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.
THANK YOU VERY MUCH
Thank you, but are you doing the MVT for the derivative? Because my question is in terms of integrals, and when I do it for integrals I get a different answer than -3
To determine if the Mean Value Theorem for Integrals applies to the function f(x) = x^3 - 4x on the interval [-1, 1], we need to check if the function meets 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 number c in (a, b) such that the average value of the function on [a, b] equals its instantaneous rate of change at c.
In this case, the function f(x) = x^3 - 4x is continuous on the interval [-1, 1] since it is a polynomial function. However, we also need to check if it is differentiable on the open interval (-1, 1).
To determine differentiability, we need to take the derivative of f(x). The derivative of f(x) = x^3 - 4x is f'(x) = 3x^2 - 4.
We can see that f'(x) = 3x^2 - 4 is defined for all real numbers and thus is also defined on the open interval (-1, 1). Therefore, f(x) = x^3 - 4x is differentiable on (-1, 1).
Since the function f(x) = x^3 - 4x is continuous on [-1, 1] and differentiable on (-1, 1), the Mean Value Theorem for Integrals applies.
To find the x-coordinates of the point(s) guaranteed to exist by the theorem, we need to find the average value of f(x) on the interval [-1, 1] and then find the x-coordinate(s) where this average value equals the instantaneous rate of change at that point.
The average value of f(x) on the interval [-1, 1] is given by:
f_avg = (1 / (b - a)) * ∫[a, b] f(x) dx
For our interval [-1, 1], the average value becomes:
f_avg = (1 / (1 - (-1))) * ∫[-1, 1] (x^3 - 4x) dx
Calculating the integral, we have:
f_avg = (1 / 2) * ∫[-1, 1] (x^3 - 4x) dx
f_avg = (1 / 2) * [((1/4)x^4 - 2x^2)] from -1 to 1
f_avg = (1 / 2) * [((1/4)(1)^4 - 2(1)^2) - ((1/4)(-1)^4 - 2(-1)^2)]
f_avg = (1 / 2) * [(1/4 - 2) - (1/4 - 2)]
f_avg = (1 / 2) * [(-7/4) - (-7/4)]
f_avg = (1 / 2) * [0]
f_avg = 0
Now, we need to find the x-coordinate(s) where the instantaneous rate of change (the derivative) is equal to the average value (f_avg).
Setting f'(x) = f_avg = 0, we have:
3x^2 - 4 = 0
Solving for x^2, we get:
3x^2 = 4
x^2 = 4/3
Taking the square root, we have:
x = ±√(4/3)
So, the x-coordinates of the point(s) guaranteed to exist by the Mean Value Theorem for Integrals are x = ±√(4/3).
To determine if the Mean Value Theorem for Integrals applies to the function \(f(x) = x^3 - 4x\) on the interval \([-1, 1]\), we need to check if the function meets a certain criterion.
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 a value \(c\) in \((a, b)\) such that the definite integral of \(f(x)\) from \(a\) to \(b\) is equal to \(f(c)\) times the length of the interval \([a, b]\), which is \(b - a\). In other words, if \(f(x)\) satisfies the necessary conditions, there must exist a point within the interval where the instantaneous rate of change (the derivative) is equal to the average rate of change (the slope of the secant).
To apply the theorem, we need to check if the function \(f(x) = x^3 - 4x\) is continuous on the interval \([-1, 1]\) and differentiable on the open interval \((-1, 1)\).
1. Continuity: The function \(f(x) = x^3 - 4x\) is a polynomial, which is continuous everywhere. Therefore, it is continuous on the interval \([-1, 1]\).
2. Differentiability: To check if the function is differentiable on the open interval \((-1, 1)\), we need to take its derivative.
\(f'(x) = 3x^2 - 4\)
The derivative exists for all values of \(x\) since it is a polynomial. Therefore, the function \(f(x) = x^3 - 4x\) is differentiable on the open interval \((-1, 1)\).
Since the function satisfies the necessary conditions, the Mean Value Theorem for Integrals applies to \(f(x) = x^3 - 4x\) on the interval \([-1, 1]\). The theorem guarantees the existence of at least one point \(c\) in \((-1, 1)\) such that the instantaneous rate of change (the derivative) at \(c\) is equal to the average rate of change from \(-1\) to \(1\) (the slope of the secant). This point \(c\) is the x-coordinate of the point(s) guaranteed to exist by the theorem.
To find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem, we need to find the value(s) of \(c\) that satisfy the equation \(f'(c) = \frac{f(1) - f(-1)}{1 - (-1)}\).
Substituting the values into the equation:
\(3c^2 - 4 = \frac{(1^3 - 4 \cdot 1) - ((-1)^3 - 4 \cdot (-1))}{2}\)
Simplifying further:
\(3c^2 - 4 = \frac{-3 - (-3)}{2}\)
\(3c^2 - 4 = 0\)
Solving for \(c\):
\(3c^2 = 4\)
\(c^2 = \frac{4}{3}\)
\(c = \pm \sqrt{\frac{4}{3}}\)
So, the x-coordinates of the point(s) guaranteed to exist by the Mean Value Theorem for Integrals on the interval \([-1, 1]\) for the function \(f(x) = x^3 - 4x\) are \(c = \sqrt{\frac{4}{3}}\) and \(c = -\sqrt{\frac{4}{3}}\).
The MVT applies to all polynomials, since they are continuous and differentiable everywhere.
The slope of the line joining the endpoints of the graph is
(f(1)-f(-1))/(1 - (-1)) = -3
Since f'(x) = 3x^2-4
you want to find the value c such that
3c^2 - 4 = -3