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

To find th extreme values for the function, first we need to determin th first derivative:

f'(x) = 2x - 5

Now we need to find th critical value which is th function's zero.

==> 2x - 5 = 0

==> 2x = 5

==> x= 5/2

Then the function has an extreme values when x= 5/2

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

= 25/4 - 25/2 + 3

= (25 - 50 + 12) /4

= 13/4

Then the function has a minimum value at f(5/2) = 12/4

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

We proceed to show that the right side of this equation can be a square expression minus a constant.

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

We choose , k sich that 2kx = -5x. So that k = -5.

Therefore f(x) = (x-5/2)^2 - (5/2)^2+3 = 0.

f(x) = (x-5/2)^2 - 25/4 + 3 = 0,

f(x) = (x-5/2)^2 - 13/4 = 0.

Therfore f(x) = a square expression minus 13/4. The right side is minimum when square expression (x-5/2)^2 is zero. Otherwise the squre expression being positive, f(x) > = -13/4 for all x.

So f(x) is minimum when x= 5/2 and the minimum f(x) = f(5/2) = -13/2. So x = is the value when f(x) at the lower extreme. f(x) grows to infinity as x--> infinity. As such, there is no upper extreme value.