What are the extreme values of the function f(x) = 2x-3x^2 + 5.

Given the parabola f(x) = 2x- 3x^2 +
5.

We need to determine the extreme values of the
function.

We know that the extreme values are the values of
f(x) such that the function has maximum or minimum
points.

To find extreme values, we need to differentiate
f(x) and determine the critical values.

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

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

The critical values are the derivative's
zeros.

==> -6x + 2 =
0

==> -6x =
-2

==> x = -2/-6 =
1/3

Then the function has an extreme value when x=
1/3

==> f(1/3) = -3 ( 1/3)^2 + 2( 1/3) +
5

                 = -1/3 + 2/3 +
5

                  = ( -1 + 2 + 15) /3 =
16/3

Since the sign of x^2 is negative, the the function
has a maximum point.

Then, the maximum point
is f(1/3) = 16/3.