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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?
(3 votes)
• 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).
(7 votes)
• 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.
(4 votes)
• 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.
(2 votes)
• 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.
(3 votes)
• 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
(2 votes)
• can you have two c's that are differentiable and continuous over some inteval in a function?
(0 votes)
• hey, why we didn't write that f(x) is differentiable over (0,2) instead of [0,2].
(0 votes)
• Why no teaching newton's method.
(0 votes)

## Video transcript

- [Instructor] The table gives selected values of the differentiable function f. All right, can we use the mean value theorem to say that there is a value c such that f prime of c is equal to five and c is between four and six? If so, write a justification. Well it meets, to meet the, to use the mean value theorem, you have to be differentiable over the open interval and continuous over the closed interval, so it seems like we've met that. Because if you're differentiable over an interval, you're definitely continuous over that interval. It's saying that it's just a generally differentiable function f I guess over any interval. But the next part is to say, all right, if that condition is met, then the slope of the secant line between four comma f of four and six comma f of six, that some at least one point in between four and six will have a derivative that is equal to the slope of the secant line. And so let's figure out what the slope of the secant line is between four comma f of four and six comma f of six. And if it's equal to five, then we could use the mean value theorem. If it's not equal to five, then the mean value theorem would not apply. And so let's do that. f of six minus f of four, all of that over six minus four is equal to seven minus three over two, which is equal to two. So two not equal to five. So mean value theorem doesn't apply. All right let's, I'll put an exclamation mark there for emphasis. All right let's do the next part. Can we use the mean value theorem to say that the equation f prime of x is equal to negative one has a solution? And now the interval's from zero to two. If so, write a justification. All right so let's see this. So if we were to take the slope of the secant line, so f of two minus f of zero, all that over two minus zero. This is equal to negative two minus zero, all of that over two, which is equal to negative two over two, which is equal to negative one. And so and we also know that we meet the continuity and differentiability conditions, and so we could say and since f is generally differentiable, generally differentiable, differentiable, it will be differentiable, differentiable, and, and continuous over the interval from zero to two, and I'll say the closed interval. You just have to be differentiable over the open interval, but it's even better I guess if you're differentiable over the closed interval because you have to be continuous over the closed interval. And so, and since f is generally differentiable, it will be differentiable and continuous over zero two. So the mean value theorem tells us, tells us, that there is an x in that interval from zero to two such that f prime of x is equal to that secant slope, or you could say that average rate of change, is equal to negative one. And so I could write, yes, yes, and then this would be my justification. This is the slope of the secant line or the average rate of change, and since f is generally differentiable, it will be differentiable and continuous over the closed interval. So the mean value theorem tells us that there is an x in this interval such that f prime of x is equal to negative one. And we're done.