Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x3 - 16x on the interval [-1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.
To determine if the Mean Value Theorem for Integrals applies to the function f(x) = x^3 - 16x on the interval [-1, 1], we need to check if the function satisfies the necessary conditions.
The Mean Value Theorem for Integrals states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that the average value of the function on [a, b] is equal to the instantaneous rate of change at c.
In this case, the function f(x) = x^3 - 16x is a polynomial function, and polynomials are continuous and differentiable on the entire real number line. Therefore, the function satisfies the necessary conditions for the Mean Value Theorem for Integrals.
To find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem, we can calculate the average value of the function on the interval [-1, 1] and then find the x-coordinate(s) where the instantaneous rate of change is equal to the average value.
The average value of the function f(x) on the interval [a, b] is given by:
average value = (1 / (b - a)) * ∫[a, b] f(x) dx
In this case, a = -1, b = 1, and f(x) = x^3 - 16x. So the average value of f(x) on [-1, 1] is:
average value = (1 / (1 - (-1))) * ∫[-1, 1] (x^3 - 16x) dx
We can now integrate f(x) over the interval [-1, 1] to find the average value:
average value = (1 / 2) * ∫[-1, 1] (x^3 - 16x) dx
Taking the integral of f(x) = x^3 - 16x, we get:
average value = (1 / 2) * [((1/4) * x^4) - (8x^2)] evaluated from -1 to 1
Evaluating the integral at the limits:
average value = (1 / 2) * [(1/4 - 8) - (1/4 + 8)]
average value = (1 / 2) * [(-31/4) - (33/4)]
average value = (1 / 2) * (-64/4)
average value = (1 / 2) * (-16)
average value = -8
Therefore, the average value of f(x) on the interval [-1, 1] is -8.
To find the x-coordinate(s) where the instantaneous rate of change is equal to the average value, we need to set up the equation:
f'(c) = average value
Taking the derivative of f(x) = x^3 - 16x, we get:
f'(x) = 3x^2 - 16
Setting f'(c) equal to the average value, we have:
3c^2 - 16 = -8
Simplifying the equation:
3c^2 - 8 = 0
Now, we can solve this quadratic equation for c. By factoring, we get:
(3c - 4)(c + 2) = 0
This gives us two possible values for c:
1. 3c - 4 = 0, which yields c = 4/3
2. c + 2 = 0, which yields c = -2
Therefore, the x-coordinates of the points guaranteed to exist by the Mean Value Theorem for Integrals are c = 4/3 and c = -2.
To determine if the Mean Value Theorem for Integrals applies to a function on a given interval, we need to check two conditions:
1. The function must be continuous on the interval [-1, 1].
2. The function must be integrable on the interval [-1, 1].
Let's check these conditions for the function f(x) = x^3 - 16x on the interval [-1, 1].
1. Continuous Function:
To check if the function is continuous on the interval [-1, 1], we need to verify that it is continuous at every point within the interval and at the endpoint values.
The function f(x) = x^3 - 16x is a polynomial function, and polynomial functions are continuous for all real numbers. Therefore, f(x) = x^3 - 16x is continuous on the interval [-1, 1].
2. Integrable Function:
To check if the function is integrable on the interval [-1, 1], we need to determine if it satisfies certain integrability conditions. Fortunately, f(x) = x^3 - 16x is a polynomial function, which means it is integrable on any closed interval.
Since the function f(x) = x^3 - 16x satisfies both conditions, we can conclude that the Mean Value Theorem for Integrals applies to the function on the interval [-1, 1].
According to the Mean Value Theorem for Integrals, there must exist at least one point c in the interval [-1, 1] such that the average value of the function f(x) over the interval [-1, 1] is equal to the value of f'(c), where f'(c) represents the derivative of f(x) at c.
The average value of f(x) over the interval [-1, 1] is given by:
1/(b-a) * ∫[a,b] f(x) dx
For the interval [-1, 1], the average value is:
1/(1-(-1)) * ∫[-1,1] (x^3 - 16x) dx
Now, we can find the average value using integration. After calculating the average value, we can equate it to f'(c) to find the x-coordinates of the point(s) guaranteed to exist by the theorem.
Since f is continuous and differentiable on the interval, MVT applies.
So, now you just have to find the point(s) c where
f'(c) = [f(1)-f(-1)]/[1-(-1)]