Find the absolute maximum and minimum values of `f(x)=2x^3-3x^2-12x` on interval [-2,3].

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lemjay | High School Teacher | (Level 3) Senior Educator

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`f(x)=2x^3-3x^2-12x`

To start, take the derivative of f(x).

`f'(x)=6x^2-6x-12`

Then, set f'(x) equal to zero.

`0=6x^2-6x-12`

To simplify, divide both sides by 6.

`0=x^2-x-2`

Then, factor.

`0=(x-2)(x+1)`

Set each factor to zero and solve for x.

`x-2=0`                   and                      `x+1=0`

    `x=2`                                                    `x=-1`

Substitute values of f(x) to determine y.

`x=-1 ` ,        `y=2x^3-3x^2-12x=2(-1)^3-3(-1)^2-12(-1)=7`

`x=2`  ,          `y=2x^3-3x^2-12x=2(2)^3-3(2)^2-12(2)=-20`

Note that the value of y when x=-1 is greater than the value of when x=2.

Hence, at interval [-2,3], the absolute maximum is the point (-1,7) and the absolute minimum is (2,-20).

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