Consider $f:\mathbb{R}\to\mathbb{R} \in C^2$ such that $f''(x)+xf'(x) = \cos(x^3f'(x)) \;\;\forall x \in \mathbb{R}$ Which of the following are correct?
(i) If $f$ has a critic point $x_0$, then it is a local minimum.
(ii) $\exists r>0$ such that $f$ is concave upwards in $(-r,r)$.
(iii) $f$ can have at most one critic point
(iv) $f$ is odd
(v) $f(x)=0$ has at most two solutions.
Here is my attempt:
(i) If $f'(x_0)=0$, then $f''(x_0) = \cos(0) = 1>0$, so it is a local minimum. True.
(ii) By continuity, as $f''(0)>0$, then there is a neighborhood $B(0,r)$ where $f''(x)>0 \forall x \in (-r,r)$. True.
(iii) Suppose $f'(x_0)=f'(x_1)=0$ for $x_0>x_1$. Then there must be a point $c \in (x_0,x_1)$ such that $f''(c)<0$, since both $x_0$ and $x_1$ are local minima and $f$ is continuous. True. (How can I write this precisely?)
(iv) Since $0$ is the only critic point of $f$, then $\lim_{|x|\to\infty} f(x) = \infty$, so it can't be odd. False. (How can I write this precisely?)
(v) Since $0$ is the only critic point and is not odd, then it can only cross $x$-axis one time left at $0$ and one time right at $0$, at most. True. (How can I write this precisely?)
I think I solved this exercise based on geometric intuition, but I'd like to be rigorous. Please help me formalize these ideas.