Pearl Carney

2021-12-04

Find the absolute maximum and minimum values of the following function over the indicated interval, and indicate the x-values at which they occur.

$f\left(x\right)=\frac{1}{3}{x}^{3}+\frac{5}{2}{x}^{2}-6x+8;[-9,3]$

The absolute maximum value is$B\otimes$ at x= $B\otimes$

The absolute minimum value is$B\otimes$ at x= $B\otimes$

The absolute maximum value is

The absolute minimum value is

Vincent Diaz

Beginner2021-12-05Added 19 answers

Step 1

Solution:

Given$f\left(x\right)=\frac{1}{3}{x}^{3}+\frac{5}{2}{x}^{2}-6x+8\text{}[-9,3]$

To find Absolute maximum and minimum

Formula:

$\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$

Step 2

$f\left(x\right)=\frac{1}{3}{x}^{3}+\frac{5}{2}{x}^{2}-6x+8\text{}[-9,3]$

Now differentiating with respect to x on both side

${f}^{\prime}\left(x\right)={x}^{2}+5x-6$

${x}^{2}+5x-6=0$

x=1, x=-6

when x=1$f\left(1\right)=\frac{1}{3}+\frac{5}{2}-6+8=4.83$

x=-6 f(-6)=62

x=-9 f(-9)=21.2

x=3 f(3)=21.5

at x=1 the f(x) is absolute minimum

and the minimum value is 4.83

at x=-6 the f(x) is absolute maximum

and the maximum value is 62

Solution:

Given

To find Absolute maximum and minimum

Formula:

Step 2

Now differentiating with respect to x on both side

x=1, x=-6

when x=1

x=-6 f(-6)=62

x=-9 f(-9)=21.2

x=3 f(3)=21.5

at x=1 the f(x) is absolute minimum

and the minimum value is 4.83

at x=-6 the f(x) is absolute maximum

and the maximum value is 62

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