Why can't you have more turning points than the degree?

Question

I get that each degree can correspond to a factor. $x^5=(x+a)(x+b)(x+c)(x+d)(x+e)$ and that results in 4 turning points, so the graph can "turn around" and hit the next zero. Why can't a curve have more turning points than zeros?

In the graph below, there are 4 roots, so degree is 4, but way more turning points than 4. What gives? Are those additional turning points represented by imaginary roots?

Each time there is a turning point, the derivative has to be zero, right? In terms of polynomials, your graph cannot represent a fourth degree polynomial function, why not? — imranfat, Jan 19 '16 at 16:33
"There are four [real] roots, so the degree is at least $4$..." — The Chaz 2.0, Jan 19 '16 at 16:39
The question is marked algebra-precalculus. Does that mean you do not want answers that involve calculus? — Austin Mohr, Jan 19 '16 at 16:41
Here is an idea not using the derivative of $p(x)$. I have difficulty in formulating my idea formally so I won't give it as an answer, but here goes: if $p(x)=0$ has more turning points than roots, than $p(x)=a$ would have more roots than the degree of $p(x)$. — MasB, Jan 19 '16 at 17:14
If you added a certain amount to your function (shifting it upwards), the degree wouldn't change but you'd get $10$ roots. — Akiva Weinberger, Jan 19 '16 at 18:55
@BernardMasse I was thinking along the same lines, but in general I think you may have to allow for $cp(x) = a$ for some small constant $c$. This would pull the turning points close together so that they cross the same horizontal line. — Austin Mohr, Jan 20 '16 at 00:16

score 3 · Accepted Answer · answered Jan 19 '16 at 17:17

The problem is that you are confusing real zeros of a polynomial with the degree. These are not the same. The degree of a single variable polynomial is the highest power the polynomial has.

Your hand drawn graph has only 4 real roots, but if it was a polynomial it must have more complex roots. You could not make all those turning points without this been true. You may not be aware of complex numbers.

Although you mention this as precalculus, this does become clearer with calculus, where you find the turning points ( "local maxima and minima" ) by equating the derivative of the polynomial to zero. The derivative of an n-th degree polynomial is an (n-1)th degree polynomial, so their can be as many as (n-1) turning points. However, the derivative's roots need not all be real, and in that case the original polynomial would have fewer real local maxima and minima than n-1.

So the problem is equating the number of real roots with the degree. You can really only know the degree by knowing the highest power the polynomial has. This is not always immediately obvious from the shape of a graph. You also need to be aware of the possibility of complex roots.

score 1 · Answer 2 · answered Jan 19 '16 at 16:41

You can use calculus to prove this. There is a "turning point" when the derivative of the polynomial is $0$ and the second derivative is non-$0.$ The derivative of your quintic function is a quartic function. There can be a maximum of $4$ zeroes in a quartic polynomial, so there can be at most $4$ "turning points," if the second derivative at each of these points is non-zero.

David E Speyer · Answer 3 · 2016-01-19T19:18:42.710

1

A quintic cannot have more than $4$ turning points, for the reason K. Jiang explains. On the other hand, a quintic (or even a cubic) can have more turning points than zeroes. For example, $x(x-1)(x-2)(x-3)(x-4)+2$ has four turning points, but only three zeroes.

edited Jan 19 '16 at 19:18

answered Jan 19 '16 at 16:48

David E Speyer

61,724

A polynomial may have more turning points than real zeroes (unless you take "zero" to mean "real root" already). – Austin Mohr Jan 19 '16 at 16:50
You're talking about quintics in your explanation, but your example is a sextic. – Akiva Weinberger Jan 19 '16 at 18:17

Emilio Novati · Answer 4 · 2016-01-19T20:00:13.557

I suppose that you want a polynomial function $y=p(x)$. As noted in the other answers the turning points are roots of the first derivative of the function so, since in your graph I count $13$ turning points, the derivative must have at least $13$ real roots. This means that the derivative $y'=p'(x)$ is a polynomial of, at least, degree $13$. So the function $y=p(x)$ have to be a polynomial of degree $14$ ( at least).

Now, from the graph, we see that this polynomial has $4$ real roots (the intersection points with the $x$ axis). If these are the only real roots, than all the other $10$ roots are complex numbers.

But note that a $14$ degree polynomial has $15$ parameters (the coefficients) in its equation. If you want find the polynomial from given turning points and roots, you have, in your case , $17$ conditions, and these gives a linear system with $14$ unknowns (the coefficients) and $17$ equation, that can have a solution only if the equations are not (all) linearly independent. This means that we cannot chose the turning points and the roots randomly, but they have to be linked by some condition.

To have a solution for a free set of $17$ conditions you have to start with a $16$ degree polnomial.

Narasimham · Answer 5 · 2016-01-20T11:08:25.540

If there is a fifth degree polynomial there are five roots, some real some complex conjugates, adding to total five roots. And four (one less) turning points.

EDIT1;

It is from the IVT and mean value theorems that are operating.Between two roots there is a maximum turning point,at its tangent is parallel to line joining roots. With two roots, one turning ... five roots four turnings. The maxima / minima tally (one less) occurs due to number of roots irrespective of real or complex .

The graph you drew has a degree much higher.For example in $ y= 1 + a \sin x + \sin b x $ it has infinitely many turning points and infinitely many roots. They cannot be fully indicated as in the domain that you have pictured. Here is a graph for $ (a = 1.1, b = 0.7 ). $

Why can't you have more turning points than the degree?

5 Answers5