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Inverse functions

  1. Jun 6, 2005 #1
    find the inverse function of:
    f: [-1/2,infin) ---> R, Where F(x) = 3sqaureroot 2x+1

    i got f(x) = 3squareroot 2x+1
    x = 3squareroot 2(f-1)+1
    f-1 = (x^3-1)/2 where (infin,-1/2]

    is this right?
    because when i graph it, it doesnt really look like the inverse

    also i just ran into another problem
    find the inverse function of:
    f : [-1,infin) ---R, where f(x) = x^2 + 2x - 8

    i did:
    f(x) = x^2 - 2x - 8
    x = (f-1)^2 - 2(f-1) - 8
    x+8 = (f-1)^2 - 2(f-1)
    and then everything goes wrong here

    edit: ARGGGGH just more and more problems, is there any forumla or way of working out the inverse of Log's and exponentials? i cant find any in my books

    Last edited: Jun 6, 2005
  2. jcsd
  3. Jun 6, 2005 #2


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    No, nothing that you're doing there looks even close. You seem to really be at a loss to manipulate equations correctly. The basic skill you seem to be missing is that of transposing simple equations. Practice of this would be a good place for you to start.

    For example, can you manipulate a simple equation like f = 9c/5 + 32 to express c in terms of f (this is called transposing btw). This is the thing you really need to lean before you can do the above problems (or just about anything else in maths actually).
  4. Jun 6, 2005 #3
    The inverse of the log function is the exponential.
    the inverse of the exponential is the log function.

    For example

    [tex] log_3 y = x [/tex]

    Inverse function is [itex] 3^n [/itex]

    [tex] 3 ^ {log_3 y} = 3 ^ x [/itex]

    [tex] y = 3^x [/tex]
  5. Jun 6, 2005 #4


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    Gold Member

    Rewrite x2-2x-8-f=0.

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