### Video Transcript

The graph of the first derivative π prime of a continuous function π is shown. State the π₯-coordinates of the inflection points of π.

Weβre given a graph of the first derivative π prime of a function π, which weβre told is continuous. We need to use this graph to determine the π₯-coordinates of all of the inflection points for our function π. Letβs start by recalling what we mean by the inflection points for our function π. We say that a point π is an inflection point for the function π if π is continuous at the point π and π changes concavity at π. So, there are two parts to checking a point π is an inflection point of π. We need to check that π is continuous at π, and we need to check that it changes concavity at π.

We might be worried at the point where π₯ is equal to six. If we look at our graph, it shows that π prime is not continuous when π₯ is equal to six. In fact, π prime is not even defined when π₯ is equal to six. However, remember, weβre told in the question that our function π is a continuous function. This means itβs continuous for all values of π₯ in its domain. So, if π is a continuous function, this means the only thing we need to check to find all the inflection points is the points where π changes concavity.

In fact, weβre not actually told if our function π is defined when π₯ is equal to six. Weβre not told enough information to determine one way or the other. Weβll assume π₯ is equal to six is in the domain of our function π of π₯ since weβre not told otherwise. So, to find all the inflection points of our function π, we now need to determine all the points where π changes concavity. So, letβs recall what we mean by the concavity of the curve π¦ is equal to π of π₯.

We say that this is concave upward on an interval if on this interval all of our tangent lines lie below the curve. And we also need to recall another way of saying this is saying that the slope of our tangent lines are increasing on this interval. But weβre not given a graph of π¦ is equal to π of π₯; weβre given a graph π¦ is equal to π prime of π₯. So, we need to think what does this mean for the curve π¦ is equal to π prime of π₯.

To do this, we need to recall what we mean by π prime of π₯. π prime of π₯ is equal to the slope of the tangent line at π₯. So, if our curve is concave upward on an interval means that the slope of the tangent lines are increasing and π prime of π₯ tells us the slope of our tangent line, that means that on these intervals, π prime of π₯ must be increasing. So, to find the intervals where π¦ is equal to π of π₯ is concave upward, we just need to check where π prime of π₯ is increasing; in other words, its curve is moving upward.

And in fact, we can do exactly the same to find the intervals where our curve is concave downward. We recall we say that a curve is concave downward on an interval if on this interval its tangent lines lie above the curve. And we also recall this is the same as saying the slope of the tangent lines is decreasing. And of course, if the slope of our tangent lines is decreasing on this interval, then this means π prime of π₯ must also be decreasing on this interval.

So, we can now determine the concavity of the curve π¦ is equal to π of π₯ by instead looking at the curve π¦ is equal to π prime of π₯. And remember, our points of inflection will be the points where our curve changes concavity, in other words, where it changes from concave upward to concave downward, or vice versa. So, letβs now look at the curve π¦ is equal to π prime of π₯ to determine where itβs increasing and where itβs decreasing. Letβs start with π₯ is equal to zero.

We can see that our curve is moving downward. In other words, itβs decreasing. And in fact, it decreases all the way down to π₯ is equal to three. And then, we can see when π₯ is greater than three, π prime of π₯ is increasing; its curve is moving upward. So, when π₯ is equal to three, π prime of π₯ moves from being a decreasing function to an increasing function. And we know this means that π¦ is equal to π of π₯ has switched from being a concave upward to a concave downward function. And this means there must be a point of inflection when π₯ is equal to three.

And we can carry on in exactly the same way. We can see that π¦ is equal to π prime of π₯ is still increasing; it increases all the way up to π₯ is equal to six. But now, we need to wonder what happens when π₯ is equal to six. If we look to the right of π₯ is equal to six, we can see that our function π prime of π₯ is still increasing; the curve is moving upward. So both to the left of π₯ is equal to six and to the right of π₯ is equal to six, π prime of π₯ is increasing. So, π prime of π₯ did not switch from increasing to decreasing or vice versa.

This means to the left of π₯ is equal to six, our curve is concave upward. And to the right of π₯ is equal to six, our curve is also concave upward. So, there cannot possibly be a point of inflection at π₯ is equal to six. The concavity of our curve did not change. And if we continue doing this, we can see for all of the rest of the values of π₯, all the way up to π₯ is equal to 10, our function π prime is increasing. And this means weβve checked all of our values of π₯ and we only found one inflection point when π₯ was equal to three.

Therefore, by looking at intervals where the graph of the first derivative π prime of a continuous function π is increasing or decreasing, we were able to find all of the π₯-coordinates of the inflection points for our function π. We only found one inflection point when π₯ was equal to three.