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

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### 2 Answers

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**

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

The extreme value of the function is at x = c for which f'(c) = 0 . Further , if f"(c) > 0 , then f(c) is the mimum.

Therefore we find f'(x) and equate it to zero and solve x.

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

f'(x) = 6x -5

So f'(x) = 0 gives: 6x =5. Or x = 5/6.

f"(x) = (6x-5)' = 6. So f"(5/6) = 6 which is > 0.

Therefore f(x) is minimum at x = 5/6) .

Minimum f(x) = f(5/6) = 3(5/6)^2 -5(5/6)+3 = 25/12 -25/6 +3 = 25/12-50/12 +36/12 = 11/12.

Th