Find the values of c that satisfy the Mean Value Theorem for f(x)=6/x-3 on the interval [-1,2].
Is it no value of c in that interval because the function is not continuous on that interval???
It is continuous between -1 and +3. The blow up is at x = 3.
y = 6/(x-3)
dy/dx = -6/(x-3)^2
now slope from -1 to +2
y(-1) = 6/-4 = -3/2
y(2) = 6/-1 = -6
delta x = 2+1 = 3
so mean slope = (-6 + 3/2)/3 = -2 + 1/2 = -1.5
now where does derivative = -1.5 ?
-3/2 = -6/(x-3)^2
(x-3)^2 = 4
x-3 = +/- 2
x = 5 or x = 1
only x = 1 is on the interval from -1 to 2
To apply the Mean Value Theorem, we need to check two conditions:
1. The function f(x) must be continuous on the closed interval [-1, 2], and
2. The function f(x) must be differentiable on the open interval (-1, 2).
Let's check if these conditions are satisfied for f(x) = 6/x - 3 on the interval [-1, 2].
1. Continuity: The function is not defined at x = 3, so it is not continuous on the entire interval (-1, 2). However, if we remove the point x = 3 from the interval (-1, 2), the function becomes continuous.
2. Differentiability: The function f(x) = 6/x - 3 is differentiable for all x ≠ 0, including the interval (-1, 2).
Since our function satisfies the second condition, we can proceed to find the values of c that satisfy the Mean Value Theorem. However, since the function is not continuous on the entire interval [-1, 2], the Mean Value Theorem does not apply.
To determine the values of c that satisfy the Mean Value Theorem for the function f(x) = 6/(x-3) on the interval [-1,2], we need to check if the function is continuous on that interval.
The Mean Value Theorem states that for a function f(x) that is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), there exists at least one value c in (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 this case, the given function f(x) = 6/(x-3) is not continuous on that interval because it is undefined at x = 3, which is contained in the interval [-1,2]. Therefore, we cannot apply the Mean Value Theorem to this function on the interval [-1,2].
Hence, there are no values of c that satisfy the Mean Value Theorem for f(x) = 6/(x-3) on the interval [-1,2].