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Math Help - more differntiable

  1. #1
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    more differntiable

    let f(X) = x^2 sin(1/x) for x is not equal to 0 and f(0)= 0

    A) use the chain rule and the product rule to show that f is differentiable at each c is not equal to 0 and find f'(c)
    B) show that f is differentiable at x = 0 and show that f(0)=0
    C) show that f' is not continuous at x = 0
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by learn18 View Post
    let f(X) = x^2 sin(1/x) for x is not equal to 0 and f(0)= 0

    A) use the chain rule and the product rule to show that f is differentiable at each c is not equal to 0 and find f'(c)
    at x = c not= 0, f(x) is continuous and hence differentiable.

    so f(c) = (c^2)sin(1/c)
    => f ' (c) = 2csin(1/c) + (-c^-2)(c^2)cos(1/c)
    => f ' (c) = 2csin(1/c) - cos(1/c)

    B) show that f is differentiable at x = 0 and show that f(0)=0
    we must show that f is continuous at x = 0

    note that |(x^2)sin(1/x)|<or= x^2 for all x

    so we have: -x^2 <or= (x^2)sin(1/x) <or= x^2
    => lim{x-->0+} -x^2 <or= lim{x-->0+}(x^2)sin(1/x) <or= lim{x-->0+}x^2
    => 0 <or= lim{x-->0+}(x^2)sin(1/x) <or= 0
    so by the squeeze theorem:

    lim{x-->0+}(x^2)sin(1/x) = 0
    similarly, we can show, lim{x-->0-}(x^2)sin(1/x) = 0

    thus f is continuos at x = 0, so by the definition of continuity we have f(0) = lim{x-->0}(x^2)sin(1/x) = 0


    C) show that f' is not continuous at x = 0
    this seems trivial to me, but i may be wrong. at x = 0, 1/x is undefined, so f ' (x) = 2xsin(1/x) - cos(1/x) is undefined and is not continuous. you can take the left and right hand limits to see this, show that they are different or something
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  3. #3
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    Quote Originally Posted by Jhevon View Post

    note that |(x^2)sin(1/x)|<or= x^2 for all x

    so we have: -x^2 <or= (x^2)sin(1/x) <or= x^2
    => lim{x-->0+} -x^2 <or= lim{x-->0+}(x^2)sin(1/x) <or= lim{x-->0+}x^2
    => 0 <or= lim{x-->0+}(x^2)sin(1/x) <or= 0
    so by the squeeze theorem:

    lim{x-->0+}(x^2)sin(1/x) = 0
    similarly, we can show, lim{x-->0-}(x^2)sin(1/x) = 0

    thus f is continuos at x = 0, so by the definition of continuity we have f(0) = lim{x-->0}(x^2)sin(1/x) = 0
    But this shows that f is continous at 0 not differenciable at 0
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    But this shows that f is continous at 0 not differenciable at 0
    there are three things that can cause a function f to be non-differentiable, is there not?

    if:

    (1) f is undefined
    (2) f is discontinuous
    (3) f has a sharp turn

    case 3 doesn't happen, and case (1) doesn't happen because of the way the function is defined (by definition, f(0) = 0, so it's not undefined there as the problem is phrased), so i thought i could just take care of case (3) and be done with it, which is what i did

    and i forgot to say "this is my 13th post!"
    darn it, i missed it again!
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  5. #5
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    Quote Originally Posted by learn18 View Post
    let f(X) = x^2 sin(1/x) for x is not equal to 0 and f(0)= 0

    B) show that f is differentiable at x = 0 and show that f(0)=0
    First f(x) is defined on an open interval containing 0.
    Thus we can consider the derivative limit:

    lim(x-->0) (f(x) -f(0))/(x-0) = lim(x-->0) f(x)/x

    lim(x-->0) x^2 sin(1/x)/x = lim (x-->0) x*sin(1/x)

    The limit exists because:

    -1<=sin(1/x) <= 1 then

    -x<= x*sin(1/x) <= x for x>0
    x<= x*sin (1/x) <= -x for x<0
    In both cases the limit as x-->0 is 0 by Squeeze Theorem.

    Thus, f(x) is differenciable at 0, and furthermore the derivative is 0.
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