anonymous
  • anonymous
The one to one function f is defined by: f(x)=x/7x+2 Find the inverse of f. Then, give the domain and range of f^-1 using interval notation. f^-1(x)= Domain (f^-1)= Range (f^-1)=
Mathematics
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jamiebookeater
  • jamiebookeater
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apoorvk
  • apoorvk
for finding inverse of a function y=f(x) you need to write x in terms of y (rearrange the equation), then interchange y and x. that ll be f inverse of x. hope this helps. post what you think should be the answer.
anonymous
  • anonymous
That's the problem though, I don't know how to switch the equation
apoorvk
  • apoorvk
okay i ll show you below. uh-oh. do you mean f=x/(7x+2) up there? if yes then, |dw:1332632228201:dw| (hmm tricky one. you hafta use hook or crook to get it) nope now you can carry on further. :)) and domain and range a problem?

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anonymous
  • anonymous
is the inverse 2x/1-7x
apoorvk
  • apoorvk
exactly!!!!! but use brackets when quoting equations here so that we can understand better. :)
anonymous
  • anonymous
okay, umm.. the domain.. (-infinity,-2/7)U(-2/7,infinity)
anonymous
  • anonymous
and range.. (-infinity,1/7)U(1/7,infinity)?
apoorvk
  • apoorvk
umm.. you mean 1/7 instead of -2/7 over there in domain? cause at 1/7 the denominator is 0, and the function value shoots up to infinity, and so gets undefined.
apoorvk
  • apoorvk
what you posted as range would be the domain.
anonymous
  • anonymous
so the domain is: (-inf,1/7)U(1/7,inf)? then whats the range cause thats what i put for the range?
apoorvk
  • apoorvk
yeah thats domain. (i hope you know what domain and range are basically). now for range let me work that out its a bit complex here.
anonymous
  • anonymous
i know what they are, math isnt my greatest subject lol!
apoorvk
  • apoorvk
range should be.. umm...'cause the graph will be like: |dw:1332633941226:dw| sorry i got really puzzled in between so took me some time. yeah so what you posted as domain should be the range!!! basically interchange them!
apoorvk
  • apoorvk
hope its clear now?

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