 Previously in this lecture 32, we talked about the mean value theorem and we talked about the assumptions of the mean value theorem that the function must be continuous on the closed interval a to b, it must be differentiable on the open interval a to b. But why do these assumptions matter? I want to show you two functions which would violate the mean value theorem with the exception of the assumptions weren't satisfied so it's not really a violation. Let's consider the function f of x equals the absolute value of x and so this is just our standard absolute value function for which the graph looks like a v, something like this. It should be symmetric with respect to the y-axis but this is close enough of a drawing here. So consider its derivative, mind you, it's going to be the function x over the absolute value of x which equals 1 when x is positive and equals negative 1 when x is negative. It is undefined at zero because you have that sharp point, the derivative is undefined at that moment. But basically this line over here has a positive one slope, this line over here has a negative one slope, that gives us the derivative. So let's consider the interval, let's consider the interval negative one to positive one right here. So if you take that interval, so let's think about our function right here, you have the point one comma one on our function. Over here you have the point negative one comma one, because if you take the absolute value of negative one you're going to get a positive one. So f of one minus f of negative one, that's going to be one minus one which is zero. In the denominator you're going to get one minus negative one which is actually one plus one which is two, zero over two is equal to zero. So this right here is our average rate of change delta y over delta x on the interval negative one to one, like so. So the average rate of change is equal to zero. But the derivative is never equal to zero right, because the derivative is only equal to one and negative one, those only two options you get. So the mean value theorem doesn't apply to this example, what's wrong with it? Well if you go from negative one to one, the function is continuous on the interval. The problem of course is right here at the vertex of the absolute value function. The function is not differentiable at that point, so the lack of differentiability then allows for the derivative to disagree with the average rate of change. So the mean value theorem doesn't apply and guarantee a point where it has a horizontal tangent line because it's not differentiable. So it's not differentiable on the interval negative one to one. So differentiability on the interval is significant, but what about the continuity, right? Because differentiability applies continuity, so if you want it to be continuous from negative one to one, it's really just the endpoints that matter. So why does continuity matter? Well let's consider the following example here. Let's take the piecewise function f of x equals zero when x doesn't equal zero and one when x equals zero. So this is sort of a curious little function. Basically what you get here is your function, your function is just the x-axis. At x equals zero, take that point away. At x equals zero, you have this remote point, and at x equals zero, it's actually equal to one. So you get y equals one when x equals zero. Like so, or maybe we'll just label it like this. You get zero comma one. That's our function graph right here. It's zero everywhere else. Well let's consider the interval zero to one in that situation. So the average area change, which we can see right here, the delta y over delta x, as x goes from zero to one. You're going to take f of one minus f of zero over one minus zero. One minus zero is easy enough. That's just a one. If you take f of one minus f of zero, well f of one is going to be zero because one's not zero. I know it. It seems weird. You flip it around, but yeah, that's how our function is defined. It's one when x is zero and zero when x is not zero. So f of one, that's equal to zero, and f of zero is equal to one, zero minus one is negative one. So we see that the average area change is going to equal a negative one. Okay. But is the derivative equal to negative one? Well our function, if we go, if we're considering the interval, right, zero to say one, right, we go from zero to one. That if we look from zero, if we go from zero to one, open interval, we don't look at the end points. Then f here is, it is in fact differentiable on this interval. We're talking about the open interval here, zero to one. It is differentiable because if we're not looking at the remove point, right, then in this situation it's just a flat line. It is differentiable. In fact, f prime of x here is going to equal zero so long x is less than one, but greater than zero, right. The derivative is going to equal zero, but we're guaranteed that somewhere the derivative should equal negative one. It's not going to happen. The derivative of our function is going to equal zero except for f prime of zero which is going to be undefined because the function was discontinuous at that value. So we see that even if the function is differentiable between the end points, if there's a discontinuity that lives on an endpoint, we can still frustrate the mean value theorem. It's not that the mean value theorem is false. The mean value theorem doesn't apply because the assumptions are not satisfied. That's why annuity or differentiability because if either assumption is lost, then there's no guarantee point C coming from the mean value theorem. So we need to make sure whenever we use the mean value theorem that the assumptions of the mean value theorem are checked because otherwise we can be making nonsensical statements. We don't.