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.
I think one point is zero.
Thank you!
To determine if the Mean Value Theorem for Integrals (MVT) applies to the function f(x) = x^3 - 9x on the interval [-1, 1], we need to check if the function satisfies the conditions of the theorem.
The conditions for MVT are as follows:
1. f(x) must be continuous on the closed interval [a, b].
2. f(x) must be differentiable on the open interval (a, b).
For f(x) = x^3 - 9x, the function is a polynomial and therefore continuous and differentiable everywhere.
Since f(x) is continuous on the closed interval [-1, 1] and differentiable on the open interval (-1, 1), the conditions for the Mean Value Theorem for Integrals are met.
According to the MVT, there exists at least one point c in (-1, 1) such that the derivative of f(x) at c is equal to the average rate of change of f(x) over the interval [-1, 1].
To find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem, we can find the average rate of change of f(x) on the interval [-1, 1] and find the point(s) where the derivative of f(x) is equal to this average rate of change.
The average rate of change of f(x) over the interval [-1, 1] is given by:
average rate of change = (f(1) - f(-1)) / (1 - (-1))
Let's calculate this:
f(1) = 1^3 - 9(1) = -8
f(-1) = (-1)^3 - 9(-1) = 10
average rate of change = (-8 - 10) / (1 - (-1)) = -18 / 2 = -9
To find the x-coordinate(s) of the point(s) guaranteed by the MVT, we need to find the value(s) of c in (-1, 1) such that the derivative of f(c) is equal to -9.
The derivative of f(x) = x^3 - 9x is:
f'(x) = 3x^2 - 9
To find the x-coordinate(s) of the point(s) guaranteed by the MVT, we set f'(c) = -9:
3c^2 - 9 = -9
Simplifying the equation:
3c^2 = 0
Solving for c:
c^2 = 0
Taking the square root:
c = 0
Therefore, according to the Mean Value Theorem for Integrals, there is one point c in (-1, 1) where f'(c) = -9. In this case, the x-coordinate of the point guaranteed to exist is c = 0.
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 if it is integrable (meaning it has a finite integral on that interval).
1. Continuity: The function f(x) = x^3 - 9x is a polynomial, which is continuous everywhere. Therefore, it is continuous on the interval [-1, 1].
2. Integrability: A function is integrable if it is continuous on a closed interval, which is the case here. So, f(x) = x^3 - 9x is integrable on [-1, 1].
Since the function is continuous and integrable on the interval [-1, 1], the Mean Value Theorem for Integrals applies.
Now, let's find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem.
According to the Mean Value Theorem for Integrals, there is at least one point c in [-1, 1] such that the integral of f(x) over [-1, 1] is equal to f(c) times the length of the interval [-1, 1].
The integral of f(x) over [-1, 1] is given by:
∫[f(x) dx] from -1 to 1 = ∫[(x^3 - 9x) dx] from -1 to 1
Evaluating the integral, we get:
∫[(x^3 - 9x) dx] from -1 to 1 = [(x^4/4 - (9x^2)/2)] from -1 to 1
= [(1/4 - 9/2) - ((1/4) - (9/2))]
= [(-17/4) - (-17/4)]
= 0
Since the integral is equal to zero, this means that f(c) times the length of the interval [-1, 1] is also zero. The length of the interval is 1 - (-1) = 2.
So, we have f(c) * 2 = 0.
This implies f(c) = 0.
To find the x-coordinates of the point(s) guaranteed to exist, we need to solve the equation f(x) = x^3 - 9x = 0.
Factoring out x, we get:
x(x^2 - 9) = 0
Setting each factor equal to zero, we have:
x = 0 or x^2 - 9 = 0
Solving x^2 - 9 = 0, we get:
x^2 = 9
x = ±3
Therefore, the x-coordinates of the point(s) guaranteed to exist by the Mean Value Theorem for Integrals are x = 0 and x = ±3.
well, f(x) is continuous, so the MVT applies.
(f(1)-f(-1))/(1-(-1)) = (-8-8)/2 = -8
So, you want c where f'(c) = -8
3c^2-9 = -8
c = ±1/?3
Note that at f'(0) = -9, so it is too steep. You were just guessing, there, right? Just to show that our c values work, note that f(x) is odd, so if one works, the other does as well.
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E3-9x,+y%3D-8(x-1%2F%E2%88%9A3)-26%2F(3%E2%88%9A3)