[SOLVED] Verifying Inverse of this Function

Sep 2009
148
2
\(\displaystyle g(x)=\frac{2x-1}{x+3}\)

My Attempt At The Inverse

Let \(\displaystyle g(x)=z\)

\(\displaystyle (x+3)(z)=2x-1\)

\(\displaystyle xz+3z=2x-1\)

\(\displaystyle 1+3z=2x-xz\)

\(\displaystyle 1+3z=x(2-z)\)

\(\displaystyle \frac{1+3z}{2-z}=x\)

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

Where did I make a mistake?

Thanks!
 

mr fantastic

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

My Attempt At The Inverse

Let \(\displaystyle g(x)=z\)

\(\displaystyle (x+3)(z)=2x-1\)

\(\displaystyle xz+3z=2x-1\)

\(\displaystyle 1+3z=2x-xz\)

\(\displaystyle 1+3z=x(2-z)\)

\(\displaystyle \frac{1+3z}{2-z}=x\)

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

Where did I make a mistake?

Thanks!
Why do you think there's a mistake. There is no mistake.
 
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