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Q. 9

Expert-verified

Calculus

Found in: Page 692

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

If $f (x) = 4 x^{3} - 5 x^{2} + 6 x + 1 a n d P_{3} (x)$ is the third Taylor polynomial for f at −1, what is the third remainder $R_{3} (x)$ ? What is $R_{4} (x)$ ? (Hint: You can answer this question without finding any derivatives.)

The required values are $R_{3} (x) = 0 a n d R_{4} (x) = 0$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information

The given function is $f (x) = 4 x^{3} - 5 x^{2} + 6 x + 1$

Step 2. Calculation

The formula to calculate the remainder is $R_{n} (x) = \frac{f^{n + 1} (c)}{(n + 1)!} {(x - x_{0})}^{n + 1}$

Substitute n as 3 to find the third remainder of the function.

$R_{3} (x) = \frac{f^{3 + 1} (c)}{(3 + 1)!} {(x - x_{0})}^{3 + 1} = \frac{f^{4} (c)}{4!} {(x - x_{0})}^{4}$

Since the given degree has highest degree 3 which implies that $f^{4} (c) = 0$

Hence, $R_{3} (x) = 0$

Similarly,

localid="1649315178808" $R_{4} (x) = \frac{f^{4 + 1} (c)}{(4 + 1)!} {(x - x_{0})}^{4 + 1} = \frac{f^{5} (c)}{5!} {(x - x_{0})}^{5} = \frac{0}{5!} {(x - x_{0})}^{5} = 0$

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