Find the POSITIVE value of x that satisfies the mean value theorem for
f(x)=sin(x) on the closed interval
[-3pi/2, 3pi/2].
please help I have no idea how to solve this problem
sin 3 pi/2 = sin 270 = -1
sin - 3 pi/2 = +1
change in sin x = -1 - +1 = -2
change in x = 3 pi/2+3pi/2 = 3 pi
so
slope from left to right = -2 / 3pi
= -.2122
so where does the derivative of sin x = -.2122 ?
d/dx (sin x) = cos x
so cos x = -.2122
x = 102.25 degrees = 1.78 radians
To find the positive value of x that satisfies the mean value theorem, we need to understand the theorem and its conditions.
The Mean Value Theorem 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 value c in the open interval (a, b) such that the derivative of f(x) at c is equal to the average rate of change of f(x) over [a, b].
In order to apply the Mean Value Theorem, we first need to check if the given function f(x) = sin(x) satisfies its conditions:
1. Continuity: The function f(x) = sin(x) is continuous on all real numbers, so it is continuous on the closed interval [-3π/2, 3π/2].
2. Differentiability: The function f(x) = sin(x) is differentiable on all real numbers, including the open interval (-3π/2, 3π/2).
Since the function satisfies the conditions of the Mean Value Theorem, we can proceed to find the value of c.
To find c, we need to find the average rate of change of f(x) over the interval [-3π/2, 3π/2], which is given by:
Average rate of change = [f(b) - f(a)] / (b - a)
Plugging in the values: a = -3π/2, b = 3π/2, f(a) = sin(-3π/2) = -1, and f(b) = sin(3π/2) = 1:
Average rate of change = [1 - (-1)] / (3π/2 - (-3π/2))
= [2] / (6π/2)
= 2/π
Now, we need to find a value of c such that the derivative of f(x) at c is equal to this average rate of change:
f'(c) = 2/π
To find this value of c, we differentiate the function f(x) = sin(x) to get f'(x) = cos(x):
cos(c) = 2/π
Since we are looking for a positive value of x, we need to find the smallest positive value of c that satisfies cos(c) = 2/π. To do this, we can use the inverse cosine function:
c = arccos(2/π)
Using a calculator, we find:
c ≈ 0.292
Therefore, the positive value of x that satisfies the Mean Value Theorem for f(x) = sin(x) on the closed interval [-3π/2, 3π/2] is approximately x ≈ 0.292.