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    (Original post by emx_eco)
    ooooooh therefore the domain is x > 3. I get it now thank you!!

    Do you think you could help me with part b) -


    f(x) = e^(2x) + 3 where x is any real number
    g(x) = ln(x-1) where x is any real number and x > 1

    Find fg(x) and state its range
    Have you made a start?
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    (Original post by _gcx)
    Have you made a start?
    so at the moment I have:

    fg(x) = eln(x-1)^2+ 3
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    (Original post by emx_eco)
    so at the moment I have:

    fg(x) = eln(x-1)^2+ 3
    Not quite.

    You should have have fg(x) = e^{2\ln(x-1)} + 3. You can write this as \left(e^{\ln(x-1)}\right)^2 + 3. Do you know what to do from there?

    (unless you meant \ln[(x-1)^2] or \left(e^{\ln(x-1)}\right)^2 + 3, then you're fine)
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    (Original post by _gcx)
    Not quite.

    You should have have fg(x) = e^{2\ln(x-1)} + 3. You can write this as \left(e^{\ln(x-1)}\right)^2 + 3. Do you know what to do from there?

    (unless you meant \ln[(x-1)^2], then you're fine)
    how can you write it in that format? what is the rule?
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    (Original post by emx_eco)
    how can you write it in that format? what is the rule?
    a^{bc} \equiv (a^b)^c. You also have b\ln(a) \equiv \ln(a^b).
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    (Original post by _gcx)
    a^{bc} \equiv (a^b)^c. You also have b\ln(a) \equiv \ln(a^b).
    oh okay I understand that now. So whats comes next?
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    (Original post by emx_eco)
    oh okay I understand that now. So whats comes next?
    You should be fine that e^{2\ln(x-1)} + 3 = (x-1)^2 + 3. Are you familiar with how to proceed?
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    (Original post by _gcx)
    You should be fine that e^{2\ln(x-1)} + 3 = (x-1)^2 + 3. Are you familiar with how to proceed?
    how do these two things equal each other?
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    (Original post by emx_eco)
    how do these two things equal each other?
    e^{\ln(x-1)} = x-1
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    (Original post by _gcx)
    e^{\ln(x-1)} = x-1
    how come the e cancels out? (sorry for so many questions!)
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    (Original post by _gcx)
    e^{\ln(x-1)} = x-1
    Does it??
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    (Original post by Ano9901whichone)
    Does it??
    yes... it is given in the question that x>1. Please don't confuse the OP, even if your comment was in jest.
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    (Original post by emx_eco)
    how come the e cancels out? (sorry for so many questions!)
    because \ln is the inverse of e^x A property of inverse functions is that f(f^{-1}(x)) = f^{-1}(f(x)) = x.
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    (Original post by _gcx)
    because \ln is the inverse of e^x A property of inverse functions is that f(f^{-1}(x)) = f^{-1}(f(x)) = x.
    ohhhhh! So now that you have (x-1)^2 +3 you just sketch it with the minimum point (1,3) meaning the range is f(x) > 1? (thats meant to be the bigger than or equal to sign with the line underneath)
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    (Original post by emx_eco)
    ohhhhh! So now that you have (x-1)^2 +3 you just sketch it with the minimum point (1,3) meaning the range is f(x) > 1? (thats meant to be the bigger than or equal to sign with the line underneath)
    Close, that's the domain. (x>1, remember that \ln 0 is undefined) The range is fg(x) > 3. Can you see why?
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    (Original post by _gcx)
    Close, that's the domain. (x>1, remember that \ln 0 is undefined) The range is fg(x) > 3. Can you see why?
    oh wait yh, my mistake! Thanks so much for all your help xxx
 
 
 
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