Determine if the Mean Value Theorem for Integrals applies to the function f(x) = 5 - x^2 on the interval 0 to sqrt 5 . If so, find the x-coordinates of the point(s) guaranteed by the theorem.
To determine if the Mean Value Theorem for Integrals applies to the function f(x) = 5 - x^2 on the interval [0, √5], we need to check two conditions:
1. The function f(x) must be continuous on the interval [0, √5].
2. The function f(x) must also be integrable on the interval [0, √5].
Let's check the first condition: continuity. The function f(x) = 5 - x^2 is a polynomial function, which means it is continuous everywhere. Therefore, it is continuous on the interval [0, √5].
Now let's check the second condition: integrability. Since the function f(x) = 5 - x^2 is a polynomial function, it is integrable on any closed interval. Therefore, it is integrable on the interval [0, √5].
So, the Mean Value Theorem for Integrals does apply to the function f(x) = 5 - x^2 on the interval [0, √5].
According to the Mean Value Theorem for Integrals, there exists at least one value "c" in the interval [0, √5] such that the average value of the function is equal to the instantaneous rate of change at that point.
To find the x-coordinate(s) of the point(s) guaranteed by the theorem, we need to find the average value of the function on the interval [0, √5]. The average value of the function f(x) on the interval [0, √5] is given by:
average value = (1 / (b - a)) * integral of f(x) from a to b
In this case, a = 0 and b = √5. Plugging in these values, we get:
average value = (1 / (√5 - 0)) * integral of (5 - x^2) from 0 to √5
Simplifying further, we have:
average value = (1 / √5) * integral of (5 - x^2) from 0 to √5
To find the x-coordinate(s) where the instantaneous rate of change is equal to the average value, we need to find the solution(s) to the equation f'(c) = average value. However, since f(x) = 5 - x^2, we need to find f'(x) first. Differentiating f(x) with respect to x, we get:
f'(x) = -2x
Setting f'(c) = average value, we have:
-2c = (1 / √5) * integral of (5 - x^2) from 0 to √5
Solving this equation will give us the value(s) of c, which represents the x-coordinate(s) of the point(s) guaranteed by the Mean Value Theorem for Integrals.