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Derivative of Inverse Function

  1. Feb 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose ##f(x) = x^5 + 2x + 1## and ##f^{-1}## is the inverse of function f. Evaluate ##f^{-1}(4)##

    solution: 1/7

    2. Relevant equations
    ##(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}##

    3. The attempt at a solution
    I attempted to use my calculator's solve function to get the solution of ##4 = 5x^4 + 2##, so the value of x when the derivative is equal to 4. I then took the inverse of this. Sort of still grasping in the dark right now tho.
     
  2. jcsd
  3. Feb 16, 2017 #2

    Simon Bridge

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    Your problem statement says you have to:
    Note: ##f^{-1}(f(x))=x \implies f^{-1}(4) = x:f(x)=4##
    Why were you computing a derivative?
     
  4. Feb 16, 2017 #3

    BvU

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    There's something fishy in the combination problem statement ##\leftrightarrow ## solution 1/7 . James ?
     
  5. Feb 16, 2017 #4
    My mistake guys,

    The problem statement asks for ##(f^{-1})^{'}(4)##
     
  6. Feb 16, 2017 #5

    PeroK

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    The relevant equation you posted is not that useful, as it involves ##f'##. You need something similar involving ##(f^{-1})'##.
     
  7. Feb 16, 2017 #6

    BvU

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    So you don't want to solve for 'derivative = 4' but you want to evaluate the derivative at the point where the function value is 4 !

    [edit] perok: he had a ##\ ## ' ##\ ## missing in
    ##(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}## ; should've been $$
    (f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}$$
     
  8. Feb 16, 2017 #7
    Hmmm. ##f^{'} = 5x^4+2## and ##f^{'}(4)= 1282##

    So I have the derivative of of the function itself at a certain x value. But I'm not sure how I can translate that into finding the derivative of the inverse at the same x value.
     
  9. Feb 16, 2017 #8

    BvU

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    Again: you don't want to evaluate the derivative at the point x = 4 but at the point f-1(4)
     
  10. Feb 16, 2017 #9

    BvU

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    Do you understand the relationship $$
    (f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}\ \ \rm ?$$
     
  11. Feb 16, 2017 #10
    I understand this relationship, but I do not know how I can evaluate ##f^{'}(4)## since I can't figure out how to get the inverse of the original function. Plugging in for ##y=x^5 + 2x + 1##, I can't figure out how I would be able to solve for x in terms of y.
     
  12. Feb 16, 2017 #11

    PeroK

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    You could always just take a guess!
     
  13. Feb 16, 2017 #12

    BvU

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    Try a few numbers, or do bisection ( f(0) = 1, f(2) >> 4 so...)

    And again (again) :
    You don't want that. You want ##f'(f^{-1}(4))## !
     
  14. Feb 16, 2017 #13

    Ray Vickson

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    Plot the graph of ##y = x^5+2x+1## over some modest ##x##-range; that will give you a good feeling for approximately where you have the solution of ##f(x)=4.## Alternately you can try using the "rational root theorem" on the equation ##x^5 + 2x - 3 = 0.## (Once you get the solution you will wonder why you did not see it right away!)
     
  15. Feb 17, 2017 #14
    Thanks guys, I guess I just really needed to sleep on it, judging by my sloppiness. Coming back today, I can see where you all were going:

    ##(f^{-1}(x))^{'} = \frac{1}{f^{'}(f^{-1}(x))}##

    First thing, we solve the interior function on the RHS, or ##4 = x^5 + 2x +1##. I used the numerical solver on the calculator, but I see now that 1 could be obvious if you take a step back for a second. After that, we're no longer working with inverses, the derivative function ##f^{'}(x) = 5x^4 + 2## taken at x = 1. One over that and boom.

    Thanks again for your patience.
     
  16. Feb 18, 2017 #15

    BvU

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    Important learning experience: brains are a lot like muscles, they need to recover from time to time or else things go sour. Sometimes dropping your pencil and doing some R&R is the quickest way to achieve some progress when stuck with a problem.
     
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