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Inverse Functions

  1. Oct 1, 2014 #1

    462chevelle

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    1. The problem statement, all variables and given/known data
    Verify that f has an inverse <- prof told us not to worry about this. Then use the function f and the given real number a to find (f^-1)'(a).

    f(x) = Cos(2x), 0<=x<=pi/2 where a=1

    2. Relevant equations
    1/(f'(g(x))) where g(x)=f^-1(a)
    d/dx(cos(2x)) = -2sin(2x)

    3. The attempt at a solution
    Ok, I know the derivative of the function.
    So i take cos(2x)=1 and I get x=0.
    then use the derivative of cos(2x) and use the formula
    1/(-2sin(2*0)) = undefined.
    I feel like i must be missing some detail or making a mistake.
    If I am not, what does it mean whenever this is undefined? is there something i can do algebraically or with a limit to make this work?
    Thanks

    P.S. sorry about the ugly math, I've tried to learn latex and it never seems to stick since I don't post many math questions.
     
  2. jcsd
  3. Oct 1, 2014 #2

    Mark44

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    To find the derivative of the inverse of Cos(2x), you need the inverse of your function, which I don't see anywhere in your post. For f(x) = cos(2x), with x in [0, ##\pi/2##], what is f-1?
     
  4. Oct 1, 2014 #3

    462chevelle

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    Well its not exactly asking for the derivative of the inverse like that. we havent gotten to inverse trig functions yet. It says find f^-1(a) and it gives us a. so if a=f(x) i can find some value. for f^-1(a) and use my formula. thats how the book words it.
     
  5. Oct 1, 2014 #4

    WWGD

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    You can use the fact that if f has an inverse ##f^{-1}##( at least in a given interval) , then ## f(f^{-1}(x)=x=f^{-1}f(x)##.
    Now, if you assume the inverse is differentiable, this is like finding the derivative of fog for two functions f,g. Can you see how to do it?
     
  6. Oct 1, 2014 #5

    462chevelle

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    Ya, if i use the chain rule on cos(2x) i get -sin(2x)*2=-2sin(2x)
    then if i use the rule 1/(f'(g(x))) i get
    1/(-2sin(2x)(g(x))
    g(x)=f^-1(a)
    if f^-1(a)=0 because cos2x=1, x=0.
    and f^-1(a)=g(x) then i get undefined.
    or am i missing your point?
     
  7. Oct 1, 2014 #6

    Mark44

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    Here a = 1, so cos(2 * ?) = 1, keeping in mind the restricted domain.
     
  8. Oct 1, 2014 #7

    462chevelle

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    the only place in the domain it can be 1 is at 0. that i know of.
     
  9. Oct 1, 2014 #8

    Mark44

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    I like to keep the x's and y's separate, and I believe it makes inverses a lot more understandable. If y = f(x), then x = f-1(y).

    So x = f-1(f(x)) and y = f(f-1(y))
    For a function f that has an inverse, the graphs of y = f(x) and x = f-1(y) are exactly the same. Many precalc texts add in what I consider unnecessary steps in their coverage of how to find the inverse, such as the whole "switch x and y" business that makes it easy for students to misunderstand the relationship between a function and its inverse.

    When inverses are actually used, such as in changing the order of integration in iterated integrals, the idea is to take an equation y = f(x) and solve for x in terms of (as the inverse function of) y.
     
  10. Oct 1, 2014 #9

    Mark44

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    Yes.
     
  11. Oct 1, 2014 #10

    462chevelle

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    Yes, that is the way i learned inverse functions and i have to say, when the prof wrote what he did on the board it took about an hour of doing problems for it to really click in my head. so is the formula i'm using not going to work in this instance? it worked for all other polynomial and rational functions i used it on. just not this one.
     
  12. Oct 1, 2014 #11

    Mark44

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    1/(f'(g(a))) looks OK to me.

    We've established that g(1) = 0. Now what is f'(0)?
     
  13. Oct 1, 2014 #12

    462chevelle

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    f'(0)=-2sin(2x) where x=0
    So that would mean -2sin(2*0)
    which would be 0.
    making it 1/0. undefined.
     
  14. Oct 1, 2014 #13

    WWGD

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    I think you should be computing $$ \frac {-1}{2sin2(cos^{-1}(x))}$$
     
  15. Oct 1, 2014 #14
    Never mind sorry

    actually yeah, it will still give him a -1/0

    Only justification I see is that the derivative of the inverse function at one will not be defined anyways.
     
  16. Oct 1, 2014 #15

    WWGD

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    Ah, right. Maybe it is a typo on the book and a is supposed to be 0 instead of 1, or some other value other than 1?

    OP: Maybe do an online search for typos on your book?
     
  17. Oct 1, 2014 #16

    462chevelle

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    if its supposed to be 0 instead of 1 i would get something like -1/1.8xxxxx something like that. which is also a weird answer. prof finally emailed back and said he doesnt understand what im trying to ask and come by office hours. he said if f'(g(x))=0 then the function is not differentiable at that point. well, i knew that but i just figured i had to have made a mistake or something.
     
  18. Oct 1, 2014 #17

    WWGD

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    Ah, sorry.
     
  19. Oct 1, 2014 #18
    If f(x)= cos(2x) then the inverse function would be arccos(y)/2 and the derivative of that would be -1/2sqrt(1-y^2) that is not defined at 1
     
  20. Oct 1, 2014 #19
    The question asks "Then use the function f and the given real number a to find (f^-1)'(a)." That is why I am using 1 and it is not diffrentiable at 1. That is why I assume there is an error.
     
  21. Oct 1, 2014 #20

    462chevelle

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    Ya, sleeping is going to be difficult with this on my mind. I feel like i should just keep undefined as my answer and forget about it but it just doesnt feel right.
     
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