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Common feature of some functions

  1. Jun 10, 2007 #1
    Hello!
    I have a very interesting task, however, at the very beginning I should notice something but... I have a small problem with that ;-].

    We have functions:
    ->sin x,
    ->5x
    ->4x7 − 5x3 + 8x
    ->tan(2x)

    I should find their common feature, then find their derivatives and check what do they have in common....

    They have different domains, different ranges...
    I plotted these function and I have observed:
    ->f(0)=0 for all these functions
    ->f(0) does not equal 0 for all their derivatives...

    Do you think that this might be the point? Is that a rule that when a function intersects (0,0), its derivative does not?

    Or there is something more sophisticated that I cannot see?

    Thank you for some hints!
    Greetings,
    Theriel
     
  2. jcsd
  3. Jun 10, 2007 #2

    matt grime

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    f(x)=0?

    What are f(1) and f(-1) in all the above?
     
  4. Jun 10, 2007 #3
    For functions:
    sin(1)=0.841
    sin(-1)=-0.841
    5x=5
    5x=-5
    4x^7 − 5x^3 + 8x=7
    4x^7 − 5x^3 + 8x=-7
    tan(2)=-2.185
    tan(-2)=2.185

    For derivatives:
    cos(1)=0.5403
    cos(-1)=0.5403
    5
    5
    28x^3-15x^2+8=21
    28x^3-15x^2+8=21
    2/cos^2x=11,55
    2/cos^2x=11,55

    Hence we see that when a function is symmetric about (0,0) then its derivative is symmetric about y-axis? :biggrin: Or there is still something more to be found?
     
    Last edited: Jun 10, 2007
  5. Jun 10, 2007 #4

    matt grime

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    So what was the common feature of the functions given? That they have *what* kind of symmetry? (Rotational, reflective?) The first list of functions are examples of ODD functions. They satisfy f(-x)=-f(x). Now, can you show that their derivatives must be EVEN functions?
     
  6. Jun 10, 2007 #5
    So:
    There is an odd function:
    f(-x)=-f(x)

    We want to show that its derivative is even, hence:
    f'(-x)=f'(x)

    f'(x) = lim{h->0} (f(x+h) - f(x))/h
    f'(-x)=lim{h->0} ((f(-x+h) - f(-x))/h = lim{h->0} (f(-x+h) - f(x))/h

    The problem is how to show that:
    f(-x+h)=f(x+h)
    knowing that f(-x)=-f(x)... #-/
     
  7. Jun 10, 2007 #6

    matt grime

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    YOu're over complicating it. (Though what you've done does work.)

    What is the derivative of f(2x), f(x^2), and now f(-x) (chain rule)...
     
  8. Jun 10, 2007 #7
    Yeah, but the point is that later I will have to generalize and prove the situation. That is why I started doing that part, because proving it for these examples (f(2x), f(x^2) etc.) is easy ;-]. So... could you refer to my small problem in previous post, please?
     
  9. Jun 10, 2007 #8

    matt grime

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    What? In what way is differentiating both sides of

    f(x)=-f(-x)

    i.e. where f is an odd function not 'doing the general situation'?

    Your previous small problem

    can't be done since it is not true that f(-x+h)=f(x+h).
     
  10. Jun 10, 2007 #9
    Sorry, I didn't get what you have said #-/. In my task I have to generalize my findings. I found that:

    For an odd function "f":
    f(-x)=-f(x)

    The derivative is even:
    f'(-x)=f'(x)

    And now the problem is how to prove it. Showing that only for the functions mentioned previously (x^2, tan(2x) etc.) will not be generalization of the situation.
     
  11. Jun 10, 2007 #10

    matt grime

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    I'm going to repeat this again f is odd means

    f(x)=-f(-x)

    now for the love of god differentiate both sides of that expression. It is a completely general proof - there is no mention of what f is other than an odd function. Just use the chain rule unless you really must do it from first principles. If you must do it from first principles, then make sure you get the signs right in your limit and it all drops out.

    [f(x+h)-f(x)]/h = -[f(-x-h)-f(-x)]/h

    now set k=-h, and let k tend to zero and what do you get?
     
    Last edited: Jun 10, 2007
  12. Jun 10, 2007 #11
    Oh, that was much simpler than I expected... Once again, I am really impressed by the effort and time you spend for helping people solve their mathematical problems... Thank you very much for that!

    Greetings,
    Theriel
     
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