Question 199967
Is the function {{{f(x)=(3/2)x-1}}} ??? If so, then....



{{{f(x)=(3/2)x-1}}} Start with the given function



{{{y=(3/2)x-1}}} Replace f(x) with "y"



{{{x=(3/2)y-1}}} Swap x and y



{{{x+1=(3/2)y}}} Add 1 to both sides.



{{{2(x+1)=3y}}} Multiply both sides by 2.



{{{(2/3)(x+1)=y}}} Divide both sides by 3.



So the inverse is *[Tex \LARGE f^{-1}(x)=\frac{2}{3}\left(x+1\right)]



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Or....



Is the function {{{f(x)=3/(2x-1)}}} ??? If so, then....



{{{f(x)=3/(2x-1)}}} Start with the given function



{{{y=3/(2x-1)}}} Replace f(x) with "y"



{{{x=3/(2y-1)}}} Swap x and y



{{{x(2y-1)=3}}} Multiply both sides by {{{2y-1}}}.



{{{2xy-x=3}}} Distribute



{{{2xy=3+x}}} Add "x" to both sides.



{{{y=(3+x)/(2x)}}} Divide both sides by {{{2x}}}.



So the inverse is *[Tex \LARGE f^{-1}(x)=\frac{3+x}{2x}]