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Let $f:\left]-1,\infty\right[ \to \mathbb{R}$ be defined by $f(x) := \frac{1}{1+x \cdot |x|} .$ I want to prove that $f$ is not bounded above.

Here is my attempt: I assume that $f$ is upper bound by $k$:

$\exists k \gt 0 : f(x) \le k \qquad x \in \left]-1, \infty\right[ $

$\Rightarrow \qquad \frac{1}{1+x\cdot\left|x\right|} \le k .$

But at this point I don't know how to proceed. I can't find the contradiction.

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    How can I finish my approach?2011-11-25

3 Answers 3

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If $x<0$, then

$ \frac{1}{1+x|x|}=\frac{1}{1-x^2}. $

You can make $\frac{1}{1-x^2}$ as large as you wish by selecting $x$ to be sufficiently close to $-1$.

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    @mcb: Is this a specific assignment for a class demanding that you use contradiction? With my formula above you can explicitly find an $x$ such that $f(x)$ is larger than $M$ for any M>0.2011-11-26
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We show that $f$ has no upper bound on $(-1,0)$, because we work with negative values you can simplify your equation to

$f(x)=\frac{1}{1-x^2}$

assume that $f$ is bounded by some $k > 2$ then we have that

$-\frac{\sqrt{k}}{\sqrt{k+1}} \in (-1,0)$ but $f(-\frac{\sqrt{k}}{\sqrt{k+1}})=k+1$ a contradiction.

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    Certainly its negative and certainly the denominator is bigger than the nominator.2011-11-25
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x|x|=-1 has one root that's x=-1 so lim f(x)=1/(1+x|x|) when x-->-1 goes to ∞