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If f(x)=2x-5 and g(x)=(1+3x)/(5-2x) compute g(f^-1(z))
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well first solve y=2x-5 for x
y=(x+5)/2
cool \[f^{-1}(x)=\frac{x+5}{2} \\ \text{ so } f^{-1}(z)=\frac{z+5}{2}\]
\[g(\frac{z+5}{2})=?\]
just replace all of those x's in that g thingy with (z+5)/2
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i got 17+3z/(-2z)
\[\frac{1+3\frac{z+5}{2}}{5-2 \frac{z+5}{2}} \\ \frac{1+\frac{3(z+5)}{2}}{5-(z+5)} \\ \frac{1+\frac{3(z+5)}{2}}{5-z-5}\] did you get that far?
\[\frac{1+\frac{3(z+5)}{2}}{-z}\] yep what you have seems right good job
thanks again
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