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Math Help - Discuss continuity and differentiability of f(x)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Discuss continuity and differentiability of f(x)?

    A function f is defined by:
    f(x) = \arccos\left(Sgn\left(\frac{2[x]}{3x - [x]}\right)\right) where Sgn(x) = \left\{ \begin{array}{cc}\frac{|x|}{x}, &\mbox{if}\ x\neq 0\\0, &\mbox{if}\ x = 0\end{array}\right. and [.] denotes the greatest integer function.

    Discuss the continuity and differentiability of f(x) at x = \pm1
    Last edited by fardeen_gen; April 28th 2009 at 12:29 AM.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    A function f is defined by:
    f(x) = \arccos\left(Sgn\left(\frac{2[x]}{3x - [x]}\right)\right) where Sgn(x) = \left\{ \begin{array}{cc}\frac{|x|}{x}, &\mbox{if}\ x\neq 0\\0, &\mbox{if}\ x = 0\end{array}\right. and [.] denotes the greatest integer function.

    Discuss the continuity and differentiability of f(x) at x = \pm1
    For continuity, I'm going to use the limit definition.

    At x=1, we have:

    \lim_{x\to 1^-}f(x) = sgn\left(\frac{2(0)}{3x-0}\right) = 0. ~~~\arccos(0)=\frac{\pi}{2}

    \lim_{x\to 1^+} f(x)= sgn\left(\frac{2(1)}{3x-1}\right) = sgn(2/2) = 1. ~~~\arccos(1) = 0

    f(1) = sgn\left(\frac{2(1)}{3x-1}\right) = sgn(2/2) = 1. ~~~\arccos(1) = 0

    So f(x) is not continuous at x=1.

    At x=-1, we have:

    \lim_{x\to -1^-}f(x) = sgn\left(\frac{2(-2)}{3x-(-2)}\right) = sgn(-4/-1)=1. ~~~\arccos(1)=0

    \lim_{x\to -1^+}f(x) = sgn\left(\frac{2(-1)}{3x-(-1)}\right) = sgn(-2/-2) = 1. ~~~\arccos(1) = 0

    f(-1) = sgn\left(\frac{2(-1)}{3x-(-1)}\right) = sgn(-2/-2) = 1. ~~~\arccos(1) = 0

    So f(x) is continuous at x=-1.
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  3. #3
    Super Member redsoxfan325's Avatar
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    It follows that f is not diffrentiable at x=1, because it is not continuous.

    At x=-1, we have: \lim_{h\to 0}\frac{sgn\left(\frac{2[-1+h]}{3(-1+h)-[-1+h]}\right)-sgn\left(\frac{2[-1]}{3(-1)-[-1]}\right)}{h} = \lim_{h\to 0}\frac{sgn\left(\frac{2[-1+h]}{3(-1+h)-[-1+h]}\right)-1}{h}.

    Looking at \lim_{h\to 0^+} and assuming 0<h<\frac{2}{3} (I think this is OK to assume because we want h\to 0), we know that [h-1]=-1 and 3h-2<0, so as h becomes less than \frac{2}{3}, we have:

    \lim_{h\to 0^+}\frac{sgn\left(\frac{-2}{3h-2}\right)-1}{h} = \lim_{h\to 0^+}\frac{sgn\left(\frac{-2}{<0}\right)-1}{h} = \lim_{h\to 0^+}\frac{1-1}{h} = 0

    Looking at \lim_{h\to 0^-} and assuming -1<h<0, we know that [h-1]=-2, and we have:

    \lim_{h\to 0^-}\frac{sgn\left(\frac{-4}{3h-1}\right)-1}{h} = \lim_{h\to 0^-}\frac{sgn\left(\frac{-4}{<0}\right)-1}{h} = \lim_{h\to 0^-}\frac{1-1}{h} = 0

    So I think f(x) is differentiable at x=-1, and furthermore that f'(-1)=0 and the tangent line is g(x)=1.

    I'm only moderately confident about this response.
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