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
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) =
1 answer