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

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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.**

f(x) = (x^2-2)/(2x-1).

We need to find the extreme value, if any, for the function f(x).

First, we need to determine the first derivative of the function.

==> We will use the quotient rule to find the derivative.

==> Let f(x) = u/ v such that.

u = x^2 -2 ==> u' = 2x

v = 2x -1 ==> v' = 2

==> f'(x) = u'v - uv'/ v^2

= 2x(2x-1) - ( 2(x^2 -2))/ (2x-1)^2

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

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

= 2(x^2 -x + 2) /(2x-1)^2

Now we will find the critical values which is the derivative zeros.

==> (x^2 -x + 2 ) = 0

==> ( x-2)(x+1) = 0-

==> x =2 and x = -1

Then the function has 2 extreme values at x= -1 and x= 2

**==> f(-1) = 1/5**

**==> f(2) = 2/3**

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