Worked example: finding relative extrema | AP Calculus AB | Khan Academy

So we have G of X being equal to X to the fourth minus X to the fifth. What we want to do, without having to graph G, is figure out what X values G has a relative maximum.

Just to remind us what's going on in a relative maximum, let me draw a hypothetical function right over here. A relative maximum is going to happen, so you can visually inspect this; okay, that looks like a relative maximum. It's kind of the top of a mountain or top of a hill. These all look like relative maxima.

What's in common? The graph, the function, is going from increasing to decreasing at each of those points. It's going from increasing to decreasing, increasing to decreasing, at either of the points. You could say that the first derivative is going from positive to negative. If you look at this interval right over here, G prime is greater than zero, and then over the next interval, when you're decreasing, G prime would be less than zero.

So what we really need to think about is when does G prime, so let me see—relative; we care about the relative maximum point. That's essentially asking when does G prime go from positive to negative, from G prime greater than zero to G prime less than zero. The values that we could look at are the points, are our critical points.

Critical points are where G prime is either zero or it is undefined. So let's think about it: where is G prime of X equal to zero? G prime of X is equal to zero when we take G prime of X using the power rule:

G prime of X equals 4X to the third minus 5X to the fourth.

Setting this equal to zero, we can factor out an X to the third, right here. So we have X to the third times (4 - 5X) equal to zero. This is going to happen when X is equal to zero.

Let me not skip steps, so this is going to happen when X to the third is equal to zero or 4 - 5X is equal to zero. For X to the third equaling zero, well, that's only going to happen when X is equal to zero. For four minus five X equaling zero, we will add 5X to both sides: you get four is equal to 5X. Divide both sides by 5, you get 4/5 is equal to X.

So here these are the two places where our derivative is equal to zero. Now, are there any places where our derivative is undefined? Well, our function right over here is just a straight-up polynomial, and our derivative is another polynomial; it is defined for all real numbers.

So these are our two critical values. Now let's think about what G prime is doing on either side of these critical values, and I'll draw a little number line here to help us visualize this. There we go, a little bit of a number line.

Let's see; we care about zero, and we care about 4/5. So let's say this is negative 1, this is 0, this is 1. We have one critical point; let me do this in magenta. We have one critical point here at x equals zero, and then we have another critical point; I will do this at x equals 4/5. So 4/5 is right around there.

Now, let's think about what G prime is doing in these intervals. These critical points are the only places where G prime might switch signs. Let's first think about this: let me pick some colors I haven't used yet.

So let's think about the interval from negative infinity to zero. This is the open interval from negative infinity to zero. We could just plug in a value; let's try negative 1. Negative 1 is pretty straightforward to evaluate.

So let's see: G prime of negative 1 is equal to 4 times negative 1 to the third power minus 5 times negative 1 to the fourth power. So that's going to be 4 times negative 1 minus 5 times 1.

So, let's see—this is going to be negative 4 minus 5, which is negative 9. So right over here, G prime is equal to negative 9. We know over this whole interval, since it's to the left of this critical point, G prime of X is less than 0.

Therefore, our function itself is decreasing over this interval. We know we need to go from increasing to decreasing, so you could already say, "Well, we can't go from increasing to decreasing at this critical point because we're already decreasing to the left of it."

But anyway, let's think about what's happening in the other intervals. In the interval between 0 and 4/5—so that interval right over there—is between 0 and 4/5. Well, let's just sample a number there; let's say 1/2 might be brilliant forward.

So we can evaluate G prime of 1/2: G prime of 1/2 is equal to 4 times (1/2) to the third power. (1/2) to the third power is 1/8, so it's 4/8, or just 1/2, minus 5 times (1/2) to the fourth, which is 5/16.

So this is equal to 8/16 minus 5/16, which is equal to 3/16. But the important thing is it's equal to a positive value.

So in this light blue interval right here, between 0 and 4/5, G prime of X is greater than zero, so we know our function is increasing.

Now, let's see what's happening to the right of this, and the easiest value to try out would just be 1. So let's try out x equals 1; it's in that interval.

So when x equals 1, I'll just write G prime of 1 is equal to 4 minus 5, which is equal to negative 1. So G prime of X is less than 0. G prime of X is less than 0, which means our function is decreasing here.

So we could say G is decreasing here; the function itself is decreasing because our derivative is negative. Then our function is increasing here because our derivative is positive, and then our function is decreasing here.

So at what critical point are we going from increasing to decreasing? We're doing that at x equals 4/5.

So we have a relative maximum at x equals 4/5. If they said, "Well, where do we have a relative minimum point?" Well, that's going to happen at x equals 0; we're going from decreasing to increasing.

But we've answered their question: where do we find a relative maximum point?