Find the extreme values for the function:f(x) = 3x^2 - 5x + 3

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

First we
will differentiate .

f'(x) = 6x -
5 

We will find the deicative's
zeros.

==> 6x - 5 =
0

==> x= 5/6

Then the
function has an extreme value when x= 5/6

==> f(5/6)
= 3*(5/6)^2  - 5(5/6) +3

                  = 3*25/36  -
25/6 + 3

                   = (75 - 150 +
108)/3

                   =  33/3 =
11

Since the factor for x^2 is positive, then
the function has MINIMUM value at f(5/6) = 11