Given a function $f$ satisfies the equation
$xf''(x) + 3x(f'(x))^2 = 1 -e^{-x}$
then
(a)$f$ has a minimum at $x = a\ne 0$
(b)$f$ has a minimum at $x =2$
(c)$f$ has maximum or minimum at $x = 0$
(d) If $f(0) = f'(0) = 0$ then $f(x) \le Ax^2$ for all $x \ge 0$ and $A > 0$
So, here is what I did
$f'(x) = \sqrt{\dfrac{1 - e^{-x} - xf''(x)}{3x}}$, when $x \ne 0$
Since range of $\sqrt{x}$ is $[0, \infty)$,
$f(x)$ is monotonically increasing for $x \ne 0$.
So, clearly option (a) and (b) are False
But I am clueless on how to tackle the case when $x =0$
and I have no idea how to prove\disprove option (d)
Can anyone help me here ?
Thank you.
Note that $c(x)=\frac{1-e^{-x}}x$ can be extended to a smooth function by setting $c(0)=1$. Then $f''(x)+3f'(x)^2=c(x)$ is a smooth ODE on the domain $\Bbb R^2$, without any singularities.
The DE can be considered a Riccati equation for $f'$. Set $g=e^{3f}$, then $f'=\frac{g'}{3g}$, and $g$ as a monotonous transformation has the same local extrema as $f$. The differential equation for $g$ is a linear second order DE $$ g''(x)=3c(x)g(x). $$ In both formulations, the full range of initial conditions is available at $x=0$ or $x=2$, so that there is nothing that can force $f$ to have an extremum there.
As $0< c(x)\le 1$ for $x\ge 0$, the growth of any solution $g$ is limited by a multiple of $e^{\sqrt3 x}$, which gives a linear bound for $f$. Combined with the local quadratic growth under the initial conditions $f(0)=f'(0)=0$, this results in a global quadratic bound.