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Find the formula for the inverse function

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data


    The function f is defined by y=f(x) = 3ln4x 0.01<=x<=1

    a)Solve for x in terms of y and hence find the formula for the inverse function f^-1(x)

    b)Write down the domain of f^-1

    c)Plot f from x =0.01 to x=1 and than plot f^-1 on the same axes but only for domain values of x given by the range of f

    d)Describe the geometric relationship that you can see between f and f^-1

    plse assist as iam very confused with functions.....!!


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 7, 2008 #2

    Hootenanny

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    Re: equation2

    You seem to have missed out an important section.
     
  4. Dec 7, 2008 #3
    Re: equation2

    a)Solve for x in terms of y and hence find the formula for the inverse function f^-1(x)

    The rule for the inverse function is given by f^-1(x)
    how to make y the subject.

    b)Write down the domain of f^-1

    I suggest that i a graph of y = f(x) over and use it to find the range of f ???

    c)Plot f from x =0.01 to x=1 and than plot f^-1 on the same axes but only for domain values of x given by the range of f

    d)Describe the geometric relationship that you can see between f and f^-1
    are they reflections of each other in the line y = x.??
     
  5. Dec 7, 2008 #4
    Re: equation2

    plse assist to answer
     
  6. Dec 7, 2008 #5

    Hootenanny

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    Re: equation2

    Okay for part (a) we have:

    y = 3ln(4x)

    Can you solve for x?
     
  7. Dec 7, 2008 #6
    Re: equation2

    for a)
    is it this way:


    x=1/4e^(y/3)
     
  8. Dec 7, 2008 #7
    Re: equation2

    plse assist is this right?
     
  9. Dec 7, 2008 #8

    Hootenanny

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    Re: equation2

    Yes, that is correct. So now you can write the inverse function as

    f-1(x) = 1/4exp(x/3)

    So, what about the domain of the inverse function?
     
  10. Dec 7, 2008 #9
    Re: equation2

    is the domain from 0.2508<=x<=0.3489

    by using 1/4 * e^(0.01/3) and 1/4 * e^(1/3)

    plse assist to check
     
  11. Dec 7, 2008 #10

    Hootenanny

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    Re: equation2

    No, that is not correct.

    HINT: Think about the range of f(x).
     
  12. Dec 7, 2008 #11
    Re: equation2

    yes the range is from 0.01<=x<=1
    therefore i plug in to the f^-1 and got

    0.2508<=x<=0.3489
     
  13. Dec 7, 2008 #12

    Hootenanny

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    Re: equation2

    No, that is the domain for f(x) and why would you want to plug it into f-1(x)?
     
  14. Dec 9, 2008 #13
    Re: equation2

    i can't seem to figure out the domain...plse assist

    b)Write down the domain of f^-1
     
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