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If $f(x)$ and $g(x)$ are functions with no inflection points, can $h(x)=f(x)\cdot g(x)$ have an inflection point?

Edit: I experimented a bit with a few functions (like $x^2\cdot x^2$) in a graphing calculator but I couldn't find a good example. I was also running into trouble with the third derivative test in some situations.

I was doing exercises and noticed $-x^4+4x^3-6x^2$ doesn't have an inflection point. It can also be rewritten as $x^2(-x^2+4x-6)$. This is a product of 2 parabolas and parabolas have no inflection points. If that fact is sufficient to show that the product also doesn't have an inflection point, it would save a lot of time when doing problems like this.

Milo
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    Think of $x \cdot x^2$. – user64494 Jun 21 '21 at 14:48
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Michael Jesurum Jun 21 '21 at 14:49
  • @user64494 oh, of course. Thank you. – Milo Jun 21 '21 at 14:55
  • @MichaelJesurum I see. I thought brevity would be better. Thank you! – Milo Jun 21 '21 at 14:58

1 Answers1

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Sure! An inflection point means the second derivative is changing sign, so

$$h''(x) = 0$$

at some particular $x$, but not in general. To ensure that $f(x), g(x)$ have no inflection points, let's just make them quadratic:

$$f(x) = f_2 x^2 + f_1 x + f_0$$

$$g(x) = g_2 x^2 + g_1 x + g_0$$.

Then, expanding around $x=0$, we find

$$h''(x) \approx 2 (f_2 g_0 + f_1 g_1 + f_0 g_2) + 6(f_2 g_1 + f_1 g_2) x$$

where we need the first term to vanish but the coefficient of $x$ to be nonzero. One possible choice is

$$f(x) = g(x) = x^2 + x -1/2$$.

Plotting verifies visually that indeed $f(x)g(x)$ has an inflection point at $x=0$ but $f(x), g(x)$ do not.

David
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