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) =

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

To find the absolute extrema of the function f(x) = √(25 - x^2) on each given interval, we need to find the minimum and maximum values of the function within those intervals. We can do this by finding the critical points and evaluating the function at the endpoints of each interval.

(a) [-5, 5]

To find the critical points, we need to find where the derivative of the function equals zero or does not exist. Let's first find the derivative of f(x):

f'(x) = d/dx [√(25 - x^2)]
= (1/2) * (25 - x^2)^(-1/2) * (-2x)
= -x / √(25 - x^2)

The derivative exists for all x within the interval [-5, 5]. Now we evaluate the function at the critical points and the endpoints:

f(-5) = √(25 - (-5)^2) = √(25 - 25) = 0
f(5) = √(25 - 5^2) = √(25 - 25) = 0

Therefore, the minimum and maximum values within the interval [-5, 5] are both zero.

(b) [-5, 0)

Since √(25 - x^2) is not defined for x < -5, we only need to consider the left endpoint -5.

f(-5) = √(25 - (-5)^2) = √(25 - 25) = 0

Therefore, the minimum value within the interval [-5, 0) is zero. To find the maximum value, we need to evaluate the function at the critical point, which occurs when the derivative is zero or does not exist. However, in this interval, the derivative is always defined, so there are no critical points. Therefore, the maximum value does not exist (DNE).

(c) (-5, 5)

Since the interval does not include the endpoints -5 and 5, we need to find the critical points by finding where the derivative equals zero or does not exist.

f'(x) = -x / √(25 - x^2)

Setting the derivative equal to zero:

-x / √(25 - x^2) = 0

This equation is satisfied at x = 0. Now we evaluate the function at this critical point:

f(0) = √(25 - 0^2) = √(25) = 5

Therefore, the minimum value within the interval (-5, 5) is 5. To find the maximum value, we evaluate the function at the endpoints:

f(-5) = √(25 - (-5)^2) = √(25 - 25) = 0
f(5) = √(25 - 5^2) = √(25 - 25) = 0

Therefore, the maximum value within the interval (-5, 5) is also 0.

(d) [1, 5)

To find the critical points, we need to find where the derivative equals zero or does not exist.

f'(x) = -x / √(25 - x^2)

Setting the derivative equal to zero:

-x / √(25 - x^2) = 0

This equation is satisfied at x = 0. Now we evaluate the function at this critical point:

f(0) = √(25 - 0^2) = √(25) = 5

Therefore, the minimum value within the interval [1, 5) is 5. To find the maximum value, we evaluate the function at the endpoint:

f(1) = √(25 - 1^2) = √(25 - 1) = √(24)

Therefore, the maximum value within the interval [1, 5) is √(24).