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

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.