Is there an algebraic way to determine the number of turning points of a polynomial from its factorization?

Question

This question came from a precalculus course:

Let $f(x) = x^5 + x^3 - 6x$. Find all the real zeros of $f$ with corresponding multiplicities and determine the number of turning points of the graph of the function.

I believe students are expected to factor $f(x)$ into $x(x^2 - 2)(x^2 + 3)$ and determine the zeros are $x = 0, \sqrt{2}, -\sqrt{2}$ with multiplicity one each.

I guess students are supposed to use graphing software to look at the graph and see that there's only two turning points. Is there any other way to do it? We know there are at most 4, since $f$ has degree 5, and at least 2 between the $x$-intercepts, so there are either 2 or 4.

In particular, is there anything about the irreducible quadratic $(x^2 + 3)$ in the factorization which suggests whether there are additional turning points? By messing around, I found that e.g. $x(x^2-2)(x^2-7x+12.5)$ has 4 turning points, but no additional zeros.

I am aware that you can find the number of solutions to $f'(x) = 0$, but as this is a precalculus course, I'm interested in algebraic methods.

Answer 1

Because the zeros aren't turning points, and the number of turning points on $(-\infty, -\sqrt{2})$ and $(\sqrt{2}, \infty)$ is even, and on $(-\sqrt{2}, 0)$ and $(0,\sqrt{2})$ is odd, we concluded that the number of turning points positives (and negatives) is odd.

Note too that this function is odd, then if $x$ is a turning point ergo $-x$ also will be. Therefore the number of turning points positives and negatives are the same.

Then the only possibility is two turning points.

Answer 2

Just to be clear: a turning point is a point where the polynomial changes from increasing to decreasing (or vice-versa)? If so, we should be a bit more careful -- the number of solutions to $f'=0$ is not the answer either. It is the number of odd-multiplicity roots of $f'$. And how did you determine the number of turning points is at most 1 less than the degree of the polynomial? Anyway, in my opinion, taking the formal derivative of a polynomial is very algebraic, and it should be the answer, at least indirectly.

Notice that if a turning point happens on the $x$-axis, you can detect it immediately. The turning points on the $x$-axis are the roots of $f$ with even multiplicity. But every turning point is on a (unique) line $y=h$, so we can detect it by finding it as an even-multiplicity root of $f(x)-h$. So the number of turning points is $$\sum_{h\in \mathbb{R}} \#\{x_0\in \mathbb{R} : x_0 \text{ is an even-multiplicity root of } f(x)-h\}.$$ That's an answer that uses no limits or continuity or anything like that. But let's look at the derivative as something algebraic, not referring to continuity or limits if even just for motivation. I understand this is complicated for a precalculus course, but I'm more trying to motivate introducing the formal derivative without any continuity or limits, and I understand I'm sort of working in reverse, and it probably shouldn't be presented this way to precalculus students. For a real polynomial $f(x)$ of degree $n$ with roots $a_1,\dots,a_n$ (counting repeats and complex roots), lets define $$df(x) := \sum_{i=1}^n \frac{f(x)}{x-a_i}.$$ You can see how this might be useful to find repeated roots of $f$. In fact, it's pretty easy to see that if $a$ is a root of $f$ with multiplicity $k\ge 2$, then $a$ is a root of $df$ with multiplicity $k-1$. For $0\le d\le n-1$, the coefficient of $x^d$ in $$\frac{f(x)}{x-a_i} = \prod_{j\ne i} (x-a_j)$$ is $(-1)^{n-1-d}$ multiplied by the sum of all products of $n-1-d$ of the roots excluding $a_i$. You can see that for each term $(-1)^{n-1-d}a_{i_1}\dots a_{i_{n-1-d}}$, with $1\le i_1<\dots<i_{n-1-d}\le n$, in the coefficient of $x^{d+1}$ in $f(x)$, there are exactly $d+1$ matching products of roots in all of the terms of the $x^d$ coefficients of the $f(x)/(x-a_i)$, namely one for each $a_i$ with $i\notin \{i_1,\dots,i_{n-1-d}\}$. This tells us the coefficient of $x^d$ in $df(x)$ is equal to $d+1$ times the coefficient of $x^{d+1}$ in $f(x)$. In other words, if $$f(x) = c_nx^n + \dots + c_1x + c_0,$$ then $$df(x) = nc_nx^{n-1}+\dots+2c_2x + c_1.$$ Therefore, $d(f-h)(x) = df(x)$ for any $h\in \mathbb{R}$. So by our earlier reasoning, if $a$ is a root of $f(x)-h$ with multiplicity $k\ge 2$, then $a$ is a root of $df(x)$ of multiplicity $k-1$. Conversely, if $a$ is a root of $df(x)$ with multiplicity $k\ge 1$, then pick $h=f(a)$ so that $a$ is a root of $f(x)-h$, so that $f(x)-h = (x-a)^\ell g(x)$ for some $\ell\ge 1$ and $g(x)$ with $g(a) \ne 0$. You see then that $$df(x) = d(f-h)(x) = \ell(x-a)^{\ell-1}g(x)+(x-a)^\ell dg(x) = (x-a)^{\ell-1}(\ell g(x)+(x-a)dg(x))$$ (the product rule step follows directly from the definition), and so since $g(a) \ne 0$, we must have $\ell = k+1$.

Therefore, the turning points of $f$ are exactly the odd-multiplicity real roots of $df$.