[SOLVED] Verifying Inverse of this Function

**unstopabl3** · May 8th 2010, 12:20 PM

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

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!

**mr fantastic** · May 8th 2010, 02:11 PM

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.

**unstopabl3** · May 9th 2010, 02:52 AM

Okay, great then!

Thanks for the verification!

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