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Math Help - Mean Value Theorem

  1. #1
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    Mean Value Theorem

    can someone please explain step by step what to do? thanx!!

    Use analytic methods to find a) the local extrema, b) the intervals on which the function in increasing, and c) the intervals on which the function is decreasing.

    y = 4 - the square root of (x+2)
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  2. #2
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    Quote Originally Posted by turtle View Post
    can someone please explain step by step what to do? ...

    Use analytic methods to find a) the local extrema, b) the intervals on which the function in increasing, and c) the intervals on which the function is decreasing.
    y = 4 - the square root of (x+2)
    Hello, turtle,

    1. Sketch the graph of this function (see attachment)

    2. to a) Local extrema exist if the first derivative equals zero and the 2nd derivative is not zero. So you need first the derivative of your function:

    y = f(x)=4-\sqrt{x+2}. The domain of f is D_f=[-2;+\infty). Because you always subtract a positive number from a constant the function is decreasing. There can't be any local extrema. Therefore the range of the function is R_f = [4; -\infty)

    \frac{df}{dx}=-\frac{1}{2} \cdot (x+2)^{-\frac{1}{2}}=-\frac{1}{2\cdot \sqrt{x+2}}
    Notice that now the domain of the derivative isn't the domain of the function: D_{f'}=(-2;+\infty)
    f'(x) \neq 0 that means there is only an absolute maximum at x = -2, f(-2) = 4 (remark: In German such a maximum is called absolute border extremum. I don't know the corresponding expression in English)

    3. to b) f'(x) < 0 for all x \in D_{f'} thus the function doesn't increase but is decreasing over the complete domain.

    EB
    Attached Thumbnails Attached Thumbnails Mean Value Theorem-wrzl_fkt1.gif  
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