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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$(f^{-1}\displaystyle \circ f)(x)=f^{-1}(f(x))=\frac{1}{f(x)+1}$
$=\displaystyle \frac{1}{\dfrac{1-x}{x}+1}$
$=\displaystyle \frac{1}{\dfrac{1-x+x}{x}}=\frac{1}{\dfrac{1}{x}}$
$=x$
$(f\displaystyle \circ f^{-1})(x)=f(f^{-1}(x))=\dfrac{1-f^{-1}(x)}{f^{-1}(x)}$
$=\displaystyle \frac{1-\dfrac{1}{x+1}}{\dfrac{1}{x+1}}$
$=\displaystyle \frac{\dfrac{x+1-1}{x+1}}{\dfrac{1}{x+1}}=\frac{\dfrac{x}{x+1}}{\dfrac{1}{x+1}}$
$=x$
$(f^{-1}\circ f)(x)=x$ and $(f\circ f^{-1})(x)=x$,
so the given function is the inverse of f.
