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Math Help - Maxima and Minima

  1. #1
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    Maxima and Minima

    Find all of the critical numbers of the function f(x)=|x^2+4x| on the open interval (-3,3).
    Last edited by Velvet Love; October 22nd 2009 at 03:37 PM. Reason: Forgot absolute value
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  2. #2
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    Quote Originally Posted by Velvet Love View Post
    Find all of the critical numbers of the function f(x)=|x^2+4x| on the open interval (-3,3).

    \frac{d}{dx} |u| = \frac{u}{|u|} \cdot \frac{du}{dx}
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    Thanks. I just didn't know what the derivative of an absolute value function was.
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    Differentiation is not the whole story for this problem, though – there is a critical point at which the function is not differentiable.

    A quick sketch of the curve should show that x^2+4x<0 for -3<x<0 and x^2+4x>0 for 0<x<3 and f(x) has a nondifferentiable critical point at x=0. Therefore a better solution is to rewrite the function as follows:

    f(x)\ =\ \begin{cases}-x^2-4x & -3<x\leqslant0\\<br />
x^2+4x & 0\leqslant x<3\end{cases}

    You can then use differentiation on the open intervals (-3,\,0) and (0,\,3) separately to find the other critical point(s).
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