# 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) =

1. 👍 0
2. 👎 0
3. 👁 279
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

1. 👍 0
2. 👎 0

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