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???
the only place where your function f(x) = 6/(x-3) is not continuous is at x= 3
your interval does not reach that value, so recall your definition of the Mean Value Theorem.
A series of YouTube clips that are done by Sal Khan offer really simple explanations.
Here is the one that deals with this topic.
http://www.youtube.com/watch?v=bGNMXfaNR5Q
To find 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 verify two conditions:
1. Continuity: The function f(x) must be continuous on the interval [-1,2].
2. Differentiability: The function f(x) must be differentiable on the interval (-1,2).
Let's check these conditions one by one:
1. Continuity: We need to determine if f(x) is continuous on the interval [-1,2]. To do this, we need to check if there are any points where the function is not defined or has a discontinuity.
In this case, f(x) = 6/(x - 3) is not defined when x = 3, as the denominator becomes zero. Therefore, the function is not continuous on the interval [-1,2] because it has a discontinuity at x = 3.
Since the function is not continuous on the interval, the Mean Value Theorem does not apply, and there is no value of c that satisfies the conditions for the Mean Value Theorem on the interval [-1,2].