Prove that $f$ has an inflection point at zero if $f$ is a function that satisfies a given set of hypotheses Prove that if $f$ is a function such that $f'(x) > 0$ $\forall x \neq 0 \wedge f'(0) = 0 \wedge f''$ is a continuous one to one function on some open interval $(a, b): a < 0 < b$ then $f$ must have an inflection point at zero.
I think that should the function $g$ be set to $g(x) = x^3$ then $g$ satisfies these hypotheses. The derivative of $g$ is $g'(x) = 3x^2$, which has $g'(x) > 0$ $\forall x \neq 0 \wedge g'(0) = 0$. The second derivative of $g$ is $g''(x) = 6x$ which is a linear function $\implies g''$ is a continuous one to one function $\forall x \in \mathbb{R}$. It is true that $g$ changes sign (has an inflection point) at $x = 0$. So, in this case, all these things make a reasonable amount of intuitive sense, but clearly these hypotheses being true for $x^3$ isn't proof for the general case. I'm having a bit of trouble understanding what this would imply for a general function that fits all of these characteristics.
It seems that for a function $f$ to satisfy these hypotheses it must be some function $f(x) = x^n ... + \text{ }c$ where $n: n \geq 3$ and $n$ is some odd integer and $c$ is a constant. This would imply that $f'(x)$ is of even-degree, and $f''(x)$ is of odd-degree.
It's a bit confusing to be honest.
 A: Alex has prodded me to a more elementary solution, so here is a second try: Suppose $f''(0)>0.$ Then by continuity of $f'',$ $f''> 0$ in a neighborhood of $0.$ This implies $f'$ is strictly increasing near $0.$ But $f'(0)=0.$ Thus $f'<0$ to left of $0$ and $f'>0$ to the right - at least near $0.$ That is a contradiction. Same idea if we assume $f''(0)<0.$ Thus $f''(0)=0.$ Now since $f''$ is given to be continuous and 1-1 near $0,$ $f''$ is either strictly increasing or strictly decreasing near $0.$ In either case, $f''$ changes sign at $0,$ which is the desired conclusion.
A: WLOG, $f(0)=0.$ Because $f'(0)=0,$ Taylor says $f(x) = f''(0)x^2/2 +o(x^2).$ Suppose $f''(0)>0.$ Then $f$ has a strict local minimum at $0.$ But the condition on $f'$ implies $f$ is strictly increasing. That's a contradiction. Similarly, $f''(0)<0$ is ruled out. Therefore $f''(0)=0.$ Now $f''$ is given to be one-to-one and continuous in a neighborhood of $0.$. Thus $f''$ is strictly increasing or strictly decreasing there. In either case, $f''$ changes sign at $0,$ giving the desired point of inflection.
