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$f : [0, \infty) \to [0,\infty) d$ and $f(x) =(\sqrt[3]{x}+x)\sqrt{x}$.

How should I go, if I wanted to show that it is bijective?

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    Please make sure I've correctly formatted the question (I'm not sure what that "d" is doing, in particular, but I suspect I've got the function right). – 2017-01-22
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    To prove it is injective simple inequalities suffice. To prove it is surjective I recommend you first prove it is continuous, and then use the intermediate value theorem. – 2017-01-22

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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.

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    Monotonicity follows more simply from the fact that the function is the sum and product of increasing functions. – 2017-01-22
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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 $y_0>0$.