= Solution
\b[It is <surjective> but not <injective>; there is no global <inverse function>.] Noninjectivity follows from
$$
f(3/2)=\frac3{\sqrt2}=f(3).
$$
For surjectivity, restrict to $x<1$, where
$$
f(x)=\frac{x}{\sqrt{1-x}},\qquad
f'(x)=\frac{1-x/2}{(1-x)^{3/2}}>0.
$$
This restriction is a <continuous function>, tends to $-\infty$ as $x\to-\infty$ and to $+\infty$ as $x\to1^-$. The <intermediate value theorem> therefore supplies a preimage for every real number. The additional branch $x>1$ destroys global injectivity but cannot destroy surjectivity.
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