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.
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It is a cubic polynomial (odd degree), continuous etc..goes to $+\infty$ as well as to $-\infty$ so.. – Anurag A Jul 21 '17 at 11:44
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4That would depend on what domain and especially what codomain you have. You can always select a codomain that makes a function non-surjective and also always select a codomain that makes it surjective. – skyking Jul 21 '17 at 11:48
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@skyking What is the codomain $C$ that makes the function $(-1,1)\ni x\mapsto \frac{x}{1-x^2}\in C$ non-surjective? – Pedro Jul 21 '17 at 13:27
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@Pedro To make the function surjective select $C = f(D)$ where $D$ is the domain, to make it non-surjective select instead $C = f(D)\cup{f(D)}$ (which is a strict superset of $f(D)$). – skyking Jul 21 '17 at 16:19
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@skyking I was thinking in a subset $C\subset \mathbb R$ but I did not write it AND your (perfectly right, as I can see now) original comment does not assume this restrictive condition. Thanks. – Pedro Jul 21 '17 at 19:58
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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.
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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.
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As a side note, if someone know the correct mathjax syntas for the definition of $f$ that would be greatly appreciated. I get close to it using matrix but it bothers me :p – Furrane Jul 21 '17 at 12:07
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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:
MPW
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