# How can I prove that $f(x):=\frac{1}{1+x\cdot\left|x\right|}$ is not bounded above?

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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I believe what you are trying to show is that the function $f(x) = \frac{1}{1 + x|x|}$ has no upper bound on the interval $(-1,\infty)$ (and yes I hate the notation $]a,b[$, despite how overloaded the parentheses notation is!). Anyways, is that what you are trying to show? – JavaMan Nov 25 '11 at 20:46
Could you be a bit more precise about what it is you're trying to prove, please? You want your $f$ to be an upper bound for what? – Henning Makholm Nov 25 '11 at 20:46

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Graph the function, or if that is difficult have software do it for you. You will find that $f(x)$ is awfully big just to the right of $x=-1$. You should be able to translate that intuitive knowledge into a proof of the level of formality appropriate for your course. – André Nicolas Nov 25 '11 at 20:52
@JavaMan : Yes. – mcb Nov 25 '11 at 20:57
@André Nicolas: I think I have to prove it by finding a contradiction. – mcb Nov 25 '11 at 20:59
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## 2 Answers

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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I think I have to prove this by pointing out a contradiction. – mcb Nov 25 '11 at 21:12
@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 $M>0$. – Joe Johnson 126 Nov 26 '11 at 2:01

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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How did you know it would be between -1 and 0? – mcb Nov 25 '11 at 21:43

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Certainly its negative and certainly the denominator is bigger than the nominator. – Listing Nov 25 '11 at 21:47
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