Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x^3 − 9x 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 − 9x on the interval [-1, 1], we need to check if the function is continuous on that interval and differentiable on the open interval (-1, 1).
1. Continuity: A function is continuous on a closed interval if and only if it is continuous on each point of the interval. So, we need to check if f(x) = x^3 − 9x is continuous on [-1, 1].
- The polynomial function x^3 − 9x is continuous everywhere because it is a combination of power functions, which are continuous for all values of x.
- Therefore, f(x) = x^3 − 9x is continuous on [-1, 1].
2. Differentiability: A function is differentiable on an open interval if its derivative exists for all values of x in that interval. Let's find the derivative of f(x) = x^3 − 9x and check if it exists on (-1, 1).
- The derivative of f(x) is given by f'(x) = 3x^2 - 9.
- The derivative 3x^2 - 9 exists for all values of x.
- Therefore, f(x) = x^3 − 9x is differentiable on (-1, 1).
Since the function f(x) = x^3 − 9x is continuous on [-1, 1] and differentiable on (-1, 1), the Mean Value Theorem for Integrals applies to it.
To find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem, we need to find the average rate of change or the average value of f(x) on the interval [-1, 1]. This is done by finding the definite integral of f(x) on the interval [-1, 1] and dividing it by the length of the interval.
The Mean Value Theorem for Integrals states that there exists at least one value c in the interval (-1, 1) such that f'(c) = (1/(b-a)) * ∫[a,b] f(x) dx, where [a, b] is the interval [-1, 1].
We can find the definite integral of f(x) on the interval [-1, 1] as follows:
∫[-1, 1] (x^3 - 9x) dx = [(x^4)/4 - (9x^2)/2] | [-1, 1]
= [(1^4)/4 - (9*1^2)/2] - [((-1)^4)/4 - (9*(-1)^2)/2]
= [(1/4) - (9/2)] - [(1/4) - (9/2)]
= [-8.75] - [-8.75]
= 0
Since the definite integral over the interval [-1, 1] is equal to 0, we can set up the equation:
0 = (1/(1-(-1))) * ∫[-1, 1] (x^3 - 9x) dx
= (1/2) * ∫[-1, 1] (x^3 - 9x) dx
To find the x-coordinate(s) of the point(s) guaranteed by the Mean Value Theorem, we need to solve the equation:
0 = ∫[-1, 1] (x^3 - 9x) dx
To solve this equation, we can find the definite integral of f(x) on the interval [-1, 1], which we already know is zero. Therefore, we have:
0 = 0
Since 0 = 0 is true, it means that all x-coordinates within the interval (-1, 1) satisfy the equation, and there exists at least one point c in (-1, 1) where f'(c) = 0.
In conclusion, the Mean Value Theorem for Integrals applies to the function f(x) = x^3 − 9x on the interval [-1, 1], and the x-coordinates of the point(s) guaranteed to exist by the theorem are all values within the interval (-1, 1).