inverse function Q (1 Viewer)

RishBonjour

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v) Assume x=N, when N is not in the domain chosen in part ii) but still in the domain for f(x)
find f^-1(f(N))

f(x) = x+1/x
inverse = x/2 +1/2(x^2-4)^(1/2)

domain found in part ii is x>=2

I've seen a similar question before in success one, but that was a parabola. not sure how to get this.
 
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Timske

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What is your question? Finding the inverse of f(x) = x+(1/x) ?
 

RishBonjour

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Sanjeet

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v) Assume x=N, when N is not in the domain chosen in part ii) but still in the domain for f(x)
find f^-1(f(N))

thats the question.

its similar to this :
http://community.boredofstudies.org/showthread.php?t=283792&highlight=inverse+functions

similar to the question Aysce asked - i understand the symmetry there. but this one, I'm not sure




BTW: have 3u test tomorrow with this topic included. so please try! Or I will fail so hard
Your question is extremely unclear, but I'll have a go

 

bleakarcher

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x+(1/x)=N+(1/N)
Nx^2+N=(N^2+1)x
Nx^2-(N^2+1)+N=0
x=(N^2+1)+/-sqrt[(N^2+1)^2-4N^2] /2N
x=(N^2+1)+/-sqrt[(N^2-1)^2] /2N
x=(N^2+1)+/-(N^2-1) /2N
x=N or N^2+1-(N^2-1) /2N
x=N or 1/N
Hence, f(N)=f(1/N). Since N<1, 1/N>1 and therefore lies in the domain x>=1.
=> f^(-1)[f(N)]=f^(-1)[f(1/N)]=1/N
 

Chowder_Head

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x+(1/x)=N+(1/N)
Nx^2+N=(N^2+1)x
Nx^2-(N^2+1)+N=0
x=(N^2+1)+/-sqrt[(N^2+1)^2-4N^2] /2N
x=(N^2+1)+/-sqrt[(N^2-1)^2] /2N
x=(N^2+1)+/-(N^2-1) /2N
x=N or N^2+1-(N^2-1) /2N
x=N or 1/N
Hence, f(N)=f(1/N). Since N<1, 1/N>1 and therefore lies in the domain x>=1.
=> f^(-1)[f(N)]=f^(-1)[f(1/N)]=1/N
wrong
 

krispykareem

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Yeah buddy i was like who's that chick in the DP and realised it was you LOL!
 

Fus Ro Dah

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View attachment 25735



v) Assume x=N, when N is not in the domain chosen in part ii) but still in the domain for f(x)
find f^-1(f(N))

f(x) = x+1/x
inverse = x/2 +1/2(x^2-4)^(1/2)

domain found in part ii is x>=2

I've seen a similar question before in success one, but that was a parabola. not sure how to get this.
From where did you get this question? it looks exactly like a past HSC Q7.
 

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