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# Finding relative extrema (first derivative test)

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.2 (EK)

## Video transcript

In the last video we saw that if a function takes on a minimum or maximum value, min max value for our function at x equals a, then a is a critical point. But then we saw that the other way around isn't necessarily true. x equal a being a critical point does not necessarily mean that the function takes on a minimum or maximum value at that point. So what we're going to try to do this video is try to come up with some criteria, especially involving the derivative of the function around x equals a, to figure out if it is a minimum or a maximum point. So let's look at what we saw in the last video. We saw that this point right over here is where the function takes on a maximum value. So this critical point in particular was x naught. What made it a critical point was that the derivative is 0. You have a critical point where either the derivative is 0 or the derivative is undefined. So this is a critical point. And let's explore what the derivative is doing as we approach that point. So in order for this to be a maximum point, the function is increasing as we approach it. The function is increasing is another way of saying that the slope is positive. The slope is changing but it stays positive the whole time which means that the function is increasing. And the slope being positive is another way of saying that the derivative is greater than 0 as we approach that point. Now what happens as we pass that point? Right at that point the slope is 0. But then as we past that point, what has to happen in order for that to be a maximum point? Well the value of the function has to go down. If the value of the function is going down, that means the slope is negative. And that's so another way of saying that the derivative is negative. So that seems like a pretty good criteria for identifying whether a critical point is a maximum point. So let's say that we have critical point a. We are at a maximum point if f prime of x switches signs from positive to negative as we cross x equals a. That's exactly what happened right over here. Let's make sure it happened at our other maximum point right over here. So right over here, as we approach that point the function is increasing. The function increasing means that the slope is positive. It's a different positive slope. The slope is changing, it's actually getting more, and more, and more steep. Or more, and more, and more positive. But is definitely positive. So it's positive going into that point. And then it becomes negative after we cross that point. The slope was undefined right at the point. But it did switch signs from positive to negative as we crossed that critical point. So these both meet our criteria for being a maximum point. So, so far our criteria seems pretty good. Now let's make sure that somehow this point right over here, which we identified in the last video as a critical point, let's make-- and I think we called this x0, this was x1, this was x2. So this is x3. Let's make sure that this doesn't somehow meet the criteria because we see visually that this is not a maximum point. So as we approach this, our slope is negative. And then as we cross it, our slope is still negative, we're still decreasing. So we haven't switched signs. So this does not meet our criteria, which is good. Now let's come up with the criteria for a minimum point. And I think you could see where this is likely to go. Well we identified in the last video that this right over here is a minimum point. We can see that. It's a local minimum just by looking at it. And what's the slope of doing as we approach it? So the function is decreasing, the slope is negative as we approach it. f prime of x is less than 0 as we approach that point. And then right after we cross it-- this wouldn't be a minimum point if the function were to keep decreasing somehow. The function needs to increase now. So let me do that same green. So right after that, the function starts increasing again. f prime of x is greater than 0. So this seems like pretty good criteria for a minimum point. f prime of x switches signs from negative to positive as we cross a. If we have some critical point a, the function takes on a minimum value at a if the derivative of our function switches signs from negative to positive as we cross a, from negative to positive. Now once again, this point right over here, this critical point x sub 3 does not meet that criteria. We go from negative to 0 right at that point then go to negative again. So this is not a minimum or a maximum point.
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