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    How would you find the inverse of this function :

    f(x) = e^(2x) + 3

    and how would you work out the domain of its inverse?
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    Starting with the inverse,

    y=e^{2x}+3

    How would we solve for x?
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    The y at the end is supposed to be f^(-1) (x)
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    (Original post by _gcx)
    Starting with the inverse,

    y=e^{2x}+3

    How would we solve for x?
    I got x = (y-3)/2e

    is this correct?
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    (Original post by NGEO)
    x

    The y at the end is supposed to be f^(-1) (x)
    Best not to post full solutions, it is much preferred to give hints instead. You might be interested in looking at the posting guidelines
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    (Original post by emx_eco)
    I got x = (y-3)/2e

    is this correct?
    It's e^{2x} not 2ex
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    (Original post by NGEO)
    what did you do to get from line 4 to line 5?
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    (Original post by emx_eco)
    what did you do to get from line 4 to line 5?
    Are you familiar with \ln?
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    (Original post by _gcx)
    It's e^{2x} not 2ex
    I dont understand how you got that answer, could you please explain
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    (Original post by NGEO)

    The y at the end is supposed to be f^(-1) (x)
    Have you seen the guideline (sticky) about not posting full solutions?
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    (Original post by _gcx)
    Are you familiar with \ln?
    yh, the same as log base e right?
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    (Original post by emx_eco)
    I dont understand how you got that answer, could you please explain
    If we have y = e^x, how would we rearrange to make x the subject? Think back to C2 logs.
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    When you find the inverse of a function, essentially what you are doing is reflecting it in the line y=x. In order to find the 'new' equation, you want to rearrange it in terms of x=.... then just swap all the signs around like NGEO did, but did it in the first step instead.
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    (Original post by _gcx)
    If we have y = e^x, how would we rearrange to make x the subject? Think back to C2 logs.
    use a log base e on both sides?
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    (Original post by emx_eco)
    use a log base e on both sides?
    Yup, so we'd have \log_e(y) = x. In C3/C4 we normally write \log_e as \ln. Is it now clearer what ngeo did to get from e^{2y} = x-3 to 2y = \ln(x-3)?
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    (Original post by _gcx)
    Yup, so we'd have \log_e(y) = x. In C3/C4 we normally write \log_e as \ln. Is it now clearer what ngeo did to get from e^{2y} = x-3 to 2y = \ln(x-3)?
    ohh I get that part now So I have the equation (lnx-3)/2 now. How do you find the domain of this?
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    (Original post by emx_eco)
    ohh I get that part now So I have the equation (lnx-3)/2 now. How do you find the domain of this?
    Where is \ln undefined? Can we have the log of 0 or a negative number?
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    (Original post by _gcx)
    Where is \ln undefined? Can we have the log of 0 or a negative number?
    no...?
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    (Original post by emx_eco)
    no...?
    So what's the domain of \ln(x) and hence \ln(x-3)?
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    (Original post by _gcx)
    Where is \ln undefined? Can we have the log of 0 or a negative number?
    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
 
 
 
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