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Calculus 2-A8

# Calculus 2-A8 - Calculus II Calculus Early Transcendental...

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Calculus II Calculus: Early Transcendental Functions , 4th ed., by Ron Larson, Robert Hostetler, and Bruce H. Edwards (Boston: Houghton Mifflin, 2007; ISBN-10: 0-618-60624-6). Written Assignment 8 Complete the following textbook exercises for sections 9.7, 9.8, 9.9, and 9.10, and submit them to your mentor for correction and grading. Show all calculations. SECTION 9.7 Exercises Do exercises 2, 4, 14, 24, 28, 42, 46, and 58 on pages 656–658 of the textbook. In Exercises 1– 4, match the Taylor polynomial approximation of the function with the correct graph. [The graphs are labeled ( a), ( b), ( c), and ( d).] Matches graph C Plotting a few points we can see the pattern. Three relative extrema, y-axis symmetry y= (1/8)x 4 - (1/2)x 2 +1 x 1 -2 0.5078 -1.5 0.625 -1 0.8828 -0.5 1 0 0.8828 0.5 0.625 1 0.5078 1.5 1 2

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Cubic, matches graph B In Exercises 13– 24, find the Maclaurin polynomial of degree n for the function. ƒ(x) = e -x ƒ(0) = 1 ƒ’(x) = -e -x ƒ’(0) = -1 ƒ”(x) = e -x ƒ”(0) = 1 ƒ (3) (x) = -e -x ƒ (3) (0) = -1 ƒ (4) (x) = e -x ƒ (4) (0) = 1 ƒ (5) (x) = -e -x ƒ (5) (0) = -1 The Maclaurin polynomial P 5 (x) = ƒ(0) + (ƒ’(0)/1!)x + (ƒ’’(0)/2!)x 2 + (ƒ (3) (0)/3!)x 3 + (ƒ (4) (0)/4!)x 4 + (ƒ (5) (0)/5!)x 5 =1 – x + ½ x 2 – 1/6 x 3 + 1/24 x 4 – 1/120 x 5 ƒ(x) = tan x ƒ(0) = 0 ƒ’(x) = sec 2 x ƒ’(0) =1 ƒ”(x) = 2sec 2 x tan x ƒ”(0) = 0 ƒ (3) (x) =4sec 2 x tan 2 x + 2 sec 4 x ƒ (3) (0) = 2 The Maclaurin polynomial P 3 (x) = 0 + 1x + 0/2!x 2 + 2/3!x 3 = x + 2/3!x 3 = x + 1/3!x 3 In Exercises 25– 30, find the nth Taylor polynomial centered at c. ƒ(x) = x 1/3 ƒ(8) = 2 ƒ’(x) = 1/3x -2/3 ƒ’(8) =1/12 ƒ”(x) = 2/9x -5/3 ƒ”(8) = 1/144 ƒ (3) (x) =10/27x -8/3 ƒ (3) (8) = 10/27 * ½ 8 = 5/3456 The Taylor polynomial P 3 (x) = 2 + 1/12(x – 8) -1/288(x – 8) 2 + 5/20,736 (x – 8) 3 In Exercises 41– 44, approximate the function at the given value of using the polynomial found in the indicated exercise. ƒ(x) = x 2 e -x ƒ’(x) = -x 2 e -x +2xe -x ƒ”(x) = -(-x 2 e -x +2xe -x ) +2(-xe -x +e -x )
ƒ (3) (x) = x 2 e -x -2xe -x -2xe -x +2xe -x Taylor polynomial P 3 (x) = ƒ(1/5) + ƒ’(1/5)(x - 1/5) + ƒ”(1/5)/2! (x - 1/5) 2 + ƒ (3) (1/5)/3! (x - 1/5) 3 ≈x 2 – x 3 + 1/2x 4 ƒ(1/5) ≈ 0.0328 In Exercises 45– 48, use Taylor’s Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error.

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