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?

Note that $f(0) = 0$ and $\lim_{x \to \infty} f(x) = +\infty$. Thus by the intermediate value theorem, since $f$ is continuous, it must take every value in $[0,\infty)$; that is, $f$ is surjective. Next, we see $$f'(x) = \tfrac 5 6 x^{-1/6} + \tfrac 3 2 x^{1/2} > 0 \,\,\, \text{ for } x > 0$$ which shows that $f$ is strictly increasing. That is, for $x < y$, we note that $$f(x) - f(y) = \int^y_x f'(t) dt > 0 \,\,\, \implies \,\,\, f(x) > f(y).$$ Thus for unequal $x,y$, we cannot have $f(x) = f(y)$; that is, $f$ is injective.
