Consider the function $y(x)=f(x)(1-x)$ where $x\in[0,1]$, $y(0)=y(1)=0$.
Knowing that $f(x)$ is continuous differentiable everywhere, $f(0)=0$, $f(1)=c$ where $0<c<1$, $f'(x)>0$, can one claim that the maximum $x^* \in (0,1) $ is unique?
Thank you in advance
First, consider $g(x)=\dfrac{f(x)}{f'(x)}+x$. We got $g'(x)=\dfrac{(f')^2(x)-f'(x)f''(x)+(f'')^2(x)}{(f'')^2(x)}=\dfrac{(f'(x)-f''(x))^2+f'(x)f''(x)}{(f'')^2(x)}>0$. Then g is a increasing function. Supose that $x$ and $z$ are critical points for your $y$ function, by your relation, $\dfrac{f(x)}{f'(x)}+x=1=\dfrac{f(z)}{f'(z)}+z$, then $g(x)=g(z)$, hence $x=z$, as you desired.
I'm wondering about your way of answering your question. How do you conclude that $f(x)/f'(x)$ is increasing? My g function is, but $g(x)-x$ is increasing?
P.S.: the derivative of g is not greater than zero. I've made a mistake on differentiation. Haha