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find the inverse of f(x) = (7x-3/2x-1) + 2
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\[f(x)=\frac{7x-3}{2x-1}+2\]?
if this is the problem, it might be easier to start with \[f(x)=\frac{11x-5}{2x-1}\]to make the algebra go smoother. then write \[x=\frac{11y-5}{2y-1}\] and solve for y via \[2xy-x=11y-5\] \[-x+5=11y-2xy=(11-2x)y\] \[\frac{5-x}{11-2x}=y\] and so \[f^{-1}(x)=\frac{x-5}{2x-11}\]
how did you get 11x-5/2x-1?
oh i see now! are you allowed to do that?
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