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Fix an integer $m\geq 1$. Let $f:(0,1)\to \mathbf{R}$ be defined by $$f(x) = 16x^{1-m}\exp \left( \frac{8x}{1-x} \right) .$$ Can we determine a lower bound for $f$ in terms of $m$?

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The title and the body pose two different questions. Which of them is your question? –  joriki Dec 6 '11 at 12:28
    
You're right. I edited it. It's clear that the function has a minimum on a compact subset of $(0,1)$. The function explodes at the end points. I just need a good choice of $x$ to get a value as small as possible. For $m=1$, you can show that $f(x) \to 16$ as $x\to 0$. –  Ralph Tomi Dec 6 '11 at 12:43
    
Take the logarithm $g(x)=\ln f(x)$, and write the equation for its minimum, $g'(x)=0$. This yields a quadratic equation in $x$ (after multiplying through by $x(1-x)^2$). You can solve it explicitly and plug the solution back into your original equation to find the exact minimum. –  mjqxxxx Dec 6 '11 at 13:04

1 Answer 1

up vote 0 down vote accepted

Differentiating with respect to $x$ yields

$$f'(x)=f(x)\left(\frac{1-m}x+8\frac{(1-x)+x}{(1-x)^2}\right)=f(x)\left(\frac{1-m}x+\frac{8}{(1-x)^2}\right)\;.$$

This is zero if the expression in parentheses is zero, that is, for

$$8x+(1-m)(1-x)^2=0\;.$$

You can solve this quadratic equation for $x$ to find the extrema. Since the function goes to infinity for $x\to0^+$ and $x\to1^-$, exactly one of the two extrema lies in $(0,1)$. I don't think you'll get a particularly nice form for the function value at the minimum, though you can simplify it somewhat by using

$$\frac{8x}{1-x}=-(1-m)(1-x)\;.$$

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