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Given is the function $$f(x) =\frac{x}{\sqrt{1-x^2}}$$

How can I prove that this function is monotonic and thus injective?

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What is this function for $|x|\ge 1$? –  Henry Oct 11 '12 at 16:21
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3 Answers

up vote 2 down vote accepted

On the interval $[0,1)$, the numerator $x$ is increasing and the denominator $\sqrt{1-x^2}$ is decreasing. So $f(x)$ is increasing on the interval.

For $-1\le x\le 0$, use the fact that $f(x)$ is an odd function, so has symmetry across the origin. We conclude that $f(x)$ is monotone on $(-1,1)$, the natural domain of definition.

Remark: For other functions, a different approach, such as looking at the sign of the derivative, may be the appropriate one. The derivative should not be the automatic go-to technique, since the derivative may be complicated, and difficult to analyze.

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Thank you, makes perfectly sense.I hope they are going to accept this answer :) –  Marco7757 Oct 11 '12 at 17:11
    
Always depends on context: if you are expected to use the derivative, use the derivative. At very beginning of an analysis course, the derivative and its properties have not yet been introduced, so one would not want to use it. In a calculus course, a derivaive calculation would be expected, so it is safest to do it. –  André Nicolas Oct 11 '12 at 17:15
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Generally approach is to take the derivative of the function and check if the derivative is positive or negative. If positive, the function is increasing. If negative, the function is decreasing.

For this function, $f'(x)=\frac{1}{(1-x^2)^{3/2}}$ which is apparently positive on interval $(-1,1)$. So $f(x)$ is an increasing function on $(-1, 1)$.

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We observe that $f(x) =\tan({\arcsin (x))}$ so as a composition of increasing functions is increasing. enter image description here Notice in that triangle $$\arcsin(\frac{x}{1}) =\theta$$ and $$\tan({\theta})= \frac{x}{\sqrt{1-x^2}}$$

$\sin(x)$ is increasing on $[ -\frac{\pi}{2},\frac{\pi}{2}]$ so then is $\arcsin{x}$ as his inverse.

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