Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x³ - 4x on the interval [-1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.
that is not the mean value theorem for INTEGRALS
To determine if the Mean Value Theorem for Integrals applies to the function f(x) = x³ - 4x on the interval [-1, 1], we need to check two conditions:
1. Continuity: The function must be continuous on the closed interval [-1, 1].
2. Integrability: The function must be integrable on the closed interval [-1, 1].
First, let's check the continuity of f(x) = x³ - 4x. The function is a polynomial, and polynomials are continuous for all real numbers. Since the interval [-1, 1] is within the domain of f(x), the function is continuous on the closed interval.
Next, let's check the integrability of f(x) = x³ - 4x. To do this, we will verify if the function is bounded on the interval [-1, 1].
Since [-1, 1] is a closed interval, we can evaluate the function at the endpoint values: f(-1) and f(1).
f(-1) = (-1)³ - 4(-1) = -1 + 4 = 3
f(1) = (1)³ - 4(1) = 1 - 4 = -3
Therefore, the function is bounded by -3 and 3 on the interval [-1, 1].
Since the function f(x) = x³ - 4x is both continuous and bounded on the interval [-1, 1], the Mean Value Theorem for Integrals applies.
To find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem, we can use the formula:
x = (1/(b - a)) * ∫[a, b] f(x) dx
In this case, a = -1 and b = 1. The formula becomes:
x = (1/(1 - (-1))) * ∫[-1, 1] (x³ - 4x) dx
Simplifying:
x = (1/2) * ∫[-1, 1] (x³ - 4x) dx
To find the definite integral, we can evaluate it:
∫[-1, 1] (x³ - 4x) dx = [ (x⁴/4) - (2x²) ] | from -1 to 1
Substituting the upper and lower limits:
(1/2) * [ (1⁴/4) - (2(1)²) - ((-1)⁴/4) + (2(-1)²) ]
Simplifying:
(1/2) * [ (1/4) - 2 - (1/4) + 2 ]
(1/2) * [0]
0
Therefore, the x-coordinate of the point guaranteed to exist by the Mean Value Theorem for Integrals is 0.
Thank you!
f(x) = x^3 - 4x
f'(x) = 3x^2 - 4
First of all, polynomial functions are all continuous in nature, so the function in continuous in [-1,1] and differentiable in (-1,1)
Now,
f(1) = f(a) = -3, f(-1) = f(b) = 3
Now, there exists one number c such that:
f'(c) = (f(b) - f(a))/(b-a)
=> 3c^2 - 4 = (3 - (-3))/(-1-1)
=> 3c^2 - 4 = -3
=> 3c^2 = 1
=> c = ±√(1/3)