Will a function with a double root always have a (local of absolute) maximum or minimum there?

Question

I was doing some coding and wanted to code the bisection method, but since this applies to the case where $f(a)f(b)<0$, I started to think how I could generalize this to $f(a)f(b)>0$ and I thought that this would be 'codeable' because I would only need to differentiate the function and find the roots of that, but then I started to wonder whether a (differentiable) function with a double rooth will always have a maximum or minimum in that point. Is this true, or are there some special cases where there is a double rooth, but no (local or absolute) maximum/minimum?

Answer 1

Let $f$ be a function on some subset of $\mathbb{R}$, and for simplicity suppose that $f$ has a root at $0$. If $f$ has a root of order $p$ at $0$, then this means that we can write $f$ as $f(x)=x^pg(x)$, where $g(x)$ is a continuous function at zero, but does not have a root at zero. As such, we have either $g(0)>0$ or $g(0)<0$. Based on the continuity of $g$, we then know that this is true for some neighborhood of $0$, so if $|x|<\epsilon$ for some $\epsilon>0$, we have $g(x)$ is uniformly positive or negative. For the sake of convenience, assume it is positive.

Now, what happens to $f$ when $|x|<\epsilon$? If $0<x<\epsilon$, then $x^p$ is positive and $g(x)$ is positive, meaning $f(x)$ is positive; if $-\epsilon<x<0$, then $x^p$ has sign of $(-1)^p$, and $f(x)$ has the same sign. Thus, for even $p$, $f$ remains positive and for odd $p$ is switches to being negative in this region. Specifically when $x=0$, we know that $f(x)=0$, so this means that if $p$ is even, then $f(x)$ is positive on the left and right, but zero in the middle. If $p$ is odd, then the sign switches.

This tells us that if $f$ has a root at some point with an even order then the root is a local extrema. When the root has odd order greater than $1$ then it visually looks like a "flat spot", meaning that it looks like it flattens out but then continues its path. When the order is exactly $1$, this means $f$ looks just like a line through the axis. For your case, if the root is exactly a double root, then it is also a local extrema.

Answer 2

It is not true that a differentiable function with a double zero at some point will always have a local minimum or maximum at that point. If the function is allowed to have a higher-order zero, then $f(x)=x^3$ furnishes a simple counterexample.

It's harder to find a counter example for a zero of order exactly $2$. Here's one:

$$f(x)=\begin{cases}x^5\sin\frac1x,&x\neq0\\0,&x=0 \end{cases}$$

It can be shown that $f'(0)=f''(0)=0$, but clearly $f$ takes both positive and negative values in any interval containing $0$.

To confirm the statements about the derivatives, just differentiate as usual when $x\neq0$. For $x=0$, use the following simple consequence of the mean value theorem.

Let $f(x)$ be continuous on the interval $(a, b)$ and differentiable there, except perhaps at some point $a<c<b$. Suppose further that $\lim_{x\to c}f'(x)$ exists. Then $f$ is differentiable at $c$, and $f'(c)=\lim_{x\to c}f'(x)$.

That is, just take the limits of the derivatives as $x\to0$, and if they exist, they're the derivatives at $0$.

Answer 3

Interesting. This is really true under these conditions: Let $g(x)=(x-a)^2f(x)$ defined on the neighbourhood of $a$ (denoted by $U(a)$), where furthermore $f(a)>0,~f\in C^2(U(a))$, then $f$ has local minimum at $a$. This can be seen from: $$g'(x)=2(x-a)f(x)+(x-a)^2f'(x),$$ $$g''(x)=2f(x)+4(x-a)f'(x)+(x-a)^2f''(x) \\ \implies g'(a)=0,~g''(a)>0.$$ If $f(a)<0$ then you get a local maximum.