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Can you solve it?

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1. Yes, I can. But I'm not just going to give you the answer - show your working.

Edit: When I posted this I wasn't anticipating that others would 'join in' - I'm actually trying to help, so if you post your working I'll let you know where you're going wrong or give you some pokes in the right direction.
2. Is this something like a questionnaire? I think I can solve it.
3. No
4. No
5. I can, however nuodai summed it up quite well
6. Yeah I can.
7. Yep, but it would ruin the fun of Maths for you if I did it, wouldn't it?
8. (Original post by nuodai)
Yes, I can. But I'm not just going to give you the answer - show your working.

Edit: When I posted this I wasn't anticipating that others would 'join in' - I'm actually trying to help, so if you post your working I'll let you know where you're going wrong or give you some pokes in the right direction.
(a) f(x) = (1-x^2)^1/2 & f(x) = -((1-x^2)^1/2)
(b) no function exists with such a property
(c)
9. (Original post by math1234)
(a) f(x) = (1-x^2)^1/2 & f(x) = -((1-x^2)^1/2)
(b) no function exists with such a property
(c)
Your answers for (a) and (b) are correct from what I can tell. But I don't understand your solution to (c); what is , for instance?
10. (Original post by nuodai)
Your answers for (a) and (b) are correct from what I can tell. But I don't understand your solution to (c); what is , for instance?

g_1, g_2, g_3 = g_i
g(f(x))=x
11. why not you guys are now making fun of my post?(except nuodai) the reason is because you don't understand this question.
12. (Original post by math1234)

g_1, g_2, g_3 = g_i
g(f(x))=x
Right, so it's a piecewise inverse to the cubic. That's right.

13. (Original post by nuodai)
Right, so it's a piecewise inverse to the cubic. That's right.

how can we find y=f_i (x) =g_i^-1 (x) explicitly? & on what intervals g are one-one contained: (-infinity, -1) or (-1, 1) or (1, infinity)?

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