x) = e Thus, p (x) = n f and e2 (2) n! + e2 = n!e (x + 2) + (x 2!e + 2) n + … + n!e (x n + 2) 36.f(x) = x + 2x + 8x + (A) Fourth-degree Taylor polynomial p4(x) for f at is: (3) (4) p (x) = f(0) + F’(0) x + F’’(0) x + f (0) x + f (0) x4 1! 2! 3! 4! f(0) = f’(x) = 5x + 6x + 16x ; f’(0) = f”(x) = 20x + 12x + 16 ; f”(0) = 16 f (x) = 60x + 12 ;f (0) = 12 (4) f (x) = 120x Thus, p 4(x) = + 8x ; f + (4) (0) =0 2x = 2x + 8x + (B) The degree of the polynomial is 37.f(x) = x f(0) = f’(x) = f”(x) = f (x) f (x) f (x) f (x) (n) + 2x + 6x + 6x 30x + 12x = 120x + 12 = 360x = 720x = 720 ; ; ; ; ; ; f’(0) = f”(0) = f (0) f (0) f (0) f (0) 0 = = = = 12 0 720 f (x) = for n ≥ Thus, for n = 0, and 6, the nth-degree Taylor polynomial for f at has degree n 38.f(x) = x f(0) = -1 f’(x) = f”(x) = f (x) f (x) f (n) – 4x 12x = 24x = 24 (x) = ; ; ; ; f’(0) = f”(0) = f (0) f (0) 0 = = 24 for n ≥ Thus, for n = and n = 4; the nth degree Taylor polynomial for f at has degree n EXERCISE 2-1 10 39 f f(x) = ln(1 + x) f'(x) =  x 1 f"(x) = (1  x) (3) (x) = (1  x) f(0) = f'(0) = f"(0) = -1 f (3) (0) = 2 Thus, x p1(x) = x, p x -0.2 -0.1 0.1 0.2 2(x) = x - p1(x) p2(x) -0.2 -0.1 0.1 0.2 -0.22 -0.105 0.095 0.18 x p1(x) - f(x) -0.2 -0.1 0.1 0.2 0.023144 0.005361 0.00469 0.017678 , p3(x) = x p3(x) -0.222667 -0.105333 0.095333 0.182667 x + f(x) -0.223144 -0.105361 0.09531 0.182322 p2(x) - f(x) 0.003144 0.000361 0.00031 0.002322 x p3(x) - f(x) 0.000477 0.000028 0.000023 0.000345 40 (3) f(x) = ln(1 + x) f'(x) = (1 + x) f"(x) = -(1 + x) -3 (3) f(0) = f'(0) = f"(0) = -1 f (x) = 2(1 + x) f (0) = Thus, p (x) = x - x + x 3 Using a graphing utility, we find that |p3(x) - ln(1 + x)| < 0.1 for -0.654 < x < 0.910 x f(x) = e f'(x) = e f"(x) = e 41 (3) f (x) = e M (n) f (x) = e Thus, f(a) = e f'(a) = e f"(a) = e x f (3) (a) = e M x f (n) (a) = e a a a (3) p (x) = f(a) + f'(a)(x - a) + n (n) f (a) + + n! (x - a)2 2! (x - a) (a) f + (x - a) 3! n a a a a = e + e (x - a) + e (x - a) + e (x - a) + a  = e = e 42  (x  a)  2! (x  a) 2! k  a n  (x - a) k k ! 3! 1 (x  a) 3! f'(x) = x (4) f f -3 (3) (x) = (-1) (2!)x f (x) = (-1) Thus, p (x) = ln a + f n+1 ((n - 1)!)x -n 2a k  = ln a f (n) (x - a) (x - a) - a n n (1) +  k k 1 ka a a (x - a) n! n  (x  a)  (x - a) a (a) = (-1) (3!) 14 a (a) = (-1) (a) = (-1) (2!) (4) -4 (x) = (-1) (3!)x ( n) e f(a) = ln a f'(a) = a f"(a) = - -1 (3)  n! +  f(x) = ln x f + n+1 ((n - 1)!) (x - a) n a 3a - … + (-1) n+1 n na (x - a) n n 43 f(x) = (x + c) n n f(0) = c f'(x) = n(x + c) f'(0) = nc n- f"(x) = n(n - 1)(x + c) n- f"(0) = n(n - 1)c M f (x) = n!(x + c) f (n) f (x) = n! Thus, p (x) = c + nc n = c n = (n) f x + n(n  1) c x + … + 2! n! n! + x + cn-2x2 (n  1)! 2!(n  2)! n! cn-kxk  k 0 (0) = n!c k!