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Expert-verified Found in: Page 692 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # If $f\left(x\right)=4{x}^{3}-5{x}^{2}+6x+1and{P}_{3}\left(x\right)$ is the third Taylor polynomial for f at −1, what is the third remainder ${R}_{3}\left(x\right)$? What is ${R}_{4}\left(x\right)$? (Hint: You can answer this question without finding any derivatives.)

The required values are ${R}_{3}\left(x\right)=0and{R}_{4}\left(x\right)=0$

See the step by step solution

## Step 1. Given Information

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

## Step 2. Calculation

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

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

${R}_{3}\left(x\right)=\frac{{f}^{3+1}\left(c\right)}{\left(3+1\right)!}{\left(x-{x}_{0}\right)}^{3+1}\phantom{\rule{0ex}{0ex}}=\frac{{f}^{4}\left(c\right)}{4!}{\left(x-{x}_{0}\right)}^{4}$

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

Hence, ${R}_{3}\left(x\right)=0$

Similarly,

localid="1649315178808" ${R}_{4}\left(x\right)=\frac{{f}^{4+1}\left(c\right)}{\left(4+1\right)!}{\left(x-{x}_{0}\right)}^{4+1}\phantom{\rule{0ex}{0ex}}=\frac{{f}^{5}\left(c\right)}{5!}{\left(x-{x}_{0}\right)}^{5}\phantom{\rule{0ex}{0ex}}=\frac{0}{5!}{\left(x-{x}_{0}\right)}^{5}\phantom{\rule{0ex}{0ex}}=0$ ### Want to see more solutions like these? 