TITLE: Prove the existence of $c$ such that $f'(c) = 2c(f(c) - f(0))$
QUESTION [5 upvotes]: Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function s.t. $f'$ is continuous. Suppose $f'\left(\frac{1}{2}\right)=0$, prove that there is $c\in\left(0,\frac{1}{2}\right)$ s.t.
$$f'(c)=2c(f(c)-f(0))$$

REPLY [3 votes]: Consider the function
$$
g(x)=e^{-x^2}(f(x)-f(0))\tag{1}
$$
Taking the derivative, we get
$$
g'(x)=e^{-x^2}\left(f'(x)-2x(f(x)-f(0))\right)\tag{2}
$$
Notice that
$$
g(0)=0\tag{3}
$$
and
$$
g(1/2)=e^{-1/4}(f(1/2)-f(0))\tag{4}
$$
and
$$
g'(1/2)=-e^{-1/4}(f(1/2)-f(0))\tag{5}
$$
If $g(1/2)=0$, then by $(3)$ and the Mean Value Theorem, there is a $c\in(0,1/2)$ so that $g'(c)=0$.
If $g(1/2)\gt0$, then there is an $x_0$ in $(0,1/2)$ so that $g'(x_0)=2(g(1/2)-g(0))\gt0$.  Furthermore, by $(4)$ and $(5)$, $g'(1/2)\lt0$. By the intermediate value theorem, there is a $c\in(x_0,1/2)$ so that $g'(c)=0$.
If $g(1/2)\lt0$, then there is an $x_0$ in $(0,1/2)$ so that $g'(x_0)=2(g(1/2)-g(0))\lt0$.  Furthermore, by $(4)$ and $(5)$, $g'(1/2)\gt0$. By the intermediate value theorem, there is a $c\in(x_0,1/2)$ so that $g'(c)=0$.
Thus, there is a $c\in(0,1/2)$ so that $g'(c)=0$, which by $(2)$, gives us
$$
f'(c)-2c(f(c)-f(0))=0\tag{6}
$$