Is $f(x)=x^3+x$ is a surjection??

Question

I am confused in this problem.I have a function $f(x)=x^3+x$ .Is it surjection?? While checking a function onto(surjection) or not at first we have to express $x=\phi(y)$.But I failed to find any that type of relation. Please give me the solution or hint.

Answer 1

Hints:

(1) $f$ is continuous.

(2) $f$ approaches $-\infty$ as $x$ approaches $-\infty$.

(3) $f$ approaches $+\infty$ as $x$ approaches $+\infty$.

(4) Think about the Intermediate Value Theorem.

Answer 2

When you ask yourself about a surjective (or injective or bijective) function, it is crucial that you consider the domains of the function itself.

One quick example :

$$ f(x) = x^2 $$ If the domains are $f : \mathbb{R}\to \mathbb{R}^+$ then $f$ is surjective but not injective hence not bijective.

But if the domains are $f : \mathbb{R}^+\to \mathbb{R}^+$ then $f$ is bijective.

Here we're going to assume your function $f$ is defined as follow :

\begin{align} f:\mathbb R&\longrightarrow\mathbb R\\ x&\longmapsto x^3+x \end{align}

For $f$ to be surjective, we need to have :

$$\forall y \in \mathbb{R}, \exists x\in \mathbb{R} : y=f(x)=x^3+x$$

Which we can re-write as a polynome of $x$ :

$$ \forall y \in \mathbb{R}, \exists x\in \mathbb{R} : P(x) = x^3+x-y = 0 $$

And we now cubic polynomes with real coefficients have at least one root in $\mathbb{R}$ so we're done.

This method is the general way to approach such problems, but quasi answer is more elegant for this particular example.

Answer 3

It is indeed. In fact, it is a bijection (note the function is strictly increasing since $f'>0$). Here is an explicit computation of the inverse:

https://math.stackexchange.com/a/60925/113214