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Answers without the blur. Sign up and see all textbooks for free! Q. 38

Expert-verified Found in: Page 261 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Use a sign chart for ${\mathbit{f}}^{\mathbf{\text{'}}}$ to determine the intervals on which each function $\mathbit{f}$ is increasing or decreasing. Then verify your algebraic answers with graphs from a calculator or graphing utility.$\mathbit{f}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}$

Ans: Increasing interval $\mathbf{\left[}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\mathbf{\pi k}\mathbf{,}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\mathbf{\pi k}\mathbf{\right]}$

and decreasing elsewhere.

See the step by step solution

## Step 1. Given information:

$\mathbit{f}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{sinx}}$

## Step 2. Finding the derivative of the function:

$f\left(x\right)=\frac{1}{\mathrm{sin}x}=\mathrm{cos}ecx\phantom{\rule{0ex}{0ex}}{f}^{\text{'}}\left(x\right)=-\mathrm{cos}ecx.cotx\phantom{\rule{0ex}{0ex}}let{f}^{\text{'}}\left(x\right)=0\phantom{\rule{0ex}{0ex}}\therefore -\mathrm{cos}ecx.cotx=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}ecx.cotx=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sin}x}.\frac{\mathrm{cos}x}{\mathrm{sin}x}=0\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{cos}x}{{\mathrm{sin}}^{2}x}=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}x=0\phantom{\rule{0ex}{0ex}}x=\left(2k+1\right)\frac{\pi }{2}\left[wherekisanyinteger\right]\phantom{\rule{0ex}{0ex}}takingpointx=0$

## Step 3. Finding increasing and decreasing intervals:

Intervals of the given function :

${f}^{\text{'}}\left(x\right)hasx=0\phantom{\rule{0ex}{0ex}}{f}^{\text{'}}\left(0\right)=\mathrm{cos}\left(2.0\right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}0\phantom{\rule{0ex}{0ex}}=1>0\phantom{\rule{0ex}{0ex}}{f}^{\text{'}}\left(\frac{\pi }{2}\right)=\mathrm{cos}\left(2.\frac{\pi }{2}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\pi \phantom{\rule{0ex}{0ex}}=-1<0\phantom{\rule{0ex}{0ex}}$

f(x)is increasing on the interval $\mathbf{\left[}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\mathbf{\pi k}\mathbf{,}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\mathbf{\pi k}\mathbf{\right]}$

and decreasing elsewhere.

## Step 4. Verifying algebraic answers with graphs :  ### Want to see more solutions like these? 