A differentiation and double differential function proof 
This is a question which I received from my friend as he was not able to solve it. Even I and my professor too failed to solve this question. Pls someone help me with this question.
I have been able to proved it till here-
g(0) 9
 A: Suppose $f(x)$ meets the conditions of the problem. Note that since $|f(0)| \le 1$, $2\sqrt 2 \le |f'(0)| \le 3$. Now $-f(x), f(-x),$ and $-f(-x)$ also meet all the conditions. At least one of the four functions will have both derivative and second derivative $\ge 0$ at $x = 0$. Let $h(x)$ be the one with both $h'(0) \ge 0, h''(0) \ge 0$.
So $2\sqrt 2 \le h'(0) \le 3$ and $h''(0) \ge 0$. If $h'''(0) < 0$, we are done. Otherwise, Let $a = \sup \{b \ge 0 \mid h''' \ge 0 \text{ on } [0,b]\}$, so $h''' \ge 0$ on $[0,a)$.
$h''$ is increasing on this interval, and so $h''(x) \ge h''(0) \ge 0$ everywhere. This in turn means that $h'$ is increasing on the interval as well. So $h'(x) \ge h'(0) \ge 2\sqrt 2$. Since $h'$ is continuous, it is integrable and $$h(a) - h(0) = \int_0^a h'(x)\,dx \ge 2\sqrt 2 a$$
But $|h(a) - h(0)| \le |h(a)| + |h(0)| \le 2$. So
$$2 \ge 2\sqrt 2a\\\frac 1{\sqrt 2} \ge a$$
Because $h'(a) \ge 2\sqrt 2$ and continuous, there is some neighborhood of $a$ where $h' > 0$. But by the choice of $a$, there are points $c$ in that neighborhood where $h'''(c) < 0$, and therefore $h'(c)h'''(c) < 0$.
From this it follows that either $f'(c)f'''(c) < 0$ or $f'(-c)f'''(-c) < 0$, and further the point $c$ or $-c$ is in $(-b, b)$ for any $b > \frac 1{\sqrt 2}$, for which $b = 3$ certainly qualifies.
