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Q. 23

Expert-verifiedFound in: Page 692

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, ${R}_{4}\left(x\right)$

$\mathrm{cos}x$

The required answer is ${R}_{4}\left(x\right)=-\frac{\mathrm{sin}c}{120}{x}^{5}$

The given function is $f\left(x\right)=\mathrm{cos}x$

Using the Lagrange form of remainder, we have,

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

Now, we will find fifth derivative of the function.

${f}^{1}\left(x\right)=-\mathrm{sin}x\phantom{\rule{0ex}{0ex}}{f}^{2}\left(x\right)=-\mathrm{cos}x\phantom{\rule{0ex}{0ex}}{f}^{3}\left(x\right)=\mathrm{sin}x\phantom{\rule{0ex}{0ex}}{f}^{4}\left(x\right)=\mathrm{cos}x\phantom{\rule{0ex}{0ex}}{f}^{5}\left(x\right)=-\mathrm{sin}x$

Thus, we get,

${R}_{4}\left(x\right)=-\frac{\mathrm{sin}c}{120}{x}^{5}$

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