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Math Help - Derivative Functions

  1. #1
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    Derivative Functions

    --
    f(x)=\left\{\begin{matrix} x^2, & \mbox{if }x\mbox{ is rational} \\ -x^2, & \mbox{if }x\mbox{ is irrational}<br />
 \end{matrix}\right.

    For which values of x f(x) is derivative? What's the value of the derivative when exists?

    --

    How can I even begin solving such problem ?
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  2. #2
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    Quote Originally Posted by adam63 View Post
    --
    f(x)=\left\{\begin{matrix} x^2, & \mbox{if }x\mbox{ is rational} \\ -x^2, & \mbox{if }x\mbox{ is irrational}<br />
 \end{matrix}\right.

    For which values of x f(x) is derivative? What's the value of the derivative when exists?

    --

    How can I even begin solving such problem ?
    What is the derivative of x^2? Where is this derivative =x^2?


    What is the derivative of -x^2? Where is this derivative =-x^2?
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    Recall that the derivative may only exist where the function is continuous. Can you identify where this function is continuous?
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    Quote Originally Posted by Prove It View Post
    What is the derivative of x^2? Where is this derivative =x^2?
    when   x^2=2x => x=0,2

    What is the derivative of -x^2? Where is this derivative =-x^2?
    when   -x^2=-2x => x=0,2

    Quote Originally Posted by drumist View Post
    Recall that the derivative may only exist where the function is continuous. Can you identify where this function is continuous?
    *My comments are marked in red.

    Does a derivative function only exist on points where the function is continuous? Well, if so then I don't understand the connection
    .

    I believe it's something basic that I can't seem to understand.


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  5. #5
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    Perhaps there is some confusion about what we're trying to do. From the way you worded the question, I believe we are supposed to identify what the derivative of f(x) is.

    First, let's look at where f(x) is continuous. At these values, the derivative might exist. It cannot exist at values where f(x) is not continuous.

    It's important to know a few things about irrational and rational numbers. First, between every two distinct irrational numbers, there exists a rational number that lies between them.

    Also, between every two distinct rational numbers, there exists an irrational number that lies between them.

    I'm not sure how best to explain this, but the implications of this is that f(x) is not continuous except where -x^2=x^2 ~\implies~ x=0. Therefore the derivative can only be defined at that single point.

    Next we verify that \frac{d}{dx}(-x^2) = \frac{d}{dx}(x^2) at x=0.

    \frac{d}{dx}(-x^2) = -2x = -2(0) = 0

    \frac{d}{dx}(x^2) = 2x = 2(0) = 0

    These values are equal, so the derivative does exist at that point. Thus:

    f'(x)=0, \quad \mbox{if } x=0
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  6. #6
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    Quote Originally Posted by Prove It View Post
    What is the derivative of x^2? Where is this derivative =x^2?


    What is the derivative of -x^2? Where is this derivative =-x^2?
    What does the derivative being equal to the function have to do with this problem?
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  7. #7
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    Quote Originally Posted by drumist View Post
    Perhaps there is some confusion about what we're trying to do. From the way you worded the question, I believe we are supposed to identify what the derivative of f(x) is.

    First, let's look at where f(x) is continuous. At these values, the derivative might exist. It cannot exist at values where f(x) is not continuous.

    It's important to know a few things about irrational and rational numbers. First, between every two distinct irrational numbers, there exists a rational number that lies between them.

    Also, between every two distinct rational numbers, there exists an irrational number that lies between them.

    I'm not sure how best to explain this, but the implications of this is that f(x) is not continuous except where -x^2=x^2 ~\implies~ x=0. Therefore the derivative can only be defined at that single point.

    Next we verify that \frac{d}{dx}(-x^2) = \frac{d}{dx}(x^2) at x=0.

    \frac{d}{dx}(-x^2) = -2x = -2(0) = 0

    \frac{d}{dx}(x^2) = 2x = 2(0) = 0

    These values are equal, so the derivative does exist at that point. Thus:

    f'(x)=0, \quad \mbox{if } x=0
    Thank you very much !! That's a very elegant proof!
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