# [SOLVED] Verifying Inverse of this Function

• May 8th 2010, 12:20 PM
unstopabl3
[SOLVED] Verifying Inverse of this Function
$g(x)=\frac{2x-1}{x+3}$

My Attempt At The Inverse

Let $g(x)=z$

$(x+3)(z)=2x-1$

$xz+3z=2x-1$

$1+3z=2x-xz$

$1+3z=x(2-z)$

$\frac{1+3z}{2-z}=x$

$g^-1(x)=\frac{1+3x}{2-x}$

Where did I make a mistake?

Thanks!
• May 8th 2010, 02:11 PM
mr fantastic
Quote:

Originally Posted by unstopabl3
$g(x)=\frac{2x-1}{x+3}$

My Attempt At The Inverse

Let $g(x)=z$

$(x+3)(z)=2x-1$

$xz+3z=2x-1$

$1+3z=2x-xz$

$1+3z=x(2-z)$

$\frac{1+3z}{2-z}=x$

$g^-1(x)=\frac{1+3x}{2-x}$

Where did I make a mistake?

Thanks!

Why do you think there's a mistake. There is no mistake.
• May 9th 2010, 02:52 AM
unstopabl3
Okay, great then!

Thanks for the verification!