The number of critical points I'd like to know if it's possible estimate the number of critical points that a function is able to have, because, I have the follow problem
Let $f: \mathbb{R}\rightarrow\mathbb{R}$ be function of class $C^2$
$$f''(x)+xf'(x)=\cos(x^3f'(x))$$
Does $f$ admit at most a critical point?
I know that's true, but I don't know how to prove it!
Also, is there any way to discover the parity of $f$?
 A: Under the hypothesis of this question, $f(x)$ has at most one critical point.
I argue as follows:
First of all, we observe, from scrutiny of the given equation satisfied by $f(x)$,
$f''(x)+xf'(x)=\cos(x^3f'(x)), \tag 1$
that 
$f''(x_c) = 1 \tag 2$
at every critical point $x_c$ of $f(x)$. For
$f'(x_c) = 0 \tag 3$
by the definition of critical point.  Setting $x = x_c$ in (1) yields
$f''(x_c) + x_c f'(x_c) = \cos(x_c^3 f'(x_c)), \tag 4$
or
$f''(x_c) + x_c \cdot 0 = \cos (x_c^3 \cdot 0), \tag 5$
whence
$f''(x_c) = \cos 0 = 1 \tag 6$
as asserted.  
Now since $f''(x)$ is continuous, we may for every critical point $x_c$ pick a real $\mu_c$ with $0 < \mu_c < 1$
and find $\delta_c > 0$ such that
$s \in (x_c - \delta_c, x_c + \delta_c) \Longrightarrow f''(s) > \mu_c; \tag 7$
then for $x \in (x_c, x_c + \delta_c)$, 
$f'(x) - f'(x_c) = \displaystyle \int_{x_c}^x f''(s) ds \ge \int_{x_c}^x \mu_c ds = \mu_c(x - x_c) > 0; \tag 8$
by virtue of (3), this becomes
$f'(x) > 0; \tag 9$
likewise for $x \in (x_c - \delta_c, x_c)$,
$- f'(x) = f'(x_c) - f'(x) = \displaystyle \int_x^{x_c} f''(s) ds \ge \int_x^{x_c} \mu_c ds = \mu_c(x_c - x) > 0, \tag{10}$
i.e., 
$f'(x) < 0; \tag{11}$
(9) and (11) together show that for any critical point $x_c$
$x \in (x_c - \delta_c, x_c) \Longrightarrow f'(x) < 0,\; x \in (x_c, x_c + \delta_c) \Longrightarrow f'(x) > 0; \tag{12}$
we also see from (12) that the critical points of $f(x)$ are isolated, that is, there is a finite distance in $x$ between any two of the $x_c$. 
Now suppose $f(x)$ has more than one critical point; then by what we have just shown, there must exist two critical values $x_{c1} < x_{c2}$ with no critical point between them, i.e.,
$x \in (x_{c1}, x_{c2}) \Longrightarrow f'(x) \ne 0; \tag{13}$
furthermore, by (12) there are $\delta_{c1}$ and $\delta_{c2}$ such that
$x_{c1} + \dfrac{\delta_{c1}}{2} < x_{c2} - \dfrac{\delta_{c2}}{2}, \tag{14}$
$f'(x_{c1} + \dfrac{\delta_{c1}}{2}) > 0, \; f'(x_{c2} - \dfrac{\delta_{c2}}{2}) < 0, \tag{15}$
and
$x \in [x_{c1} + \dfrac{\delta_{c1}}{2},x_{c2} - \dfrac{\delta_{c2}}{2}] \Longrightarrow f'(x) \ne 0; \tag{16}$
but now we have $f'(x)$ continuous on $[x_{c1} + \dfrac{\delta_{c1}}{2},x_{c2} - \dfrac{\delta_{c2}}{2}]$ and taking values with opposite signs on the endpoints of this interval; by the intermediate value theorem, there must exist some $x_c \in [x_{c1} + \dfrac{\delta_{c1}}{2},x_{c2} - \dfrac{\delta_{c2}}{2}]$ with $f'(x_c) = 0$, in contradiction to (16); therefore, $f(x)$ has at most one critical point on all of $\Bbb R$.
Must $f(x)$ have at least one critical point, implying in light of the above it has exactly one?  Perhaps a further analysis of (1) would provide an answer to this question, but I don't have one as of this writing.  Nor do I have, at the moment, any words on the parity of $f(x)$.
Finally, there are some general results available on estimating the number and nature of critical points a function $f: \Bbb R \to \Bbb R$ may have; for example, if they are non-degenerate, meaning $f''(x_c) \ne 0$, one can see they must alternate between local maxima and minima; further results may be available if more hypotheses are placed upon $f(x)$; remember, google is your friend . . . 
