Prove $\exists c\in(0,x), \frac{f(x)}{x^2}=\frac{f"(c)}{2}$ I am given a twice differentiable $f$ in $[0,\infty)$ such that $f(0)=f'(0)=0$. I need to prove that for all $x>0$, there's a $0<c<x$ such that $\frac{f(x)}{x^2}=\frac{f"(c)}{2}$.
This equation has brought up many Ideas in my mind:
-Manipulating the equation to get $2f(x)-x^2f''(c)=0$, which could somehow be related to the derivative of a quotient.
-Seeing that $\lim_{x\rightarrow0}\frac{f(x)}{x^2}=\frac{f"(0)}{2}$ using L'Hôpital's rule
-Using Lagrage's theorem, we see there's a $0<c<x$ such that $f''(c)=\frac{f'(x)-f'(0)}{x-0}=\frac{f'(x)}{x}$, or perhaps using the theorem on $f$/both?
All of these ideas didn't get me anywhere at all, so I thank everyone who helps either by using what I brought up (unlikely) or coming up with their own solution (which can't include taylor expansions/integrals), in advance.
 A: An often-neglected property of derivatives is the intermediate value property: if $f''(c) \not= \dfrac{2f(x)}{x^2}$ for all $0 < c < x$ then either $f''(c) < \dfrac{2f(x)}{x^2}$ for all $0 < c < x$ or $f''(c) > \dfrac{2f(x)}{x^2}$ for all $0 < c < x$.
Assume the latter condition holds. We can invoke the mean-value theorem to see that for every $0 < b < x$ there is a point $c \in (0,b)$ which satisfies
$$f'(b) = f'(b) - f'(0) = f''(c)(b-0) > \frac{2 bf(x)}{x^2}.$$
Define $g(b) = f(b)-\dfrac{b^2 f(x)}{x^2}$.  Then $g(0) = 0$ and $g'(b) > 0$ for all $0 < b < x$.  Thus $g(x) > 0$ which reduces to $f(x) > f(x)$, a contradiction.
A: Recall the generalization of the mean value theorem called Cauchy's theorem:

if $f$ and $g$ are continuous over $[a,b]$, differentiable over $(a,b)$ and $g'(t)\ne0$, for $t\in(a,b)$, then there exists $\xi\in(a,b)$ such that
  $$
\frac{f(b)-f(a)}{g(b) -g(a) }=\frac{f'(\xi)}{g'(\xi)}
$$

By Cauchy's theorem applied to $[0,x]$ with $g(x)=x^2$, there is $d\in(0,x)$ such that
$$
\frac{f(x)}{x^2}=\frac{f(x)-f(0)}{x^2-0^2}=\frac{f'(d)}{2d}
$$
Again, by Cauchy's theorem applied to $f'(x)$ and $g'(x)$ on $[0,d]$, there is $c\in(0,d)$ such that
$$
\frac{f'(d)}{2d}=\frac{f'(d)-f'(0)}{2d-2\cdot0}=\frac{f''(c)}{2}
$$
This is essentially the same as the proof of the Taylor expansion with Lagrange remainder.
A: Here is the typical standard proof available in almost any textbook. In what follows $x$ is a fixed positive number and $u$ is the variable with respect to which derivatives are calculated.
Let $$g(u) = f(x) - f(u) - (x-u) f'(u), G(u) = g(u)-\frac{(x-u) ^{2}}{x^{2}}g(0)$$ then we have $G(0)=G(x) =0$ and hence there is a $c\in (0,x)$ for which $G'(c) =0$. This means that $$g'(c) + \frac{2(x-c)}{x^{2}}g(0)=0$$ or $$-f'(c) + f'(c) - (x- c) f''(c) + \frac{2(x-c)}{x^{2}}f(x)=0$$ or dividing by $2(x-c)$ we have $$\frac{f(x)} {x^{2}}=\frac{f''(c)}{2}$$
A: Well, I figured I could just use the generalized intermediate value theorem twice.
$\exists c\in(0,x),\  f''(c)=\frac{f'(y)-f'(0)}{y-0}=2\frac{f'(y)}{2y}=2\frac{f(x)-f(0)}{x^2-0^2}$
