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Answers without the blur. Sign up and see all textbooks for free! 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): $Fromthegraphidentifythelocalminimumatx=2indicatedonthechart\phantom{\rule{0ex}{0ex}}withthefirstderivativetest,\mathrm{sin}cef\text{'}\left(x\right)changessignfromnegativeto\phantom{\rule{0ex}{0ex}}positiveatx=2.\phantom{\rule{0ex}{0ex}}f\left(0\right)=\left(0-2{\right)}^{2}\left(1+0\right)\phantom{\rule{0ex}{0ex}}=4\left(localmaximum\right)\phantom{\rule{0ex}{0ex}}f\left(2\right)=\left(2-2{\right)}^{2}\left(1+2\right)\phantom{\rule{0ex}{0ex}}=0\left(localminimum\right)\phantom{\rule{0ex}{0ex}}\therefore theextremevaluesofthefunctionarex=\left\{0,4\right\}$

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