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

  • $\begingroup$ Thanks for the comment, I hope it is clearer now. $\endgroup$ – Hans Schnier Jun 13 '18 at 18:21
  • $\begingroup$ Absolute. I did the edit. Thanks for the comment. $\endgroup$ – Hans Schnier Jun 13 '18 at 18:56
  • $\begingroup$ Are you sure you don't mean local max/min? For example, when $f(x)=x^2$, $y$ has no absolute max in $(0,1)$. $\endgroup$ – philbo_baggins Jun 13 '18 at 18:59
  • $\begingroup$ I may be wrong, but according to the Rolle's Theorem in (0,1) there is at least a stationary point, moreover since f(x) is strictly increasing any value of $x \in(0,1)$ implies y(x)>y(0)=y(1)=0. I am asking if the stationary point is unique under the condition above. If the question is not clear let me know that I can edit it again. Best $\endgroup$ – Hans Schnier Jun 13 '18 at 19:17
  • $\begingroup$ I was thinking to show that $y(x)'>0$ from $(0,x_1)$ and $y(x)'<0$ from $(x_1,1)$ where $0<x_1<1$, but the $f''(x)>0$ confused me. Best $\endgroup$ – Hans Schnier Jun 13 '18 at 19:39

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


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