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

Expert Answers
lemjay eNotes educator| Certified Educator


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


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


To simplify, divide both sides by 6.


Then, factor.


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).