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Inverse of a function is not differentiable

  1. Mar 1, 2014 #1
    Please see attached.

    I am not sure whether my example of this function is correct.

    f(x) = ##sin(\frac{\pi x}{2})##
    obviously, f(x) is continuous on [-1,1] and differentiable on (-1,1)

    Inverse of f(x) will be ##\frac{2 sin^{-1}x}{\pi} ##

    and d/dx (inverse of f(x)) will be ##\frac{2}{π \sqrt{1-x^2}}##

    So not differentiable in -1,1 ??
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2014 #2

    PeroK

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    That inverse function looks differentiable to me.
     
  4. Mar 1, 2014 #3
    anyone can help and explain?
     
  5. Mar 1, 2014 #4

    SammyS

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    Your problem as an image:
    attachment.php?attachmentid=67141&d=1393706030.jpg

    The inverse of the function you propose is differentiable on (-1,1). It's derivative does approach +∞ as x → -1 or +1 . ( Why does it approach ∞ at 1 & -1 ? -- What is it about f(x) that causes this?)

    Can you think of a 1 to 1 function which maps [-1, 1] onto [-1, 1] but has a slope of zero somewhere on the interval (-1, 1) ?
     
  6. Mar 1, 2014 #5
    Thanks for your help

    so I should think of a question which has two parts?

    for example f(x)=##cos(\frac{\pi x}{2})## for x=/=0. and f(x)=1 for x=0
    so the inverse of this kind of function is not differentiable, right?
     
  7. Mar 1, 2014 #6

    PeroK

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    You're overthinking this. The derivative of the inverse is 1/ the derivative of the function, so all you need is a function whose derivative is 0. In the middle, at x = 0, for example.
     
  8. Mar 1, 2014 #7

    Yes, I have been thinking this part for 3 hours...

    Is the function f(x)=cosx satisfy the requirements?

    f'(x)=-sinx so at x=0, f'(x)=0

    derivative of the inverse = 1/sin(arcosx), so x=1 or -1, the denominator is 0

    correct?
     
  9. Mar 1, 2014 #8

    PeroK

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    cos(x) is not 1-1 on [-1, 1]

    You need a function that is 1-1, has a derivative of 0 in the middle, which implies a point of inflection.
     
  10. Mar 1, 2014 #9

    SammyS

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    Is there any reason for you to be using trig functions?
     
  11. Mar 1, 2014 #10

    i m almost crazy..may be i m an idiot
    i cannot think of a equation which satisfy these conditions

    1. 1 to 1
    2. f:[-1,1] to [-1,1]
    3. derivative is 0 in the middle
    4. has an inverse

    i now think of e^x - x
    but this kind of equation seems i m overthinking
     
    Last edited: Mar 1, 2014
  12. Mar 1, 2014 #11
    In my opinion trig function is the easiest way to satisfy [-1,1] to [-1,1]
     
  13. Mar 1, 2014 #12

    Dick

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    You can also do it with many other functions. How about a simple power?
     
  14. Mar 1, 2014 #13
    finally, x^3 satisfy all conditions, right?

    f(x)=x^3
    f'(x)=3x^2
    inverse of f(x) = x^(1/3)
    derivative of the inverse = 1/3 * x^(-2/3) , so not differentiable at x=0
     
  15. Mar 1, 2014 #14

    Dick

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    Sure, that's an easy example.
     
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