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Consider the function $g:\left(0,1\right)\rightarrow\mathbb{R}$ defined by $ g\left(x\right)=\left(1-x\right)\left(1-\frac{1}{1+f\left(x\right)}\right), $ where $f\left(x\right)$ is a continuously differentiable function that is positive and strictly increasing with $x\in\left(0,1\right)$. Can one claim that if $g$ has a maximum, this maximum is unique?

Thanks in advance,

Paul

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For example if $f(x) = 0.1 + x \text { if } 0 \le x \lt 0.1$ $f(x) = 1.15 + x \text { if } 0.1 \le x \lt 0.4$ $f(x) = 4.6 + x \text { if } 0.4 \le x \le 1$ then I think you will find there are maxima at $x = 0.1 \text{ and } 0.4$ when $f(x) = 1.25 \text{ and } 5$ respectively and $g(x)=0.5$.

It would not be difficult to make $f(x)$ continuous.

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    @Paul: When adding conditions to the question please be sure to edit them into the question itself so everyone can see it.2011-07-27
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How about $ f(x) = x + \frac{{1 - \cos (100x)}}{{100}}. $ Note that $f$ is strictly increasing on $(0,1)$, since f'(x) = 1+\sin(100x). In turn, since $f(0+)=0$, $f$ is positive on $(0,1)$. However, $g$ seems to have quite many local maxima (and minima).

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    I thank you all for your helpful answers.2011-07-28