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

The table gives selected values of the continuous function f. All right, fair enough. Can we use the Intermediate Value Theorem to say that the equation f of x is equal to 0 has a solution where 4 is less than or equal to x is less than or equal to 6? If so, write a justification.

So pause this video and see if you can think about this on your own before we do it together.

Okay, well, let's just visualize what's going on and visually think about the Intermediate Value Theorem. So if that's my y-axis there, and then let's say that this is my x-axis right over here. We've been given some points over here. We know when x is equal to zero, f of x is equal to zero. Let me draw those.

So we have that point when x is equal to two, y or f of x, y equals f of x is going to be equal to negative two, so we have a negative two right over there. When x is equal to four, f of x is equal to 3. So 1, 2, 3. I'm doing them on a slightly different scale so that I can show everything. And when x is equal to 6, f of x is equal to 7. So 3, 4, 5, 6, 7.

So right over here. Now they also tell us that our function is continuous. So one intuitive way of thinking about continuity is I can connect all of these dots without lifting my pencil. So the function might look, I'm just going to make up some stuff, it might look something anything like what I just drew just now. It could have even wilder fluctuations, but that is what my f looks like.

Now the Intermediate Value Theorem says, "Hey, pick a closed interval," and here we're picking the closed interval from 4 to 6. So let me look at that. So this is 1, 2, 3, 4. Here, this is 6 here, so we're going to look at this closed interval. The Intermediate Value Theorem tells us that, look, if we're continuous over that closed interval, our function f is going to take on every value between f of 4, which in this case is equal to 3, and f of 6, which is equal to 7.

So someone said, "Hey, is there going to be a solution to f of x is equal to say 5 over this interval?" Yes, over this interval for some x, you're going to have f of x being equal to 5. But they're not asking us for an f of x equaling something between these two values; they're asking us for an f of x equaling 0.

0 isn't between f of 4 and f of 6. And so we cannot use the Intermediate Value Theorem here. If we wanted to write it out, we could say, "f is continuous, but 0 is not between f of 4 and f of 6," so the Intermediate Value Theorem does not apply.

All right, let's do the second one. So here they say, "Can we use the Intermediate Value Theorem to say that there is a value c such that f of c equals zero and 2 is less than or equal to c is less than or equal to 4?" If so, write a justification.

So we are given that f is continuous, so let me write that down. We are given that f is continuous, and if you want to be over that interval, but they're telling us it's continuous in general. Then we can just look at what is the value of the function at these endpoints.

So our interval goes from 2 to 4. So we're talking about this closed interval right over here. We know that f of 2 is going to be equal to negative 2. We see it in that table, and what's f of 4? f of 4 is equal to 3.

So 0 is between f of 2 and f of 4. You can see it visually here; there's no way to draw between this point and that point without picking up your pen, without crossing the x-axis, without having a point where your function is equal to zero.

And so we can say, "So according to the Intermediate Value Theorem, there is a value c such that f of c is equal to zero and 2 is less than or equal to c is less than or equal to 4." So all we're saying is, "Hey, there must be a value c," and the way I drew it here, that c value is right over here where c is between 2 and 4, where f of c is equal to 0.

This seems all mathy and a little bit confusing sometimes, but it's saying something very intuitive. If I had to go from this point to that point without picking up my pen, I am going to at least cross every value between f of 2 and f of 4 at least once.