11.10 Taylor and Maclaurin Series-solutions

# 11.10 Taylor and Maclaurin Series-solutions - im(gi768...

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im (gi768) – 11.10 Taylor and Maclaurin Series – sadun – (53088) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of4)10.0points Suppose T 4 ( x ) = 8 - 6( x - 2) + 2( x - 2) 2 - 6( x - 2) 3 + 3( x - 2) 4 is the degree 4 Taylor polynomial centered at x = 2 for some function f . (i) What is the value of f (2)? 1. f (2) = 10 2. f (2) = - 9 3. f (2) = 8 correct 4. f (2) = - 8 5. f (2) = 9 Explanation: Since T 4 ( x ) = f (2) + f (2)( x - 2) + f ′′ (2) 2! ( x - 2) 2 + f (3) (2) 3! ( x - 2) 3 + f (4) (2) 4! ( x - 2) 4 , we see that f (2) = 8 . 002(part2of4)10.0points (ii) What is the value of f (3) (2)? 1. f (3) (2) = 2 2. f (3) (2) = - 2 3. f (3) (2) = - 6 4. f (3) (2) = 36 5. f (3) (2) = - 36 correct Explanation: Since T 4 ( x ) = f (2) + f (2)( x - 2) + f ′′ (2) 2! ( x - 2) 2 + f (3) (2) 3! ( x - 2) 3 + f (4) (2) 4! ( x - 2) 4 , we see that f (3) (2) = - 3! × 6 = - 36 . 003(part3of4)10.0points (iii) Use T 4 to estimate the value of f (2 . 1). 1. f (2 . 1) 7 . 4143 correct 2. f (2 . 1) 7 . 3143 3. f (2 . 1) 7 . 7143 4. f (2 . 1) 7 . 5143 5. f (2 . 1) 7 . 6143 Explanation: Since T 4 ( x ) is an approximation for f ( x ) we see that f (2 . 1) 8 - 6 10 + 2 10 2 - 6 10 3 + 3 10 4 . Consequently, f (2 . 1) 7 . 4143 . 004(part4of4)10.0points

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im (gi768) – 11.10 Taylor and Maclaurin Series – sadun – (53088) 2 (iv) Use T 4 to estimate the value of f (1 . 9). 1. f (1 . 9) ≈ - 6 . 792 2. f (1 . 9) ≈ - 6 . 992 3. f (1 . 9) ≈ - 6 . 692 4. f (1 . 9) ≈ - 6 . 592 correct 5. f (1 . 9) ≈ - 6 . 892 Explanation: The Taylor polynomial centered at x = 2 for f is given by T 3 ( f ,x ) = f (2) + f ′′ (2)( x - 2) + f (3) (2) 2! ( x - 2) 2 + f (4) 3! ( x - 2) 3 where f (2) = - 6 , f ′′ (2) = 4 , f (3) (2) = - 36 , f (4) (2) = 72 .
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