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Finding (f^-1)'a

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data

    f(x)= x^3 + 3sinx + 2cosx, a=2

    2. Relevant equations

    (f^-1)'(a)= 1 / f'(f^-1(a))

    3. The attempt at a solution

    I am lost as to where to start on this problem. Can someone get me going in the right direction? Thanks in advance.
  2. jcsd
  3. Sep 11, 2011 #2


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    A good place to start would be to try to find f^(-1)(2). Your expression for f is hard to invert in general but you can find f^(-1)(2) by guessing.
  4. Sep 11, 2011 #3
    It is not possible to solve for the inverse analytically (as far as I know at least), so think about what value of [itex]x[/itex] would yield [itex]f(x)=2[/itex]. In other words, compute [itex]f^{-1}(2)[/itex] by inspection. This is not as hard as it seems because you've been given some trig functions that are easy to deal with for certain values of [itex]x[/itex]. Then, it is trivial as it is simply plug and chug.
  5. Sep 12, 2011 #4


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    What does your Relevant Equation, (f -1)'(a)= 1 / f '(f -1(a)) tell you to do ??

    Finding f -1(a) is like solving your previous problem.

    Then find f '(x), and as lineintegral1 said, "plug & chug".
  6. Sep 12, 2011 #5
    That notation is just making it confusing and harder to deal with. May I suggest looking at it more intuitively to make it easier to deal with.



    and therefore:


    and therefore we can write:


    Ok, so now at that point, just calculate a few that you already know just to get the feel of it. Say (approx):


    so that I can write:


    Now do a few more say let x=-2,-1,0,2 and see what happens.
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