### Video Transcript

The graph of the derivative π prime of a function π is shown. On what intervals is π increasing or decreasing?

Weβre given a graph of the curve of the derivative π prime of a function π. We need to use this to determine the intervals on which the function π will be increasing or decreasing. Letβs start by recalling what it means for a function π to be increasing or decreasing on an interval. We say that a function π of π₯ is increasing on an interval if the following is true: If we take any two points in this interval, then the point with higher π₯-coordinate must have the higher output. In other words, increasing the value of π₯ on our interval will increase the output of our function. And we define decreasing in the same way. We say that a function π of π₯ is decreasing on an interval if we take any two points on our interval. Then the point with higher π₯-coordinate must have a lower output.

In other words, on this interval, taking a higher input value of π₯ will decrease our output. And itβs often easier to see this graphically. When the curve π¦ is equal to π of π₯ is moving upward, that means our outputs are getting larger. So our function is increasing. And when the curve π¦ is equal to π of π₯ is moving downward, our outputs are getting smaller. So our function is decreasing. And itβs also worth pointing out if our function stays constant, we donβt say itβs increasing or itβs decreasing. In fact, itβs doing neither. But this only helps us if weβre given the curve π¦ is equal to π of π₯. In this case, weβre only given the graph of the function π¦ is equal to π prime of π₯.

This means to answer this question, we need to think what happens to the curve π¦ is equal to π prime of π₯ when π of π₯ is increasing and what happened to the curve π¦ is equal to π prime of π₯ when π of π₯ is decreasing. To do this, weβre going to need to remember what we mean by the derivative function π prime of a function π. We need to recall that π prime measures the slope of our tangent lines to the curve π¦ is equal to π of π₯. So to find out what happens to the curve π¦ is equal to π prime of π₯ when π of π₯ is increasing or decreasing, we need to find out what happens to the slope of our tangent lines.

Letβs start with an interval where π¦ is equal to π of π₯ is increasing. Graphically, it makes sense that on this interval our slopes will all be positive. After all, our outputs are getting larger. And in fact, this is true, and we can even prove this directly from the definition of a derivative. First, letβs consider a point π₯ zero where π₯ zero is in our interval where π of π₯ is increasing. Then from the definition of a derivative, π prime of π₯ zero is equal to the limit as β approaches zero of π of π₯ zero plus β minus π of π₯ zero all divided by β.

We want to show that this limit is positive. The next thing weβll do is since β approaching zero, weβll choose β small enough so that π₯ zero plus β is also in our interval. Now, there are two options. Either β is positive or β is negative. First, if β is positive, then π of π₯ zero plus β must be greater than π of π₯ zero. This is because we chose β small enough so that π₯ zero plus β is also in our interval where π is increasing. And π₯ zero plus β will be bigger than π₯ zero. And we get a similar inequality of β is less than zero. This time, π of π₯ zero will be greater than π of π₯ zero plus β.

Again, this is because π is an increasing function on this interval. And because β is less than zero, π₯ zero plus β will be less than π₯ zero. We can then see what this does to our limit. Letβs consider the case where β is greater than zero. In our numerator inside of our limit, weβre now subtracting a bigger number from a smaller number. So our numerator is positive. And in this case, β is positive. So weβre dividing two positive numbers. And we know the quotient of two positive numbers is positive. And we get something very similar in the case where β is less than zero.

This time, weβre subtracting a bigger number from a smaller one. So our numerator will be negative. But remember, β is less than zero. So our denominator is also negative. And the quotient of two negative numbers is positive. This only proves that our limit will be greater than or equal to zero because a sequence of positive numbers can in fact approach zero. This is why we sometimes like to include points where the slope is equal to zero in our intervals. However, in this video, we wonβt be including these.

We can, in fact, make a very similar argument on intervals where our function is decreasing. On these intervals, our function will be going downwards. This means that our slopes will all be negative. We now have enough information to start determining the intervals where our function π is increasing or decreasing just from the curve π¦ is equal to π prime of π₯. We know that π prime of π₯ will be increasing on intervals where π prime of π₯ is positive. This means the curve π¦ is equal to π prime of π₯ will be above the π₯-axis. And similarly, on intervals where π of π₯ is decreasing, π prime of π₯ will be negative. This means our curve π¦ is equal to π prime of π₯ will be below the π₯-axis.

Weβre now ready to answer this question. However, thereβs one more thing we need to talk about. We want to talk about the endpoints of this curve when π₯ is equal to zero and when π₯ is equal to eight. We can see from the curve π prime evaluated at zero is equal to two, and π prime evaluated at eight is approximately equal to negative 1.5. This is represented by the filled-in circles in our diagram. So if we were to look at our curve from π₯ is equal to zero to π₯ is equal to one, we can see that the curve π¦ is equal to π prime of π₯ is above the π₯-axis. In other words, π prime of π₯ is greater than zero for all values of π₯ greater than or equal to zero and less than one.

And weβve already discussed why weβre not including the endpoint when π₯ is equal to one. In fact, itβs just mathematical convention. We could include it if we wanted to. And we could leave our answer like this. However, thereβs another piece of mathematical convention. A lot of people donβt like to include the endpoints of these intervals. And both of these answers are correct. Itβs all personal preference among mathematicians. In this video, weβre going to leave that out and leave this is as the open interval from zero to one. So, so far, weβve shown that π of π₯ is increasing on the open interval from zero to one. Letβs carry on finding more intervals where π of π₯ is increasing or decreasing.

Next, by looking at the curve π¦ is equal to π prime of π₯, we can see that this curve is entirely below the π₯-axis for all values of π₯ between one and five. So because π prime of π₯ is below the π₯-axis on the open interval from one to five, our function π of π₯ is decreasing on the open interval from one to five. Next, by looking at our curve, we can see that π prime of π₯ is positive for all values of π₯ on the open interval from five to seven. This means π of π₯ is increasing on the open interval from five to seven. And finally, we can see the curve π¦ is equal to π prime of π₯ is less than zero on the open interval from seven to eight. And weβre not including the endpoint of this curve by convention.

So weβve shown on the open interval from seven to eight, the slopes of our tangent lines will be negative. And this means that our function π of π₯ will be decreasing on this interval. Now we found all the intervals where our function π is increasing and decreasing. We can combine this into our final answer. Therefore, by looking at the intervals where our function π prime of π₯ is positive or negative, we were able to determine the intervals where the function π was increasing or decreasing. We were able to show that π will be increasing on the open interval from zero to one and the open interval from five to seven and decreasing on the open interval from one to five and the open interval from seven to eight.