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Let $f:[-1,1]\rightarrow \mathbb R$ and $f(x)=\frac {x} {1+\sqrt{1-x^2}}$. Show that $f$ is continuous, $f([-1,1]) = [-1,1]$ and $f$ has an inverse function $f^{-1}: [-1,1]\rightarrow[-1,1]$.

I think it all comes down to showing that it's monotonously increasing. If I proove that, then I can show that $f([-1,1]) = [-1,1]$ and that it has an inverse function follows from it being strictly monotonously increasing. (please correct me if I said something false!)

Do I have to show something special to prove that it's continuous? Clearly $x$ is continuous and $1+\sqrt{1-x^2}$ also has no point that could present a problem in this interval. Can I just write this?

For the monotony proof I'm a little stuck. I have the following:

$\frac {x} {1+\sqrt{1-x^2}} < \frac {x+1} {1+\sqrt{1-(x+1)^2}}$

$x\sqrt{1-(x+1)^2} < x(1+\sqrt{1-x^2})$

$x\sqrt{-x^2+2x+2} < x\sqrt{1-x^2}+1+\sqrt{1-x^2}$

I can not reduce this anymore and from here it's not clear that it's monotone. Any tips? Many thanks guys!

PS: we are not allowed to use the derivative here.

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For proving that $f$ is monotone, you should consider $x_1,x_2\in[-1,1]$ such that $x_1<x_2$ and compare $f(x_1)$ with $f(x_2)$ – Norbert Dec 18 '11 at 16:08
Regarding monotony, you have to prove that $x\leq y\ \Rightarrow \ \text{either} f(x)\leq f(y) \text{ or } f(x)\geq f(y)$. – Pacciu Dec 18 '11 at 16:09
Try setting $x=\sin \theta$ with $|\theta|\leq \pi/2$. Remember your trig identities – Thomas Andrews Dec 18 '11 at 16:10
Can I write: The numerator is increasing, which causes the functions to grow. [-1, 0] the denominator is increasing, but the numerator grows faster than the denominator, so the function grows. [0, 1], the denominator is getting smaller, which makes the function grow even faster. => it grows between [-1,1]? – Clash Dec 18 '11 at 16:12
@Clash: For $0 \le s <t \le 1$, you have pinned down the why, a little informally. More formally you can write $s/(1+\sqrt{1-s^2})<t/(1+\sqrt{1-s^2})<t/(1+\sqrt{1-t^2})$. – André Nicolas Dec 18 '11 at 17:37
up vote 3 down vote accepted

For continuity, just note (as you said) that $f$ is built up from continuous functions and that there are no domain problems.

For invertibility:

Your function is odd and has value 0 at 0. Towards showing that $f$ is one-to-one, it thus suffices to show that $f$ is monotone for $x\in (0,1]$.

If $x$ and $y$ are in $(0,1]$ with $x<y$, then $1-x^2>1-y^2$. Then $1+\sqrt{1-x^2}>1+\sqrt{1-y^2}$, and so $$ {1\over 1+\sqrt{1-x^2}}<{1\over 1+\sqrt{1-y^2}}. $$ Since $x>0$ and $y>x$:

$$ {x\over 1+\sqrt{1-x^2}}<{y\over 1+\sqrt{1-y^2}} $$

So, $f$ is monotone on $(0,1]$ and, thus, on $[-1,1]$.

Now, one can show $|f(x)|\le 1$ for all $x\in[-1,1]$. Since $f(-1)=-1$ and since $f(1)=1$, it follows from the continuity of $f$ that its range is $[-1,1]$. Thus $f^{-1}$ exists and its domain is $[-1,1]$.

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