Justification with the mean value theorem: table | AP Calculus AB | Khan Academy

The table gives selected values of the differentiable function f. All right, can we use a mean value theorem to say that there is a value c such that f prime of c is equal to 5 and c is between 4 and 6? If so, write a justification.

Well, 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, that if that condition is met, then the slope of the secant line between (4, f(4)) and (6, f(6)) suggests that at least one point in between 4 and 6 will have a derivative that is equal to the slope of the secant line. So, let's figure out what the slope of the secant line is between (4, f(4)) and (6, f(6)). If it's equal to 5, then we could use the mean value theorem. If it's not equal to 5, then the mean value theorem would not apply.

And so, let's do that: f(6) minus f(4), all of that over (6 - 4) is equal to (7 - 3) over 2, which is equal to 2. So, 2 is not equal to 5. So, the mean value theorem doesn't apply. All right, let's put an exclamation point 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 is from 0 to 2? If so, write a justification.

All right, so let's see this. If we were to take the slope of the secant line: f(2) minus f(0), all that over (2 - 0) is equal to (-2 - 0) all of that over 2, which is equal to (-2 over 2), which is equal to -1. And so, we also know that we meet the continuity and differentiability conditions.

And so, we could say that since f is generally differentiable, it will be differentiable and continuous over the interval from 0 to 2. To 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 since f is generally differentiable, it will be differentiable and continuous over (0, 2). So, the mean value theorem tells us that there is an x in that interval from 0 to 2 such that f prime of x is equal to that secant slope, or you could say that the average rate of change is equal to -1.

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. 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 -1. And we're done.