 Solution

If $0\le x<y$, 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 $\mathbb{R}$: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.