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.


iv) is false.

If $f$ were odd you would have $$f(x) = - f(-x) \implies f'(x) = f'(-x) \implies f''(x) = - f''(-x)$$ which contradicts the fact that $f'' > 0$ on a neighborhood of $0$.

iii) and v) are true.

You've showed that $f'' > 0$ in a neighborhood of any critical number $x_0$, implying that $f'$ is strictly increasing in that neighborhood. Since $f'(x_0) = 0$, this means that $f$ is increasing on some interval $(x_0,x_0+\epsilon)$ and decreasing on some interval $(x_0-\epsilon,x_0)$.

Suppose (for the sake of contradiction) that $f$ has two distinct critical numbers $x_1$ and $x_2$. Since $f$ is continuous it attains a maximum value on the interval $[x_1,x_2]$. Since $f$ is increasing on an interval $(x_1,x_1+\epsilon)$ and decreasing on an interval $(x_2-\epsilon,x_2)$ the maximum cannot occur at either $x_1$ or $x_2$, meaning that the maximum value of $f$ on $[x_1,x_2]$ occurs at an interior point $x_3$.

According to the remarks above, $f$ must be decreasing on some interval $(x_3-\epsilon,x_3) \subset [x_1,x_2]$, contrary to the fact that $f$ attains a maximum value at $x_3$. Thus $f$ cannot have two distinct critical numbers.

Furthermore, if $f$ had three distinct zeros, then according to Rolle's theorem $f'$ would have to have at least two distinct zeros, which was just ruled out.

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