Questions about the equation $f''(x)+(f'(x))^2=x$. Suppose that $f''(x)+(f'(x))^2=x$. How could we show that $f(0)$ is not a maximal value of $f(x)$, $f(0)$ is not a minimal value of $f(x)$, and $(0, f(0))$ is an inflection point of $y=f(x)$?
My partial solution: $f''(x)+(f'(x))^2=x$, we have $f''(0)=0$. If $x<0$, then $f''(x) = x - (f'(x))^2 < 0$. But we don't know the sign of $f''(x)$ when $x > 0$. Thank you very much.
Edit: $f'(0)=0$ is given.
 A: With $g(x) := e^{f(x)}$ you have
$$
   g' = f' g, \quad g'' = f''g + f'^2 g = x g \quad .
$$
Therefore at $x = 0$
$$
  g'(0) = f'(0) = 0 \quad ,
$$
for $x > 0$
$$
  g''(x) > 0 \implies g' \text{ is increasing}
 \implies g'(x) > 0 \implies f'(x) > 0 \quad ,
$$
and for $x < 0$
$$
  g''(x) < 0 \implies g' \text{ is decreasing}
 \implies g'(x) > 0 \implies f'(x) > 0 \quad .
$$
So $f$ is strictly increasing and does not have any (local) maximum or minimum.
$f'$ has a minimum at $x = 0$ which means that $f$ has an inflection point
at $(0, f(0))$.
A: You can write for $x>0$ that $f^{\prime}(x)=f^{\prime}(0)+xf^{\prime\prime}(c)=xf^{\prime\prime}(c)$ with $0<c<x$ (hence $f^{\prime\prime}(c)\to 0$ as $x\to 0$). Then
$$f^{\prime\prime}(x)=x-x^2(f^{\prime\prime}(c))^2=x(1-xf^{\prime\prime}(c)^2)$$
and there exists $a>0$ such that $f^{\prime\prime}(x)>0$ if $0<x<a$.
A: Differentiate the differential equation once, and see that $f'''(0)=1$. Therefore $f''(x)$ goes from being negative to being positive around $0$.
