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# Justification with the mean value theorem: table

Example justifying use of mean value theorem (where function is defined with a table).

## Want to join the conversation?

• Why do we use the domain as values to calculate the slope of the secant line when < & > are non-inclusive?
• Think about what the MVT actually says:
`for a function f that is - differential over (a, b)- continuous over [a, b]There exists a c in open interval (a, b) where f'(c) = avg rate of change over the closedinterval [a, b]`
So in the questions, our c is
a < c < b
or in other words
c is a member of (a, b)
But because of what the MVT says, we still use a and b to find the average rate of change (or the secant line).
• In the first example, does the MVT not apply because there's no value of f'(x) that equals the average rate of change over the interval 4 < c < 6? Or is it because there's no value of c such that f'(c)= 5 AND the average rate of change?
Also I felt like Sal was in a rush.
• There may be a value of x that causes f'(x) to equal the rate of change over the interval [4,6], but there is no value of c such that f'(c) = 5 and the rate of change. This is because you are adding the condition that f'(c) has to equal 5 while the rate of change over that interval stays the same.
• Why we use open interval in differentiabilty case??
(1 vote)
• Because if the function isn't defined outside the interval, the derivative won't exist at the endpoints. But this doesn't affect the main point of the theorem.
• In the second question, it is said to prove that f'(x) = -1 has a solution but according to the theorem this equation is satisfied but that doesn't mean x has solutions in the given interval that satisfies the above condition. Isn't it?
(1 vote)
• You seem to have misunderstood the idea here. See, the MVT simply states that for some interval [a,b], the average rate of change over the interval equals the instantaneous rate of change at some point in that interval.

For example, suppose your average speed during a car trip was 40kmph. This obviously means that at some point, you were driving at 40kmph (not necessarily at every point though).

So, same idea here. We proved that the average rate of change between the two points is -1, which means that the instantaneous rate of change at some point in that interval must also be -1