C3 Functions help Watch

becky.fm
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#1
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1d) Given that f(x) = x^3 + 2  ;  g(x) = x - 3 what is:

i) fg(x)

ii) (fg)^-1 (-6) (thats the inverse of the (fg) function by the way)



I got i) as being fg(x) = f(g(x)) = f(x - 3) = (x - 3)^3 + 2 = x^3 - 9x^2 +27x - 25
Is that right?

And i don't understand part ii).. any help would be appreciated!

Oh, and i know about how to get inverse functions.. but i dont know how to do it at all on this question
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generalebriety
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(fg)^{-1} = g^{-1} f^{-1}, which is a result you should know and can quote. It's composed as a normal function is, so (fg)^{-1}(-6) = g^{-1} f^{-1}(-6) = g^{-1} (f^{-1}(-6)).
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wanderlust.xx
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First part is right, yeah. For part two, the notation seems to be a bit off... do you mean fg^-1(x)=-6?

edit: Nevermind. Genius above beat me to it. :rolleyes:
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becky.fm
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#4
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Ok, great. I'm still a little bit confused tbh though.

So after i've got g^{-1}(f^{-1}(-6))

and g^{-1} = x + 3 ; f^{-1} = \sqrt[3]{x - 2}

then  g^{-1}(f^{-1}(\sqrt[3]{-6-2})) which doesnt work, as you can't get the cube root of -8.

Where have i gone wrong?

Sigh, this is what you get when you aren't given a textbook yet, and the library doesn't have any
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nota bene
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You would get = \sqrt[3]{x-2}+3 and putting x=-6 indeed gives a non-real answer... I don't see any mistake
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generalebriety
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#6
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(Original post by becky.fm)
then  g^{-1}(f^{-1}(\sqrt[3]{-6-2})) which doesnt work, as you can't get the cube root of -8.
(Original post by nota bene)
You would get = \sqrt[3]{x-2}+3 and putting x=-6 indeed gives a non-real answer... I don't see any mistake
Ahem... the cube root of -8 is -2.
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becky.fm
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(Original post by generalebriety)
Ahem... the cube root of -8 is -2.
Oh ****, yeah Ok, great, done it Thanks!
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generalebriety
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(Original post by becky.fm)
Oh ****, yeah Ok, great, done it Thanks!
No problem.
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