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Q. 37

Expert-verifiedFound in: Page 261

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Use a sign chart for ${\mathit{f}}^{\mathbf{\text{'}}}$ to determine the intervals on which each function $\mathit{f}$ is increasing or decreasing. Then verify your algebraic answers with graphs from a calculator or graphing utility.

role="math" localid="1648370582124" $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{.}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$

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

and decreasing elsewhere.

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{.}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$

$f\left(x\right)=\mathrm{sin}x.\mathrm{cos}x\phantom{\rule{0ex}{0ex}}{f}^{\text{'}}\left(x\right)=\mathrm{sin}x(-\mathrm{sin}x)+\mathrm{cos}x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}{f}^{\text{'}}\left(x\right)={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\phantom{\rule{0ex}{0ex}}{f}^{\text{'}}\left(x\right)=\mathrm{cos}2x\phantom{\rule{0ex}{0ex}}let{f}^{\text{'}}\left(x\right)=0\phantom{\rule{0ex}{0ex}}\therefore \mathrm{cos}2x=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=\left(2k+1\right)\frac{\pi}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\left(2k+1\right)\frac{\pi}{4}[wherekisanyinteger]\phantom{\rule{0ex}{0ex}}takingpointx=0$

Intervals of the given function :

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

and decreasing elsewhere.

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