Post a New Question

CALCULUS

posted by on .

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

  • CALCULUS - ,

    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)

  • CALCULUS - ,

    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

  • CALCULUS - ,

    Whoops !

  • CALCULUS - ,

    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!

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question