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

### 2 Answers | Add Yours

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

To find extreme values, first we need to determine the first derivative.

f'(x) = 3 - 6x

Now we need to determine the critical values which they are the derivative's zeros.

==> 3- 6x = 0

==> 3= 6x

**==> x= 2**

**Then we know that the function has an extreme value when x= 2**

**Since the sign of x^2 is negative, then the function has a MAXIMUM value at x= 2.**

**==> f(2) = 3+ 3*2 - 3*2^2**

** = 3+ 6 - 12 = -3**

**Then the function has a maximum values at f(2) = -3**

**Or the point ( 2, -3) is the maximum value for f(x) **

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.