What are the extremes of f(x)=ln(2x^2-20x+53)?

The local extremes of a function are the maximum or
minimum points and they could be found in the critical points. Critical points are the
roots of the 1st derivative of the function.

We'll
differentiate the function using the chain rule.

f'(x) =
(2x^2-20x+53)'/(2x^2-20x+53)

f'(x) = (4x -
20)/(2x^2-20x+53)

We'll solve the equation f'(x) =
0.

(4x - 20)/(2x^2-20x+53) =
0

This equation is cancelling when the numerator is
cancelling.

4x - 20 = 0

4x =
20

x = 5

The critical point of
the function is x = 5

For x<5 =>
f'(x)<0 =>  f(x) is decreasing.

For
x>5, f'(x)>0 => f(x) is
increasing.

Then x = 5 is a minimum point of the
function.

We'll substitute x by 5 in the expression of
f(x).

f(5) = ln
3

The extreme point of the function is a
minimum point and it has the coordinates (5 ; ln
3).