Find the inverse of function f(x) = x^2 + 2x - 1

[LaurenM]

I'm given f(x) = x^2 + 2x - 1 and need to find the inverse. I have gotten this far:

substituted y for f(x): y = x^2 + 2x - 1

interchanged x and y: x = y^2 + 2y - 1

solving for y:
. . .x + 1 = y^2 + 2y
. . .x + 1 = y(y + 2)
. . .(x + 1)/y = y + 2

Not sure where to go from there or if the steps I did leading up to it are the best way to go. Thanks for any help in advance!

 

[galactus]

Hello Lauren:

\(\displaystyle \L\\x=y^{2}+2y-1\)

add 1:

\(\displaystyle \L\\x+1=y^{2}+2y\)

COMPLETE THE SQUARE by adding half the coefficient of y to both sides:

\(\displaystyle \L\\x+2=y^{2}+2y+1\)

Factor:

\(\displaystyle \L\\x+2=(y+1)^{2}\)

\(\displaystyle \L\\\sqrt{x+2}=y+1\)

\(\displaystyle \L\\y=-(\sqrt{x+2}+1) \;\ or \;\ sqrt{x+2}-1\)

 

[stapel]

COMPLETE THE SQUARE....

...or plug-n-chug with the Quadratic Formula. Either way should work!

Eliz.

 

[Deleted member 4993]

x = y^2 + 2y - 1

You need to solve for 'y' - so lump 'x' with constant term.

y^2 + 2y - (x+1) = 0

Using quadratic formula (plug-n-chug - following staple's advice:

y = [-2 ± sqrt{4 + 4(x+1)}]/2 = -1 ± sqrt(x+2)