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.

$f\left(x\right)={x}^{\frac{1}{3}}(x+4)\text{}on\text{}[-27,27]$

Philip Williams

Beginner2021-12-13Added 39 answers

Step 1

Absolute maxima and minima:

$f\left(x\right)={x}^{\frac{1}{3}}(x+4),[-27,27]$

${f}^{\prime}\left(x\right)={x}^{\frac{1}{3}}\cdot \frac{d}{dx}(x+4)+(x+4)\cdot \frac{d}{dx}\left({x}^{\frac{1}{3}}\right)$

$f}^{\prime}\left(x\right)={x}^{\frac{1}{3}}+(x+4)\cdot \frac{1}{3}\cdot {x}^{\frac{-2}{3}$

$f{}^{\u2033}\left(x\right)=\frac{1}{3}{x}^{\frac{-2}{3}}+\frac{x+4}{3}\cdot \left(\frac{-2}{3}\right)\cdot {x}^{\frac{-5}{3}}$

$=\frac{1}{3}{x}^{\frac{-2}{3}}-\frac{2(x+4)}{9}\cdot {x}^{-\frac{5}{3}}>0$

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).

Absolute maxima and minima:

Step 2

To find maxima or minima,

f'(x)=0

x=-1

f(x)=-3

The absolute minima is (-1,-3).

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