The function y = x^2/(2x-1)-2/(2x-1)

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

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

To find the roots of y'

[2x(2x - 1) - (x^2 - 2)*2]/(2x - 1)^2 = 0

=> 4x^2 - 2x - 2x^2 + 4 = 0

=> 2x^2 - 2x + 4 = 0

=> x^2 - x + 2 = 0

x1 = 1/2 + sqrt (1 - 8)/2

=> x1 = 1/2 + i*(sqrt 7)/2

x2 = 1/2 - i*(sqrt 7)/2

**The roots of the derivative of y are 1/2 + i*(sqrt 7)/2 and 1/2 - i*(sqrt 7)/2**

We'll re-write the function:

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

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

Now, 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 equation f'(x) = 0 has no roots.**