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Answers without the blur. Sign up and see all textbooks for free! Q. 23

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

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

See the step by step solution

## Step 1. Given Information

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

## Step 2. Explanation

Using the Lagrange form of remainder, we have,

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

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,

${R}_{4}\left(x\right)=-\frac{\mathrm{sin}c}{120}{x}^{5}$ ### Want to see more solutions like these? 