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Math Help - Continuity and Differentiation...

  1. #1
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    Continuity and Differentiation...

    Hello,

    I need help in figuring out the continuity of these weird looking function..
    Please help!

    Q1) Find the maximum and minimum value of the function:
    f(x) = 4x^3 - x^2 - 4x + 2 on [ -1,1] and on [0,1]?



    Q2) Let f: [0, infinity), -> R be defined by

    f(x) = x sin (1/x) when x>0
    and f(x) = 0 when x = 0.
    Figure out if 'f' is continuous and differentiable at x = 0.



    Q3) Define f: [-2,2] -> R by f(x) = |x^3 - 1|.

    f: [-2,2] -> R by f(x) = |x^3| -1.
    Determine the points where f is differentiable and find the derivative at those points?
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  2. #2
    Eater of Worlds
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    Q2) Let f: [0, infinity), -> R be defined by

    f(x) = x sin (1/x) when x>0
    and f(x) = 0 when x = 0.
    Figure out if 'f' is continuous and differentiable at x = 0.
    Well, to be continuous the limit must exist.

    \lim_{x\to 0^{+}}xsin(\frac{1}{x})

    -1 \leq sin(1/x) \leq 1

    so -x \leq xsin(1/x) \leq x

    But  -x \;\ and \;\ x\rightarrow{0} \;\ as \;\ x\rightarrow{0^{+}}

    This, \lim_{x\to 0^{+}}xsin(1/x) = 0

    Let's describe the interval where it is continuous.

    If we have:

    f(x) = \begin{Bmatrix}xsin(1/x), \;\ x\neq 0\\0, \;\  x=0\end{Bmatrix}


    By the Squeezing Theorem:

    -|x|\leq xsin(1/x) \leq |x|

    And we can conclude that \lim_{x\to 0}f(x)=0

    So f is continuous on the entire real line.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Vedicmaths View Post
    Hello,

    I need help in figuring out the continuity of these weird looking function..
    Please help!

    Q1) Find the maximum and minimum value of the function:
    f(x) = 4x^3 - x^2 - 4x + 2 on [ -1,1] and on [0,1]?



    Q2) Let f: [0, infinity), -> R be defined by

    f(x) = x sin (1/x) when x>0
    and f(x) = 0 when x = 0.
    Figure out if 'f' is continuous and differentiable at x = 0.



    Q3) Define f: [-2,2] -> R by f(x) = |x^3 - 1|.

    f: [-2,2] -> R by f(x) = |x^3| -1.
    Determine the points where f is differentiable and find the derivative at those points?
    For the limit, it also might be more apparent to let

    \varphi=\frac{1}{x}

    So as x\to{0}\Rightarrow\varphi\to\infty

    So we have

    \lim_{\varphi\to\infty}\frac{\sin(\varphi)}{\varph  i}

    Now it might be a little more obvious
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Vedicmaths View Post
    Hello,




    Q2) Let f: [0, infinity), -> R be defined by

    f(x) = x sin (1/x) when x>0
    and f(x) = 0 when x = 0.
    Figure out if 'f' is continuous and differentiable at x = 0.
    We can use the limit defintion to see if it is differentiable at x=0

    \lim_{x \to 0}\frac{f(x)-f(0)}{x-0}

    \lim_{x \to 0}\frac{x\sin\left( \frac{1}{x}\right)-0}{x-0}=\lim_{x \to 0}\sin\left( \frac{1}{x}\right)


    This limit does dont exits so the function is not differentiable at x=0
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  5. #5
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    Thanks everyone for the solution.
    Could anyone please help me understanding the the third problem, the absolute one?

    Regards,
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  6. #6
    Eater of Worlds
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    I am sorry, but I beg to differ about the sine limit. It appears to be continuous at 0. The oscillations are damped by the factor x.

    The limit is 0 rather approaching from the left or right.

    sin(1/x) is not continuous, though. You can see that from the oscillation on the graph.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  7. #7
    Eater of Worlds
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    Here is the graph of sin(1/x). It is not continuous at x=0
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  8. #8
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    Quote Originally Posted by Vedicmaths View Post
    Thanks everyone for the solution.
    Could anyone please help me understanding the the third problem, the absolute one?
    Always, always graph the function!
    Look at the graph. The answer is absolutely clear.
    Attached Thumbnails Attached Thumbnails Continuity and Differentiation...-absvl.gif  
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