Finding the inverse of a function

**user_5** · November 25th 2009, 11:28 PM

I need help finding the inverse for this function

$g:[-1,\infty) \rightarrow, g(x) = x^2+2x$

What I've done so far.

Let $g(x) = y$

$y=x^2+2x$

Swap x and y, and solve for y

$x=y^2+2y$

$x=y(y+2)$

I'm stuck here, can anyone please help?

Nevermind, I got it, had to complete the square!

**rowe** · November 26th 2009, 02:11 AM

I had typed this up by the time I saw your last comment, d'oh- so I post anyways.

$g(x) = x^2+2x$

Set g(x) to y:

$y = x^2+2x$

Complete that square:

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

$y + 1 = (x+1)^2$

$\sqrt{y+1} - 1 = x$

Swap x and y, so that $y=g^{-1}(x) = \sqrt{y+1}-1$

Prove to check; it should be that f(g(x)) = g(f(x)) = x, here's an example:

$g(f(x)) = \sqrt{x^2+2x+1} - 1$

$g(f(x)) = \sqrt{(x+1)^2} - 1$

$g(f(x)) = x+1 - 1$

$g(f(x)) = x$

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