# Extreme valueVerify if the function has an extreme value and if it is true, determine the extreme of f(x) = (x^2-2)/(2x-1) ?

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### 1 Answer

We recall that a function has a local extreme for all the critical values. The critical value x is the root of the first derivative of the function.

We'll differentiate the function to get the roots 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.

**Since the function f(x) is strictly increasing over the real set of numbers, then it has no extreme values.**