What is the extreme value of the function f(x) = (x^2-2)/(2x-1)

A function has an extreme local for the critical value x.
The critical value x is the root of the first derivative of the
function.

So, we need to differentiate the function for
finding the roots of the expression of the first
derivative.

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

f'(x) = [(x^2-2)'*(2x-1)
- (x^2-2)*(2x-1)']/(2x-1)^2

f'(x) = [2x(2x-1) - 2(x^2 -
2)]/(2x-1)^2

We'll remove the
brackets:

f'(x) = (4x^2 - 2x - 2x^2 +
4)/(2x-1)^2

We'll combine like
terms:

f'(x) = (2x^2 - 2x +
4)/(2x-1)^2

We'll determine the critical values for f'(x).
For this reason we'll put f'(x) = 0.

(2x^2 - 2x +
4)/(2x-1)^2 = 0

Since the denominator is always positive,
for any value of x, only the numerator could be zero.

2x^2
- 2x + 4 = 0

We'll calculate
delta:

delta = b^2 - 4ac

We'll
identify a,b,c:

a = 2 , b = -2 , c =
4

delta = 4 - 32 = -28 <
0

Since delta is negative and a = 2>0, the
expression 2x^2 - 2x + 4 is always positive for any avlue of
a.

So, the first derivative is positive and
it is not cancelling for any value of a. The function f(x) is increasingly over R set
and it has no extreme values.