Prove a function is onto

Question

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?

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.) — Calvin Lin, Jun 26 '13 at 04:23
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 $x \to - \infty$. — user49685, Jun 26 '13 at 04:24
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. — Billy, Jun 26 '13 at 04:25
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? — Cameron Williams, Jun 26 '13 at 04:28
You will not find a nice formula for $x$ in terms of $y$. Use the Intermediate Value Theorem. — André Nicolas, Jun 26 '13 at 04:28

Nishant · Answer 1 · 2013-06-26T06:33:24.770

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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

edited Jun 26 '13 at 06:33

answered Jun 26 '13 at 05:29

Nishant

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Please use LaTeX when writing mathematics in this site. Directions in the FAQ section. – DonAntonio Jun 26 '13 at 05:48
@DonAntonio : Sorry, I am new. From next time onwards I will have this in mind. – Nishant Jun 26 '13 at 06:14
Don't worry, @Nishant: we've all been there. – DonAntonio Jun 26 '13 at 06:18
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 – Justin Nov 30 '13 at 22:35
@Nishant I think you are confusing onto and one to one – Justin Dec 02 '13 at 00:06

score 1 · Answer 2 · answered Jun 26 '13 at 05:53

$\,f(x) \;$ 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.

Prove a function is onto

2 Answers2