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Math Help - End Point Maxima and Minima?

  1. #1
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    End Point Maxima and Minima?

    Hello!

    Quick question, in my Calc class we are covering End Point Maxima and Minima.

    I understand that for lets say prove there is a max at x=-2

    f(x)=(x+2)^(3/2)+5*x^2 for x >= -2

    f'(x)=(3/2)*(x+2)^(1/2)+10*x

    You can just plug in -2 and get -20, which proves that it is a max because the function is decreasing to the right of it.

    What I do not understand is this

    f(x)=(25-x^2)^(1/2)-2*x^2 for -5 <= x <= 5

    f'(x)=(1/2)*(2x)*(25-x^2)^(-1/2)-4x

    It is asking if 5 is a max or a min

    But when you plug in 5 it DNE

    The book describes using infinity, but I do not understand how they are using it?

    Thanks for anyone who can offer any help!
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  2. #2
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    Re: End Point Maxima and Minima?

    f(x) = \sqrt{25-x^2} - 2x^2

    f'(x) = \frac{-x}{\sqrt{25-x^2}} - 4x

    note that f(x) is defined at x = \pm 5 , but f'(x) is not.

    f'(x) = 0 at x = 0

    for -5 < x < 0 , f'(x) > 0 ... f(x) is increasing on that interval. therefore, f(-5) is an endpoint minimum.

    for 0 < x < 5 , f'(x) < 0 ... f(x) is decreasing on that interval. therefore, f(5) is an endpoint minimum.

    f(0) is a maximum
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