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.
Not quite sure how to do this one at all to be completely honest. Thank you for helping!
https://www.khanacademy.org/math/differential-calculus/derivative-applications/mean_value_theorem/v/mean-value-theorem-1
the average slope is (f(1)-f(=1))/(2)=-16/2=-8 check that.
then the points are
f'=-8=3x^2-9
x^2=1/3
x=+- sqrt (1/3)
check: f'(sqrt1/3)=1-9=-8
f'(-sqrt1/3)=1-9=-8
To determine if the Mean Value Theorem for Integrals applies to a function, we need to check two conditions:
1. The function must be continuous on the closed interval [a, b].
2. The function must be integrable on the interval [a, b].
Let's check these conditions for the given function f(x) = x^3 - 9x on the interval [-1, 1]:
1. Continuity: To determine if the function is continuous on the interval [-1, 1], we need to check if it has any breaks, jumps, or holes within this interval. The function f(x) = x^3 - 9x is a polynomial, and polynomials are continuous on the entire real number line. Therefore, the function is continuous on the interval [-1, 1].
2. Integrability: To determine if the function is integrable on the interval [-1, 1], we need to check if it is well-behaved in terms of integrability. Polynomials are always integrable on any closed interval, so the function f(x) = x^3 - 9x is integrable on the interval [-1, 1].
Since the given function satisfies both conditions, we can conclude that 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 exists at least one point c in the interval [a, b] such that the function's average rate of change over the interval is equal to its instantaneous rate of change at c.
The average rate of change of a function over an interval [a, b] is given by the formula:
Average rate of change = (f(b) - f(a)) / (b - a)
In this case, a = -1, b = 1, and f(x) = x^3 - 9x. So, we have:
Average rate of change = (f(1) - f(-1)) / (1 - (-1))
To find f(1) and f(-1), we substitute the corresponding values of x into the function:
f(1) = 1^3 - 9(1) = 1 - 9 = -8
f(-1) = (-1)^3 - 9(-1) = -1 + 9 = 8
Plugging these values into the formula for average rate of change, we have:
Average rate of change = (-8 - 8) / (1 - (-1)) = -16 / 2 = -8
Now we need to find the x-coordinate(s) of the point(s) where the instantaneous rate of change equals -8.
By taking the derivative of the function f(x) = x^3 - 9x, we get:
f'(x) = 3x^2 - 9
To find the x-coordinate(s), we set the derivative equal to -8 and solve for x:
3x^2 - 9 = -8
Rearranging the equation, we have:
3x^2 = 1
Dividing both sides by 3, we get:
x^2 = 1/3
Taking the square root of both sides, we find:
x = ±√(1/3)
So the x-coordinate(s) of the point(s) guaranteed to exist by the theorem are x = ±√(1/3).
Therefore, in the interval [-1, 1], the Mean Value Theorem for Integrals guarantees the existence of at least one point with x-coordinate ±√(1/3) where the instantaneous rate of change of the function f(x) = x^3 - 9x is equal to the average rate of change (-8).