Let $$f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$$

be a real polynomial of degree $n > 0$. Using the derivative test, the values of $x$ for which the function $f(x)$ attains local minima and local maxima can be determined. However, considering this does not immediately give the values of $f(x)$ at those local minima and local maxima, I was wondering the following:

  1. Is it possible for $f(x)$ to attain a higher value at a local minimum than at a local maximum?
  2. If so, what is the lowest degree in which this can happen?

For the second question, I was thinking about degree $5$. Such a polynomial can have two local maxima and two local minima, since its derivative has degree $4$. Order the extrema in increasing value of $x$ at which it is attained. First suppose $a_n > 0$. I think it could have the first local maximum be lower than the last local minimum if the increase from the first local minimum to the second local maximum is more than the decreases from the local maxima to the next local minima combined (vice versa if $a_n < 0$).

However, I could not find an example, so I am not sure if the first question is even true.


2 Answers 2


Yes, it is possible, and the degree must be at least $5$: If $f$ has a local maximum at $x=a$ and a local minimum at $x=b$ with $f(a) <f(b)$ then $f$ has (at least) one more local minimum and one more local maximum between $a$ and $b$, so that the degree of $f'$ is at least $4$.

An example with degree $5$ (WolframAlpha plot): $$ f(x) = \frac 15 x^5 - \frac 53 x ^3 + 4x \, ,\\ f'(x) = x^4-5x^2+4 =(x+2)(x+1)(x-1)(x-2) \, . $$

You can verify that $f$ has a local maximum at $x=-2$ with $f(-2) < 0$, and a local minimum at $x=2$ with $f(2) > 0$.


Yes you are on the right track. You can construct such a polynomial with some systematic reasoning and some guessing. Let's say you want the degree $5$ polynomial $p : \mathbb R \to \mathbb R$ to have extrema at $a < b < c < d$. So its derivative $p'$ should have roots there. The obvious example is $ p'(t) = (t - a)(t - b)(t - c)(t - d) $ and so we can let $p(x) = \int_0^x (t - a)(t - b)(t - c)(t - d) \ dt$. With some experimentation you can find that letting $a = 0, b = 1, c = 3, d = 4$ gives $$ p(x) = \frac{1}{5}x^5 - 2x^4 + \frac{19}{3}x^3 - 6x^2 $$ with its local maximum at $0$ smaller than its local minimum at $4$.


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