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) =

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

I'm glad you asked! Let's find the absolute extrema of the function f(x) = √(25 - x^2) on the interval [-5, 5].

To find the minimum and maximum values, we need to evaluate the function at the endpoints and at any critical points within the interval.

At x = -5:
f(-5) = √(25 - (-5)^2) = √(25 - 25) = √0 = 0

At x = 5:
f(5) = √(25 - 5^2) = √(25 - 25) = √0 = 0

Now let's find the critical points within the interval. To do this, we need to find the derivative of f(x) and set it equal to zero:

f(x) = √(25 - x^2)
f'(x) = (2x) / (2√(25 - x^2))

Setting f'(x) = 0 gives us:
(2x) / (2√(25 - x^2)) = 0

Since the numerator is already zero, the equation simplifies to:
x = 0

Now, let's evaluate the function at x = 0:
f(0) = √(25 - 0^2) = √25 = 5

So, on the interval [-5, 5], the minimum value is (x, y) = (-5, 0) and the maximum value is (x, y) = (5, 0).

I hope that puts a smile on your face!

To find the absolute extrema of the function f(x) = √(25 - x^2) on the interval [-5, 5], we need to evaluate the function at the critical points and endpoints.

1. Critical points:
To find the critical points, we need to find the values of x where the derivative of f(x) is either 0 or undefined. Differentiating f(x) with respect to x, we have:

f'(x) = (1/2)(25 - x^2)^(-1/2)(-2x)
= -x / √(25 - x^2)

Setting f'(x) = 0, we get:
-x / √(25 - x^2) = 0

This equation is true when x = 0.

2. Endpoints:
Evaluate f(x) at the endpoints of the interval [-5, 5]:
f(-5) = √(25 - (-5)^2) = √(25 - 25) = √(0) = 0
f(5) = √(25 - 5^2) = √(25 - 25) = √(0) = 0

3. Absolute Extrema:
Now we need to evaluate f(x) at the critical points and endpoints to find the absolute extrema.

f(x) at x = -5: f(-5) = 0
f(x) at x = 0: f(0) = √(25 - 0^2) = √(25) = 5
f(x) at x = 5: f(5) = 0

Therefore, the absolute extrema on the interval [-5, 5] are:
Minimum (x, y) = (0, 0)
Maximum (x, y) = (0, 5)

To find the absolute extrema of a function, we need to find the minimum and maximum values of the function on the given interval. In this case, we are given the function f(x) = √(25 - x^2) and the interval [-5, 5].

To find the minimum and maximum values on this interval, we can follow these steps:

1. Find the critical points: These are the points where the derivative is either zero or undefined. In this case, the function is continuous for all x-values on the interval [-5, 5], so we don't have any critical points.

2. Evaluate the function at the endpoints of the interval: We need to find the value of the function at x = -5 and x = 5.

For x = -5:
f(-5) = √(25 - (-5)^2) = √(25 - 25) = √0 = 0.

For x = 5:
f(5) = √(25 - 5^2) = √(25 - 25) = √0 = 0.

Both endpoints give us the value 0.

3. Compare the values: The minimum value will be the smallest value obtained from the function. In this case, the minimum value is 0.

The maximum value will be the largest value obtained from the function. In this case, the maximum value is also 0.

Therefore, the function f(x) = √(25 - x^2) has a minimum value of 0 and a maximum value of 0 on the interval [-5, 5].