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Q. 9

Expert-verifiedFound in: Page 692

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

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

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

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

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

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" ${R}_{4}\left(x\right)=\frac{{f}^{4+1}\left(c\right)}{(4+1)!}{(x-{x}_{0})}^{4+1}\phantom{\rule{0ex}{0ex}}=\frac{{f}^{5}\left(c\right)}{5!}{(x-{x}_{0})}^{5}\phantom{\rule{0ex}{0ex}}=\frac{0}{5!}{(x-{x}_{0})}^{5}\phantom{\rule{0ex}{0ex}}=0$

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