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can you help me with some hints how to draw or construct the function

$\frac{\sqrt{1-x^2}}{x}$ ($x \in (0,1]$) ?

I really have no idea - maybe by using unit circle..?!

Vazrael
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    How much detail are you expected to draw? If this were a problem in pre-calculus I would suggest finding any intercepts and asymptotes. If this were in calculus I would recommend looking at whether the function changes concavity as well. – DMcMor Oct 11 '17 at 19:55
  • I assume that I should draw it with as much detail as possible. – Vazrael Oct 11 '17 at 19:58

1 Answers1

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To draw it with as much detail as possible, you have to completely "investigate" the function. So, let's start.

We have the function :

$$f(x) = \frac{\sqrt{1-x^2}}{x}, x \in (0,1] $$

The function intersects the x axis at :

$$ f(x) = 0 \Rightarrow \frac{\sqrt{1-x^2}}{x} = 0 \Leftrightarrow x=1 $$

It does not intercept the y axis, but infinitely approaches it, as (only checking the limit to $0^+$ since your given subspace of the function is $(0,1]$ :

$$\lim_{x\to 0^+} \frac{\sqrt{1-x^2}}{x} = \infty$$

After these, let's examine the way the function behaves :

$$f'(x) = -\frac{1}{x^2\sqrt{1-x^2}}, x\in (0,1)$$

Differentiating, we can clearly see that :

$$f'(x) < 0 \forall x\in(0,1)$$

so, our function is decreasing in $(0,1)$.

That's pretty much all you have to know, since the space you're examining the function at is only $(0,1)$ which simplifies things up.

  • Intersection with $x'x$ is at : $x=1$
  • Your function is decreasing at $(0,1]$
  • The function approaches $0$ infinitely at $+\infty$

This should be enough to make a pretty good graph of the given function.

Rebellos
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