Asked by Sagar
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) =
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) =
Answers
Answered by
Reiny
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
= (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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