Concepcion Hale

2021-12-12

Absolute maxima and minima Determine the location and value of the absolute extreme value of ƒ on the given interval, if they exist.

Philip Williams

Step 1
Absolute maxima and minima:
$f\left(x\right)={x}^{\frac{1}{3}}\left(x+4\right),\left[-27,27\right]$
${f}^{\prime }\left(x\right)={x}^{\frac{1}{3}}\cdot \frac{d}{dx}\left(x+4\right)+\left(x+4\right)\cdot \frac{d}{dx}\left({x}^{\frac{1}{3}}\right)$
${f}^{\prime }\left(x\right)={x}^{\frac{1}{3}}+\left(x+4\right)\cdot \frac{1}{3}\cdot {x}^{\frac{-2}{3}}$
$f{}^{″}\left(x\right)=\frac{1}{3}{x}^{\frac{-2}{3}}+\frac{x+4}{3}\cdot \left(\frac{-2}{3}\right)\cdot {x}^{\frac{-5}{3}}$
$=\frac{1}{3}{x}^{\frac{-2}{3}}-\frac{2\left(x+4\right)}{9}\cdot {x}^{-\frac{5}{3}}>0$
Step 2
To find maxima or minima,
f'(x)=0
${x}^{\frac{1}{3}}+\frac{x+4}{3}\cdot {x}^{\frac{-2}{3}}=0$
$3{x}^{\frac{1}{3}}+{x}^{\frac{1}{3}}+4{x}^{\frac{-2}{3}}=0$
$4{x}^{\frac{1}{3}}=-4{x}^{\frac{-2}{3}}$
${x}^{\frac{1}{3}}=-{x}^{\frac{-2}{3}}$
x=-1
f(x)=-3
The absolute minima is (-1,-3).

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