Answer
$(f^{-1}\circ f)(x)=x$ and $(f\circ f^{-1})(x)=x$,
so the given function is the inverse of f.
Work Step by Step
If a function $f$ is one-to-one, then $f^{-1}$ is the unique function for which
$(f^{-1}\circ f)(x)=f^{-1}(f(x))=x$
and $(f\circ f^{-1})(x)=f(f^{-1}(x))=x.$
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$(f^{-1}\displaystyle \circ f)(x)=f^{-1}(f(x))=\frac{3}{f(x)}-2$
$=\displaystyle \frac{3}{\frac{3}{x+2}}-2$
$=x+2-2$
$=x$
$(f\displaystyle \circ f^{-1})(x)=f(f^{-1}(x))=\frac{3}{f^{-1}(x)+2}$
$=\displaystyle \frac{3}{\frac{3}{x}-2+2}$
$=\displaystyle \frac{3}{\frac{3}{x}}$
$=x$
$(f^{-1}\circ f)(x)=x$ and $(f\circ f^{-1})(x)=x$,
so the given function is the inverse of f.
