# Real analysis

Homework Statement

Show that if $f:\mathbb{R}\to \mathbb{R}$ is a polynomial function of odd degree, then $f(\mathbb{R}) = \mathbb{R}$.

The attempt at a solution

How can I rigorously prove this? What is the most direct and concise way to prove this?

What I have is the following:

Any polynomial in $x_1, x_2, \ ... \ ,x_n$ with coefficients in $\mathbb{R}$ is continuous. Since $\mathbb{R}$ is connected, $f(\mathbb{R})$ is connected, thus we can apply the intermediate value theorem. Now any polynomial of odd degree has at least one real root thus there exists $p,q\in \mathbb{R}$, $p<q$ such that $f(p)<0$ and $f(q)>0$. Thus we can then choose $p\in \mathbb{R}$ such that $f(p)\ge 0$, $f(p) \le 0$ and obtain any $f(\mathbb{R})\in \mathbb{R}$. Therefore, $f(\mathbb{R})\in \mathbb{R}.$

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Dick
Homework Helper
Homework Statement

Show that if $f:\mathbb{R}\to \mathbb{R}$ is a polynomial function of odd degree, then $f(\mathbb{R}) = \mathbb{R}$.

The attempt at a solution

How can I rigorously prove this? What is the most direct and concise way to prove this?

What I have is the following:

Any polynomial in $x_1, x_2, \ ... \ ,x_n$ with coefficients in $\mathbb{R}$ is continuous. Since $\mathbb{R}$ is connected, $f(\mathbb{R})$ is connected, thus we can apply the intermediate value theorem. Now any polynomial of odd degree has at least one real root thus there exists $p,q\in \mathbb{R}$, $p<q$ such that $f(p)<0$ and $f(q)>0$. Thus we can then choose $p\in \mathbb{R}$ such that $f(p)\ge 0$, $f(p) \le 0$ and obtain any $f(\mathbb{R})\in \mathbb{R}$. Therefore, $f(\mathbb{R})\in \mathbb{R}.$
f is a polynomial function in one variable. What are these $x_1, x_2, \ ... \ ,x_n$ you speak of? Yes, it is the intermediate value theorem. And if you know a polynomial of odd degree has at least one real root, how did you prove that? The proof $f(\mathbb{R}) = \mathbb{R}$ should use the same ingredients.

Its wrong in placing $x_1, ... ,x_n$, so I have to use one variable?

Also, I must prove, in my proof, that a polynomial of odd degree has at least one real root?

And for $f(\mathbb{R}) = \mathbb{R}$ can you elaborate, please?

Dick
Homework Helper
Its wrong in placing $x_1, ... ,x_n$, so I have to use one variable?

Also, I must prove, in my proof, that a polynomial of odd degree has at least one real root?

And for $f(\mathbb{R}) = \mathbb{R}$ can you elaborate, please?
When they say $f:\mathbb{R}\to \mathbb{R}$ that implies f is a function of one real variable, doesn't it? So you can just write $f(x)=a_n x^n+a_{n-1} x^{n-1}...a_1 x+a_0$, it's a simple real polynomial. And I think you should think about proving any polynomial of odd degree has a root. If you really don't have to and IF you can just assume it out of blind nowhere then you don't even need the intermediate value theorem. If f(x) is a polynonomial of odd degree and c is any real number then f(x)-c is also a polynomial of odd degree, so? I would regard this as somewhat of a cheat. It wouldn't show much understanding of what's really going on.

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Office_Shredder
Staff Emeritus
Gold Member
Also, I must prove, in my proof, that a polynomial of odd degree has at least one real root?
You said it was true in your proof, can you elaborate on why you think it is?

The standard proof for that is very useful in solving this problem.

Office_Shredder and Dick -

To prove that, I got:

Let $p(x) = a_nx^n + a_{n - 1}x^{n - 1} + ... + a_0$ be a polynomial of odd degree and let $c$ be a real number such that $p(c) = 0$. Since polynomials are continuous we have that if $a_nx_n$ has opposite signs then if $a_n > 0,$ there are real numbers $x_0 < x_1$ such that $P(x_0) < 0$ and $P(x_1) > 0.$ Now if $a_n < 0,$ there exists $x_0 < x_1$ such that $P(x_0) > 0$ and $P(x_1) < 0.$ so from the intermediate value theorem there is a real number $c \in [x_0,x_1]$ such that $p(c) = 0$.

Is that correct?

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Dick
Homework Helper
Office_Shredder -

To prove that, I got:

Let $p(x) = a_nx^n + a_{n - 1}x^{n - 1} + ... + a_0$ be a polynomial of odd degree and let $c$ be a real number such that $p(c) = 0$. Since polynomials are continuous we have that if $a_nx_n$ has opposite signs then if $a_n > 0,$ there are real numbers $x_0 < x_1$ such that $P(x_0) < 0$ and $P(x_1) > 0.$ Now if $a_n < 0,$ there exists $x_0 < x_1$ such that $P(x_0) > 0$ and $P(x_1) < 0.$ so from the intermediate value theorem there is a real number $c \in [x_0,x_1]$ such that $p(c) = 0$.

Is that correct?
It's awful. You just proved what you assumed and not at all well at that. Don't you see that?

Dick - I am sorry for my ignorance. What is the standard way to prove it?

Dick
Homework Helper
Dick - I am sorry for my ignorance. What is the standard way to prove it?
I think the first thing I should ask is are you clear on what you want to prove? Can you state it in words?

Dick -

I know that every polynomial has as many roots as it has degree, so $n^{\text{th}}$ degree polynomial has $n$ roots. I also know that any complex root are paired with their complex conjugate thus they take up an even number of roots. So if I have an odd degree polynomial then only an even number of the roots can be complex therefore at least one root must be real?

Is that correct?

Dick
Homework Helper
Dick -

I know that every polynomial has as many roots as it has degree, so $n^{\text{th}}$ degree polynomial has $n$ roots. I also know that any complex root are paired with their complex conjugate thus they take up an even number of roots. So if I have an odd degree polynomial then only an even number of the roots can be complex therefore at least one root must be real?

Is that correct?
Yes, that's all correct. What you need to prove is that if f is an odd degree polynomial then $f(\mathbb{R}) = \mathbb{R}$. Which means you need to prove that for any a in $\mathbb{R}$, that there exists an b in $\mathbb{R}$ such that f(b)=a. What I would do is think about what the limits are of f(x) as x→+∞ and x→-∞ and how you can use the intermediate value theorem.

If you really want to just use the 'one real root' trick. Then think about the polynomial f(x)-a. Is it an odd degree polynomial? What does the 'one real root' tell you about that?

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Thanks, Dick. I think I know what to do now.