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Prove that the function $f(x)=x^5+2x^2+x$ is onto. My teacher did an example in class were he said $f(x)=y$ and then found the inverse of the function. He then plugged the inverse back into the function and found that it equaled y. And that was the proof that the function was onto.

This makes sense, but I don't know how to find the inverse of this function and I am wondering if there is another way to prove this without finding the inverse?

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  • $\begingroup$ Draw the graph, this follows immediately. Note that we can find the 'inverse' by flipping it about $y=x$. (However, we don't really have a function, because it is one to many.) $\endgroup$
    – Calvin Lin
    Commented Jun 26, 2013 at 4:23
  • $\begingroup$ Have you learnt about taking limits? If so, one way to go is to prove that the image of $f$ is all $\mathbb{R}$ by noticing that $f$ is continuous, and then just take the limits of $f$ when $x \to + \infty$, and $x \to - \infty$. $\endgroup$
    – user49685
    Commented Jun 26, 2013 at 4:24
  • $\begingroup$ You'll struggle to find an inverse, because this function doesn't have one! Instead, show that the function takes arbitrarily high and low values, and that it's continuous in between. $\endgroup$
    – Billy
    Commented Jun 26, 2013 at 4:25
  • $\begingroup$ To elaborate more on what @user49685 said, if you have a continuous function on $[a,b]$ and we have that $f(b) > f(a)$, then for each $y$ in $[f(a),f(b)]$, there is an $x$ such that $f(x) = y$. So suppose at $-\infty$ we have that $f$ is $-\infty$ and at $+\infty$, we have that $f$ is $+\infty$. What can we say, if anything, about whether or not $f$ is onto? $\endgroup$ Commented Jun 26, 2013 at 4:28
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    $\begingroup$ You will not find a nice formula for $x$ in terms of $y$. Use the Intermediate Value Theorem. $\endgroup$ Commented Jun 26, 2013 at 4:28

2 Answers 2

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$\,f(x) \;$ is a continuous function on the whole real line, as any other polynomial is, and since

$$\lim_{x\to -\infty}f(x)=-\infty\\\lim_{x\to \infty}f(x)=\infty$$

the intermediate value theorem tells us that for any value $\;c\in(-\infty,\infty)\,$ there exists $\,r\in\Bbb R\;\;s.t.\;\;f(r)=c\;$ .

This much is true for any real polynomial with odd degree, btw.

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You can differentiate your function and if it either always increasing or always decreasing then it will onto. In your case it is always increasing. Following is the solution : enter image description here

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    $\begingroup$ Please use LaTeX when writing mathematics in this site. Directions in the FAQ section. $\endgroup$
    – DonAntonio
    Commented Jun 26, 2013 at 5:48
  • $\begingroup$ @DonAntonio : Sorry, I am new. From next time onwards I will have this in mind. $\endgroup$
    – Nishant
    Commented Jun 26, 2013 at 6:14
  • $\begingroup$ Don't worry, @Nishant: we've all been there. $\endgroup$
    – DonAntonio
    Commented Jun 26, 2013 at 6:18
  • $\begingroup$ What if $f:\mathbb{R}\to\mathbb{R}$ and $f(x) = \tanh{(x)}$? $\tanh{(x)}$ is always increasing, but $f(x)$ wouldn't be onto $\endgroup$
    – Justin
    Commented Nov 30, 2013 at 22:35
  • $\begingroup$ @Nishant I think you are confusing onto and one to one $\endgroup$
    – Justin
    Commented Dec 2, 2013 at 0:06

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