The extreme values of a function occur at the points where the derivative is equal to 0.

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

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

6x^2 + 6x - 12 = 0

=> x^2 + x - 2 = 0

=> x^2 + 2x - x - 2 = 0

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

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

x = 1 and x = -2

f(x) = -2 for x = 1 and f(x) = 25 for x = -2.

**The extreme values are x1 = -2 and x2 = 25.**

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.**