Find a formula for the inverse of the function.

y=x^2-x, x>=(greater than or equal to) 1/2

Please give me a step by step explanation. I think my algebra is wrong... Ty

x = y^2 - y

y^2 - y + 1 = x + 1

(y-1)^2 = x + 1

y -1 = sqrt (x+1)

y = 1 + sqrt (x+1)

noticed an error from line 2 to line 3

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

y - 1/2 = ±√(4x+1)/2

y = 1/2 ± √(4x+1)/2

= (1 ± √(4x+1) )/2 , x ≥ -1/4

Whoops !

I don't understand. Why are you adding 1/4? to both sides? I don't see how you are getting a quadratic formula from this. More importantly, thank you guys for the help!

To find the inverse of a function, you need to swap the variables x and y, and then solve for y. Let's go through the steps together.

Step 1: Swap the variables:
Instead of having y = x^2 - x, we will write it as x = y^2 - y.

Step 2: Solve for y:
Rearrange the equation to isolate y:
x = y^2 - y
Now, let's bring all the terms to one side:
y^2 - y - x = 0

Step 3: Apply the quadratic formula:
To solve for y, we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = -1, and c = -x.
Substituting these values into the quadratic formula, we get:
y = [1 ± √(1 - 4(-x))]/2

Step 4: Simplify the square root:
Inside the square root, we have 1 - 4(-x). Simplifying this further:
y = [1 ± √(1 + 4x)]/2

So, the inverse function of y = x^2 - x, given that x >= 1/2, is:
f^(-1)(x) = [1 ± √(1 + 4x)]/2