Let B be set of all twice differentialbe function $ f(0)=1, f'(0)=-1$ . .. Find supremum of $ {(f''(0):f\in B})$

Question

Let B be set of all twice differentiable function $f$ such that $f: (-1,1) \to (0,\infty)$ and $ f(0)=1, f'(0)=-1$ .
We have new function $g(x)$ such that $g(x)=\frac{1}{f(x)}$ and $g(x)$ is convex on interval $(-1,1)$.
Find the supremum of the set $\{ f''(0):f \in B \} $

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My work:
If function $g(x)$ is convex that means that $g''(x)>0$ $$g'(x)=-\frac{f'(x)}{f^2(x)}$$ $$g''(x)=\frac{2f'(x)^2-f(x)f''(x)}{f^3(x)}$$ But I don't have idea how to use value of $f$ and $f'$ to get resultat.

Answer 1

Hint:

After calculating $$g^{\prime\prime} (x) = -\frac{f^{\prime\prime} (x)f(x)-2(f^\prime(x))^2}{f(x)^3}$$ By inserting the value for $x=0$ and using $g^{\prime\prime} (x)\ge0$ you arrive, after rearranging at $f^{\prime\prime} (x)\le-2$. You can now try to find $f$ such that value is attained or such that you get at least close to that value. If that works out the $\sup$ is $-2$...