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Q. 39

Expert-verifiedFound in: Page 261

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Use the first derivative test to determine the local extrema of each function in Exercises 39- 50. Then verify your algebraic answers with graphs from a calculator or graphing utility.

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}$

Ans: The local extrema value of f(x) are $x=\left\{0,4\right\}$

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}$

Rewriting the function by simplifying it,

$f\left(x\right)={(x-2)}^{2}(x+1)\phantom{\rule{0ex}{0ex}}f\left(x\right)=({x}^{2}-4x+4)(1+x)\phantom{\rule{0ex}{0ex}}f\left(x\right)={x}^{2}+{x}^{3}-4x-4{x}^{2}+4+4x\phantom{\rule{0ex}{0ex}}f\left(x\right)={x}^{3}-3{x}^{2}+4\phantom{\rule{0ex}{0ex}}f\text{'}\left(x\right)=3{x}^{2}-6x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}lets,f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}\therefore 3{x}^{2}-6x=0\phantom{\rule{0ex}{0ex}}3x(x-2)=0\phantom{\rule{0ex}{0ex}}x=\{0,2\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}substitutex=-1,1,3intof\text{'}\left(x\right)=3{x}^{2}-6x\phantom{\rule{0ex}{0ex}}f\text{'}(-1)=3{(-1)}^{2}-6(-1)\phantom{\rule{0ex}{0ex}}=9>0\phantom{\rule{0ex}{0ex}}f\text{'}\left(1\right)=3{\left(1\right)}^{2}-6\left(1\right)\phantom{\rule{0ex}{0ex}}=-3<0\phantom{\rule{0ex}{0ex}}f\text{'}\left(3\right)=3{\left(3\right)}^{2}-6\left(3\right)\phantom{\rule{0ex}{0ex}}=9>0$$Fromthegraphidentifythelocalminimumatx=2indicatedonthechart\phantom{\rule{0ex}{0ex}}withthefirstderivativetest,\mathrm{sin}cef\text{'}\left(x\right)changessignfromnegativeto\phantom{\rule{0ex}{0ex}}positiveatx=2.\phantom{\rule{0ex}{0ex}}f\left(0\right)=(0-2{)}^{2}(1+0)\phantom{\rule{0ex}{0ex}}=4(localmaximum\left)\phantom{\rule{0ex}{0ex}}f\right(2)=(2-2{)}^{2}(1+2)\phantom{\rule{0ex}{0ex}}=0(localminimum)\phantom{\rule{0ex}{0ex}}\therefore theextremevaluesofthefunctionarex=\{0,4\}$

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