calculus

Find the absolute extrema of the function (if any exist) on each interval. (If an answer does not exist, enter DNE.)
f(x) = square root of (25 − x^2)
(a)
[−5, 5]

minimum (x, y) =



(smaller x-value)
(x, y) =



(larger x-value)
maximum (x, y) =



(b)
[−5, 0)

minimum (x, y) =




maximum (x, y) =



(c)
(−5, 5)

minimum (x, y) =




maximum (x, y) =



(d)
[1, 5)

minimum (x, y) =




maximum (x, y) =

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  1. let y = √( 25-x^2)
    = (25-x^2)^(1/2)

    dy/dx = (1/2)(25-x^2)^(-1/2) (-2x)
    = -x/√(25-x^2)
    = 0 for a max/min

    thus x = 0
    and f(0) = √25 = 5

    I will let you decide if (0,5) is a max or a min
    You might want want to look at a quick sketch of the function to easily answer the other parts

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