= Solution
If $0\leq x<y$, then $\sqrt x<\sqrt y$ and $x<y$, so $\sqrt x+x<\sqrt y+y$. Thus the <function> is <strictly increasing> on its <domain> and is <injective>. Its values are nonnegative, so it is not <surjective> onto $\mathbb R$:
$$
\boxed{\text{one-to-one, but not onto}.}
$$
If $\mathbb R_+$ means $(0,\infty)$ its <image of a function> is $(0,\infty)$; if it includes zero its <image of a function> is $[0,\infty)$. Indeed, for any $u>0$, set $t=(\sqrt{1+4u}-1)/2>0$ and $x=t^2$; then $\sqrt x+x=t+t^2=u$. This convention does not change the classification.
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