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Firstly, what is a function of q? Am I correct to assume it means $f(q)$?

$w=q+1$ For this one, it is a linear function, so it has to be a function of q. But I'm not sure how to express it?

$wq=-27$ Again, $w=\frac{-27}{q}$ This is a linear function? So it has to be a function?

$sin(w) = q^2$ I am not sure for this one.

All I've learned about functions is $f(x)$ notation. So I am not sure how I am supposed to see these questions?

How do I turn the above relations into functions of q?

Jason
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  • Is there some domain given for $q$? E.g. something like $q>0$ or $q\in\Bbb R$. – Berci Aug 12 '14 at 19:43
  • No, all the question states is "Which of the following relations are functions of q:" – Jason Aug 12 '14 at 19:48
  • All of them are functions of $q$. But there should be some restrictions on $q$ for second and third one. Meanly for second one $q\neq0$ and for third one $q\in[-1,1]$. – Ömer Aug 12 '14 at 19:52

1 Answers1

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Here is an answer for $sin(w)=q^2$:

In order to express $w$ as a function of $q$, you need to apply $\arcsin$ on both sides of the equation.

This gives you $w=\arcsin(q^2)$.

This function is defined only in the domain $q\in[-1,1]$, and its range is $w\in[0,\frac{\pi}{2}]$.

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barak manos
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  • Ah yes, and it's not a function because arcsin can return multiple values, i.e. arcsin of 0 is $k\pi$ where k is an integer. – Jason Aug 12 '14 at 19:54
  • @Jason: Please read the answer a little more carefully. It's a function in the specified domain. – barak manos Aug 12 '14 at 19:55