How to show $f(x) =(\sqrt[3]{x}+x)\sqrt{x}$ is injective and surjective? $f : [0, \infty) \to [0,\infty) d$ and $f(x) =(\sqrt[3]{x}+x)\sqrt{x}$.
How should I go, if I wanted to show that it is bijective?

For injective is enough to see that:
$$f'(x)=\frac{5}{6x^{1/6}}+\frac{3x^{1/2}}{2}>0$$
so, $f$ is monotone increasing.
For surjective, take $y_0 \in (0, \infty)$ (because if $y_0=0$ take $x=0$) then
$$f(x)=x^{5/6}+x^{3/2}=y_0$$  
$p=x^{1/6}$ then
$$g(p)=p^5+p^9-y_0=0$$
The polynomial $g$ has real coefficients and odd degree so there is a odd number of real roots. 
Also there is a positive root because the product of all roots is $y_0>0$.  
