first, realize that the inverse f⁻¹ means switch x and y
so what you do is switch x and y and solve for y
so
y=f(x)
y=(x+2)/(x-3)
switch
x=(y+2)/(y-3)
times both sides by (y-3)
x(y-3)=y+2
expand
xy-3x=y+2
minus xy both sides
-3x=y-xy+2
minus 2 both sides
-3x-2=y-xy
undistribute y
-3x-2=y(1-x)
divide both sides by (1-x)
\(\frac{-3x-2}{1-x}=y\)
\(\frac{3x+2}{x-1}=y\)
\(f^{-1} \frac{3x+2}{x-1}\)
now show that f(f⁻¹(x))=x
\(\frac{x}{y}\)
\(f(f^{-1}(x))= \frac{\frac{3x+2}{x-1}+2}{\frac{3x+2}{x-1}-3}\)
\(f(f^{-1}(x))= \frac{\frac{3x+2}{x-1}+\frac{2(x-1)}{x-1}}{\frac{3x+2}{x-1}-\frac{3(x-1)}{x-1}}\)
\(f(f^{-1}(x))= \frac{\frac{3x+2+2(x-1)}{x-1}}{\frac{3x+2-3(x-1)}{x-1}}\)
\(f(f^{-1}(x))= \frac{\frac{3x+2+2x-2}{x-1}}{\frac{3x+2-3x+3}{x-1}}\)
\(f(f^{-1}(x))= \frac{\frac{5x}{x-1}}{\frac{5}{x-1}}\)
\(f(f^{-1}(x))= (\frac{5x}{x-1})(\frac{(x-1)}{5})\)
\(f(f^{-1}(x))= \frac{(5)(x)(x-1)}{(x-1)(5)}\)
\(f(f^{-1}(x))= x\)
