Prove that every even degree polynomial function $f$ has maximum or minimum in $\mathbb{R}$. (without direct using of derivative and making $f'$)

The problem seems very easy and obvious but I don't know how to write it in a mathematical way.

For example if the largest coefficient is positive, it seem obvious to me that from a point $x=a$ to $+\infty$ the function must be completely ascending. And from $-\infty$ to a point $x=b$ the function must be completely descending. If it is not like that, its limit will not be $+\infty$ at $\pm\infty$. Now, because it is continuous, it will have a maximum and minimum in $[b,a]$ so it will have a minimum (because every other $f(x)$ where $x$ is outside $[b,a]$ is larger than $f(a)$ or $f(b)$ and we get the minimum that is less than or equal to both of them) . We can also the same for negative coefficient.

But I can't write this in a formal mathematical way.

  • $\begingroup$ Think about polynomial functions of one variable... What do you do when you want to find the max/min of a function? $\endgroup$
    – Bruno Reis
    Oct 25, 2018 at 21:30
  • $\begingroup$ @BrunoReis I use derivative. But this question is from the limit chapter of a book and it is before the derivative part. I think what I said is an informal proof without using derivative but I don't know the formal mathematical writing. $\endgroup$
    – titansarus
    Oct 25, 2018 at 21:32
  • $\begingroup$ @JoséCarlosSantos . That question uses derivative but I said I don't want a solution related to derivative. I informally know the proof but can't write it mathematically. So I think my question is not a duplicate. $\endgroup$
    – titansarus
    Oct 25, 2018 at 21:38
  • 1
    $\begingroup$ @titansarus Hmm... But using derivatives, do you know that every odd degree polynomial have at least one real root? If so, when you differentiate your polynomial function with even degree, you're going to get a new polynomial function with odd degree, and that is guaranteed to have a root, that implies that you'll have max/min. Now I'll think about how to write that in a formal way using limits. $\endgroup$
    – Bruno Reis
    Oct 25, 2018 at 21:40
  • $\begingroup$ @BrunoReis. I said, I normally use derivative and use that method. But this question is asked in the book even before it defines the derivative! So it must be solved without derivative. $\endgroup$
    – titansarus
    Oct 25, 2018 at 21:41

2 Answers 2


Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. Let us assume that $a_n>0$ (the case in which $a_n<0$ is similar). Then\begin{align}\lim_{x\to\pm\infty}f(x)&=\lim_{x\to\pm\infty}a_nx^n\left(1+\frac{a_{n-1}}{a_nx}+\frac{a_{n-2}}{a_nx^2}+\cdots+\frac{a_0}{a_nx^n}\right)\\&=+\infty\times1\\&=+\infty.\end{align}Therefore, there is a $R>0$ such that $\lvert x\rvert>R\implies f(x)\geqslant f(0)$. So, consider the restriction of $f$ to $[-R,R]$. Since $f$ is continuous and since $[-R,R]$ is closed and bounded, $f|_{[-R,R]}$ attains a minimum at some point $x_0\in[-R,R]$ and, of course, $f(x_0)\leqslant f(0)$. Since outside $[-R,R]$ you always have $f(x)\geqslant f(0)$, $f$ attains its absolute minimum at $x_0$.


In an even degree polynomial $p_{\,2n}(x) \quad |\; 0<n$, we have that $$ \mathop {\lim }\limits_{x\; \to \, + \infty } p_{\,2n} (x) = \mathop {\lim }\limits_{x\; \to \, - \infty } p_{\,2n} (x) = \pm \infty $$ and it does not have other poles other than those.

Assume that the limit is $+ \infty$.

Since a polynomial is a continuous surjective function, its co-domain will be limited below, and this limit will correspond to one or more values of $x$.

The same applies, flipped over, if the limit is $-\infty$.


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