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Q. 39

Expert-verified
Found in: Page 261

Calculus

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

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

See the step by step solution

Step 1. Given Information

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

Step 2. Finding the derivative of the function

Rewriting the function by simplifying it,

$f\left(x\right)={\left(x-2\right)}^{2}\left(x+1\right)\phantom{\rule{0ex}{0ex}}f\left(x\right)=\left({x}^{2}-4x+4\right)\left(1+x\right)\phantom{\rule{0ex}{0ex}}f\left(x\right)={x}^{2}+{x}^{3}-4x-4{x}^{2}+4+4x\phantom{\rule{0ex}{0ex}}f\left(x\right)={x}^{3}-3{x}^{2}+4\phantom{\rule{0ex}{0ex}}f\text{'}\left(x\right)=3{x}^{2}-6x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}lets,f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}\therefore 3{x}^{2}-6x=0\phantom{\rule{0ex}{0ex}}3x\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}x=\left\{0,2\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}substitutex=-1,1,3intof\text{'}\left(x\right)=3{x}^{2}-6x\phantom{\rule{0ex}{0ex}}f\text{'}\left(-1\right)=3{\left(-1\right)}^{2}-6\left(-1\right)\phantom{\rule{0ex}{0ex}}=9>0\phantom{\rule{0ex}{0ex}}f\text{'}\left(1\right)=3{\left(1\right)}^{2}-6\left(1\right)\phantom{\rule{0ex}{0ex}}=-3<0\phantom{\rule{0ex}{0ex}}f\text{'}\left(3\right)=3{\left(3\right)}^{2}-6\left(3\right)\phantom{\rule{0ex}{0ex}}=9>0$

Step 3. Finding local extrema of the function on the number line(Sign chart):

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