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Thread: Show it is differentiable and compute f '

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    Member thaopanda's Avatar
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    Show it is differentiable and compute f '

    Show that the function f: R $\displaystyle \rightarrow$ R given by

    f(x): =
    $\displaystyle x^3, x \leq 0$
    0, x = 0

    is differentiable and compute f '.
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  2. #2
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    Quote Originally Posted by thaopanda View Post
    Show that the function f: R $\displaystyle \rightarrow$ R given by

    f(x): =
    $\displaystyle x^3, x \leq 0$
    0, x = 0

    is differentiable and compute f '.
    What about when $\displaystyle x > 0$?
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  3. #3
    Member thaopanda's Avatar
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    oh wow, I wasn't paying attention. Messed that one up. It's actually:

    f(x) :=
    $\displaystyle x^3, x \leq 0 $
    $\displaystyle x^2*sin(\frac{1}{x}), x > 0$

    sorry about that
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    Quote Originally Posted by thaopanda View Post
    oh wow, I wasn't paying attention. Messed that one up. It's actually:

    f(x) :=
    $\displaystyle x^3, x \leq 0 $
    $\displaystyle x^2*sin(\frac{1}{x}), x > 0$

    sorry about that
    To be differentiable, the function needs to be continuous and smooth.


    Clearly $\displaystyle x^3$ is continuous since it is a polynomial.

    Also, since $\displaystyle x^2$ and $\displaystyle \sin{\left(\frac{1}{x}\right)}$ are both continuous for $\displaystyle x > 0$, so will be their product.

    So the only place where this hybrid function might not be continuous is where the function "changes" - so at $\displaystyle x = 0$. We need to check that the left hand and right hand limits are 0.

    Clearly the left hand limit is 0. So we would need to show that $\displaystyle \lim_{x \to 0^{+}}x^2\sin{\left(\frac{1}{x}\right)} = 0$.
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