Find the extreme values of f(x) = 3+3x - 3x^2

f(x) = 3+3x-3x^2. To determine maximum or minimum of
f(x).

We rewitte the given function as below in order that
the right side becomes like 

f(x) =  3  - 3
(x^2-x))

f(x) = 3 - 3{x^2 - 2* (1/2)x +(1/2)^2 } +
3(1/2)^2, where we subtracted  (1/2)^2  and added (1/2)^2  without altering the value of
the right side.

f(x) = 3+3(1/2)^2  -3
(x-1/2)^2

f(x) = (15/4) - 3(x-1/2)^2. Here the right side
3(x-1/2)^2  is a perfect square and becomes 0 at minimum when x= 1/2. and  3(x-1/2)^2 
> 0 . for all x positive or negative.

Therefore f(x)
=  15/4 - a positive quantity.

Therefore f(x) < 15/4
for all x.

Therefore f(x) < 15/4 is the maximum when
x = 1/2.

Alternative method
:

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

By
calculus we know that when f'(c) = 0  , f(x) either maximum or minimum. Further if f " (
c) <  0 , then f(C) is maximum and if f'(c) > 0
.

So f'(x) = 0 gives: (3+3x-3x^2)' = 0. 3-6x = 0. So  x = c
= 3/6 = 1/2.

f"(x) = (3-6x)' = -6 for all x. So f"(1/2) =
-6 which is < 0.

Therefore , f(1/2) =
3+3(1/2)-3(1/2)^2 = 15/4.