TITLE: Can the product of two functions with no inflection points have an inflection point?
QUESTION [5 upvotes]: 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.

REPLY [8 votes]: 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.