(n  k)! (0) = n! n! (n  1)! + … + x cx n! n +x n! 44.Let f(x) be a polynomial of degree k, k ≥ Then f(x) = a + a 1x + a x + a 3x f'(x) = a + 2a 2x + 3a 3x2 (3) xk f(0) = a + … + ka kxk-1 f'(0) = a + 6a3x + … + k(k - 1)a kxk-2 f"(x) = 2a f + … + a k f"(0) = 2a + 2!a (3) xk-3 (x) = 6a + … + k(k - 1)(k - 2)a k In general,  m!am for m  0,1,2,K k f (m) (0) = for m  k  Since p (x) = f(0) + f'(0)x + f"(0) 2+ f 2! n f (0) = 6a = 3!a  (3) (0) x x 3! it follows that pn(x) = f(x) for all n ≥ k 45.f(x) = e f(0) = x x f’(x) = e f”(x) = e (k) x f (x) = e Therefore, p x (x) = + ; ; f’(0) = f”(0) = (k) ; f x2 + … + x10 1! 2! 10! p (x) = + x + x + … + x + 1x11 1! 2! 10! 11! 11 10 x + (0) = + … + (n) f (0) n x n! For x > 0, e x e x > p 11(x) (x) > p (x) – p 10 11 1/11 Take x = 2(11!) , then e – p x – p (x) > 10 and hence 1x 11! (x) = 10 (2(11!) 11! 1/11 11 ) = 11 11 = 2048 So, there exist values of x for which |p x x 10(x) – e | = |e 46.f(x) = = x f(1) = f’(x) = -x – p 10 (x)| ≥ 100 -1 -2 ; f’(1) = -1 = -1! or f (1) = -1 1! f”(x) = (-1)(-2)x M (k) f -3 k f (1) ; f”(x) = = 2! or f (1) = (-1) k! or (k) (1) k! = (-1) Therefore, p 12(x) = – (x – 1) + (x – 1) and |p12(x) – f(x)| = -  (x  1)  (x  1) = 2! k - (x – 1) = 11  L  (x  1) x |x – x(x – 1) + x(x – 1) - + (x – 1) 12 … x 12 , 12 + x(x – 1) – 1| If we take x = 0.001, then x = 1000 and every term involving x on the right-hand side of the above equation is positive and smaller than x Thus, |p12(x) – f(x)| ≥ 1000(1 – 13x) = 1000(1 – 0.013) = 987 So there exist values of x ≠ for which |p12(x) – f(x)| ≥ 100 EXERCISE 2-1 115 47.ln 1.1 Let f(x) = ln(1 + x) -1 = (1 + x) f’(x) =  x f”(x) = -(1 + x) (3) f (x) = 2(1 + x) (n) f Rn(x) = -3 (x) = (n – 1)!(-1) f(n 1)(t)xn 1 Note that f (n  1)! (n+1) n-1 -n (1 + x) for some t between and x n -(n+1) (t) = n!(-1) (1 + t) (n+1) n -(n+1) |f (t)| = |n!(-1) (1 + t) Therefore, |R (x)| = f(n 1)(t)xn 1 (n  1)! n and n! x n 1 < (n  1)! and hence | = n!(1 + t) -(n+1) < n! for t > x n 1 =n  n 1 R (0.1) < (0.1) n (n  1) (0.1) For n = 4, |R (0.1)| < polynomial with the lowest degree is p which has degree ln(1.1) ≈ p (0.1) = 0.1 - (0.1) ≈ 0.095308 (x) = x - x + (0.1) 3 +1x = 0.000 002 < 0.000 005, and hence the - x4 - (0.1) CHAPTER REVIEW 209 ... f’(0) = f”(0) = f (0) f (0) f (0) f (0) 0 = = = = 12 0 720 f (x) = for n ≥ Thus, for n = 0, and 6, the nth-degree Taylor polynomial for f at has degree n 38.f(x) = x f(0) = -1 f’(x) = f”(x) = f (x)... = 24x = 24 (x) = ; ; ; ; f’(0) = f”(0) = f (0) f (0) 0 = = 24 for n ≥ Thus, for n = and n = 4; the nth degree Taylor polynomial for f at has degree n EXERCISE 2-1 10 39 f f(x) = ln(1 + x) f'(x)... In general,  m!am for m  0,1,2,K k f (m) (0) = for m  k  Since p (x) = f(0) + f'(0)x + f"(0) 2+ f 2! n f (0) = 6a = 3!a  (3) (0) x x 3! it follows that pn(x) = f(x) for all n ≥ k 45.f(x)