Show there exists $c\in(0,1)$ such that $\frac{f'(1-c)}{f(1-c)}=\frac{2f'(c)}{f(c)}$

Question

I got a question like this:

Prove there exists $c\in(0,1)$ in which $\displaystyle \frac{f'(1-c)}{f(1-c)}=\frac{2f'(c)}{f(c)}$, where $f:[0,1]\rightarrow\mathbb{R}$ is a differentiable function on $[0,1]$ such that $f(0)=0$ and $f(x)>0$ for any $x\in(0,1]$.

The first thing that comes to my mind is the Cauchy MVT, I tried with $g(x)=f(x)^2$ and created the term $g'(x)=2f(x)f'(x)$ so I can have the coefficient 2, but I could not proceed. Then I tried to rearrange the terms and notice $f'(1-c)f(c)-f'(c)f(1-c)=f'(c)f(1-c)$ is what we need to prove, the terms in the LHS is actually the derivative of $h(x)=f(x)f(1-x)$, where $h(0)=h(1)=0$, but I still failed to proceed. May I look for some advice?

Answer 1

My solution:

Let $g(c)=(f(c))^2 f(1-c)$

The function $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$ since f is differentiable on (0,1).

$$g'(c)=2f(c)f'(c)f(1-c)-(f(c))^2f(1-c)$$

Since $g(0)=g(1)=0$, then by Rolle's theorem there exists a $c\in (0,1)$ such that $g'(c)=0$.

For this $c$ we have

$$2f(c)f'(c)f(1-c)=(f(c))^2f(1-c)$$

Since $f(c)>0$, dividing both sides by $f(c)$ gives

$$2f'(c)f(1-c)=f(c)f'(1-c)$$

Answer 2

Optimisation version: since $$ g(x) = \ln(f(1-x)) + 2\,\ln(f(x)) $$ tends to $-\infty$ when $x\to 0$ and $x\to 1$ and is differentiable, it has a maximum at some $c\in (0,1)$ at which $g'(c)=0$.