What is the extreme value of 2x^3+3x^2-12x+5?

To determine the extreme values of the given function,
we'll have to calculate the critical points of the
expression.

For this reason, we'll determine the first
derivative, since the critical points are the roots of the first
derivative.

f'(x) =
(2x^3+3x^2-12x+5)'

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

We'll put f'(x) = 0:

6x^2 +
6x - 12 = 0

We'll divide by
6:

x^2 + x - 2 = 0

We'll apply
the quadratic formula:

x1 = [-1+sqrt(1 +
8)]/2

x1 = (-1+sqrt9)/2

x1 =
(-1+3)/2

x1 = 1

x2 =
(-1-3)/2

x2 = -2

The extreme
values of the function are:

f(1) =
2*1^3+3*1^2-12*1+5

f(1) = 2 + 3 - 12 +
5

f(1) = -2

f(-2) =
2(-2)^3+3(-2)^2-12(-2)+5

f(-2) = -16+ 12+24 +
5

f(-2) = 25

The
extreme values of the function are: f(1) = -2 and f(-2) =
25.