Sign of average rate of change of polynomial | Algebra 2 | Khan Academy

So we are given this function h of x, and we're asked over which interval does h have a positive average rate of change. So, like always, pause this video and have a go at it before we do this together.

All right, now let's work through this together. To start, let's just remind ourselves what an average rate of change even is. You can view it as the change in your value of the function for a given change in the underlying variable. For a given change in x, we could also view this as, if we want to figure out the interval, we could say our x final minus our x initial. In the numerator, it would be the value of our function at the x final minus the value of the function at our x initial.

Now, they aren't asking us to calculate this for all of these different intervals. They're just asking us whether it is positive. If you look over here, as long as our x final is greater than x initial, in order to have a positive average rate of change, we just need to figure out whether h at x final is greater than h at x initial. If the value of the function at the higher endpoint is larger than the value of the function at the lower endpoint, then we have a positive average rate of change.

So let's see if that's happening for any of these choices. So let's see: h of 0, this endpoint is going to be equal to 0. If I just say one eighth times zero minus zero, and h of two is equal to one eighth times two to the third power, which is eight. So one eighth times eight is one minus four, so that's going to be, this is negative three.

And so we don't have a situation where h at our endpoint at our higher endpoint is actually larger. This is a negative average rate of change, so I'll rule this one out. And actually, just to help us visualize this, I did go to Desmos and graph this function, and we can visually see that we have a negative average rate of change from x equals 0 to x equals 2.

At x equals 0, this is where our function is, and at x equals 2, this is where our function is. You can see that at x equals 2, our function has a lower value. You could also think of the average rate of change as the slope of the line that connects the two endpoints on the function, and so you can see it has a negative slope. So we have a negative average rate of change between those two points.

Now, what about between these two? So h of 0 we already calculated to be 0, and what is h of 8? Well, let's see, that's one eighth times 8 to the third power. Well, if I do 8 to the third power but then divide by 8, that's the same thing as 8 to the second power, so that's going to be 64 minus 8 to the second power minus 64. So that's equal to 0.

So here we have a 0 average rate of change because this numerator is going to be 0, so we can rule that out. And you see it right over here: when x is equal to 0, our function's there; when x is equal to 8, our function is there. You can see that the slope of the line that connects those two points is zero, so you have zero average rate of change between those two points.

Now, what about choice C? So let's see, h of 6 is going to be equal to one eighth times 6 to the third power. So let's see, 36 times 6 is 216 plus 36, so that is going to be 216 minus 36. 216 is 6 times 6 times 6, and then if we divide that by 8, that is going to be the same thing as, this is 27.

Then we have 6 eighths of 36, so this is going to simplify to 3 fourths times 36 minus 36, which is going to be equal to negative 9. You could have done it with a calculator or done some long division, but hopefully what I just did makes some sense; it's a little bit of arithmetic.

At h of 6, we have our function as negative 9, and then h of 8. I'll draw a line here so we don't make it too messy. H of 8 we already know is equal to 0, so our function at this endpoint is higher than the value of our function at this endpoint. So we do have a positive average rate of change, so I would pick that choice right over there.

You could see it visually. H of 6, when x is equal to 6, our value of our function is negative 9, and when x is equal to 8, our value of our function is 0. So the line that connects those two points definitely has a positive slope, so we have a positive average rate of change over that interval.

Now, if we were just doing this on our own, we'd be done, but we could just check this one right over here. If we compare h of 0, we already know is 0, and h of 6 we already know is equal to negative 9. So this is a negative average rate of change because at the higher endpoint right over here, we have a lower value of our function, so we'd rule this out.

You see it right over here: if you go from x equals 0, where the function is, to x equals 6, where the function is, it looks something like that. Clearly, that line has a negative slope, so we have a negative average rate of change.