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## College Algebra

### Unit 10: Lesson 1

Average rate of change of polynomials

# Sign of average rate of change of polynomials

Finding the intervals where the polynomial h(x)=⅛x³-x² has a positive average rate of change.

## Video transcript

- [Instructor] 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. And 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 wanna figure out the interval, we could say our x final minus our x initial, and 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. And 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 zero, this endpoint, is going to be equal to zero. If I just say 1/8 times zero minus zero, and h of two is equal to 1/8 times two to the third power is eight. So 1/8 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 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 zero to x equals two. At x equals zero, this is where our function is, at x equals two, this is where our function is, and so you can see, at x equals two, 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 zero, we already calculated as zero, and what is h of eight? Well, let's see that's 1/8 times eight to the third power. Well, if I do eight to third power, but then divide by eight, that's the same thing as eight to the second power. So that's going to be 64 minus eight to the second power minus 64, so that's equal to zero. So here we have a zero average rate of change 'cause this numerator's going to be zero, so we can rule that out. And you see it right over here. When x is equal to zero, our function is there, when x is equal to eight, our function is there, and 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 six is going to be equal to 1/8 times six to the third power. So let's see, 36 times six is 180 plus 36, so that is going to be 216. 216 minus 36, 216 is six times six times six, and then if we divide that by eight, that is going to be the same thing as, this is 36, and then we have 6/8 of 36. So this is going to simplify to 3/4 times 36 minus 36, which is going to be equal to negative nine. 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. And so, at h of six, we have our function is negative nine, and then h of eight, I'll draw a line here so we don't make it too messy, h of eight we already know is equal to zero. 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. And you could see it visually. h of six, when x is equal to six, our value of our function is negative nine. And when x is equal to eight, our value of our function is zero. And 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 zero, we already know is zero, and h of six, we already know is equal to negative nine. 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. And you see it right over here, if you go from x equals zero where the function is to x equals six where the function is, it looks something, it looks something like that. Clearly that line has a negative slope. So we have a negative average rate of change.