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find the f-1(x) of f(x) =2x+1 over x-3
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\[y= \frac{ 2x+1}{x-3}\]
swap x and y
\[x = \frac{2y +1}{y-3}\]
now x(y-3) = 2y+1 xy -3x = 2y +1 Now get the y's together and factor xy-2y = 1+3x y(x-2) = 1+3x y= (1+3x) / (x-2)
f ^(-1) (x) = (1+3x) / (x-2)
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