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

    Discuss the continuity and differentiability of $\displaystyle f(x)$ at $\displaystyle x = \pm1$
    Last edited by fardeen_gen; Apr 27th 2009 at 11:29 PM.
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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:
    $\displaystyle f(x) = \arccos\left(Sgn\left(\frac{2[x]}{3x - [x]}\right)\right)$ where $\displaystyle Sgn(x) = \left\{ \begin{array}{cc}\frac{|x|}{x}, &\mbox{if}\ x\neq 0\\0, &\mbox{if}\ x = 0\end{array}\right.$ and [.] denotes the greatest integer function.

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

    At $\displaystyle x=1$, we have:

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

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

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

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

    At $\displaystyle x=-1$, we have:

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

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

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

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

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

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

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

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

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

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

    I'm only moderately confident about this response.
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