What is the maximum value of f(x) = (x-3)/ (2x^2 - 5)

To calculate the maximum value of the given function, first
we'll have to determine the critical points.

For this reason, we'll
determine the first derivative of the function, knowing that the critical point are the roots of
the first derivative.

Since the function is a ratio, we'll apply the
quotient rule:

f'(x) = [(x-3)'*(2x^2 - 5) -  (x-3)*(2x^2 -
5)']/(2x^2 - 5)^2

 f'(x) = [(2x^2 - 5) - 4x(x-3)]/(2x^2 -
5)^2

We'll remove the brackets:

 f'(x)
= (2x^2 - 5 - 4x^2 + 12x)/(2x^2 - 5)^2

We'll combine like
terms:

 f'(x) = (-2x^2 + 12x - 5)/(2x^2 -
5)^2

We'll put f'(x) = 0

Since the
denominator is always positive, for any value of x, we'll find the roots of the
numerator:

2x^2 - 12x + 5 = 0

We'll
apply the quadratic formula:

x1 =
[12+sqrt(144-40)]/4

The critical values of the
function are:

x1 =
(6+sqrt26)/2

x2 =
(6-sqrt26)/2

The extreme values of the
function are:

f(x1) =
((6+sqrt26)/23)/(2((6+sqrt26)/2)^2 -
5)

f(x2) =
((6-sqrt26)/23)/(2((6-sqrt26)/2)^2 - 5)