Is the mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=5x^3+3$ onto? Let $f\colon \mathbb R \to \mathbb R$ be defined by $f(x)= 5x^3+3$. Is it onto?
According to me, if $y=5x^3+3$, then $x = \sqrt[3]{(y-3)/5}$ is not an element of $\mathbb R$ for all $y \in (-\infty,3)$ so all numbers in the codomain $(-\infty,3)$ wont have pre-images.
But many say $5x^3+7$ as an odd degree equation will have at least one real root. Is it onto?
 A: Simply looking at Wolfram|Alpha shows that $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=5x^3+3$ is not just onto but also one-to-one. Let's prove that $f$ is onto though, using your choice of $x$ to do this. 
Claim: The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=5x^3+3$ is onto.
Proof. Suppose $y\in\mathbb{R}$. Then let $x=\sqrt[3]{\frac{y-3}{5}}$. We have the following:
\begin{align}
f(x) &= f\left(\sqrt[3]{\frac{y-3}{5}}\right)\\[1em]
&= 5\left(\sqrt[3]{\frac{y-3}{5}}\right)^3+3\\[1em]
&= 5\cdot\frac{y-3}{5}+3\\[1em]
&=y-3+3\\[0.5em]
&= y.
\end{align}
Thus, $f$ is onto. $\blacksquare$
A: Why do you say that $x = \sqrt[3]{\frac{y-3}5}$ is not an element of $\mathbb R$ for all $y$?
For every $b \in \mathbb R$ there is exactly one $a \in \mathbb R$ such that $a^3 = b$, we call this $a = \sqrt[3]{b}$.
Maybe you are confused because for $b < 0$ there is no $a \in \mathbb R$ such that $a^2 = b$, in other words we can't define $\sqrt b$.
But the situation is different for the third root.
A: You overlook that real cube roots (in contrast to square roots) are defined for all reals For example $\sqrt[3]{-8}=-2$ because $(-2)^3=-8$.
A: $x=\sqrt[3]{\frac{y-3}5}$ is an inverse as you claimed; keep in mind that odd radicals ($\sqrt[3]{}$, $\sqrt[5]{}$, etc.) are defined for all real numbers, whereas even ones are not (unless we introduce extensions such as $\mathbb{C}$.) Therefore all numbers in the codomain $\mathbb{R}$ have preimages given by the rule you stated, not just the ones in $\left[-3,\infty\right)$.
Perhaps this example will help clarify: consider the equation $x^3=k$. For positive $k$, this has a solution at $\sqrt[3]{k}$. For negative $k$, we have $x^3=k\leftrightarrow-x^3=-k\leftrightarrow\left(-x\right)^3=-k$ (as $\left(-1\right)^3=-1$.) $-k$ is positive, so some value of $-x$ satisfies this equation as just demonstrated; just take the negative of this to get $x$. Thus, for all nonzero $k$, a real solution to $x^3=k$ exists. The case $k=0$ is easy: let $x=0$.
After showing that $x$ is unique for all $k$, $x$ is the cube root of $k$ by definition, so all real numbers have a real cube root.
