### Video Transcript

The graph of our function π¦ equals π of π₯ is shown. At which point are dπ¦ by dπ₯ and d squared π¦ by dπ₯ squared both positive?

Looking at the graph, we see the we have five points to choose from points π΄, π΅, πΆ, π·, and πΈ. And we want to know at which one of these points dπ¦ by dπ₯ and d squared π¦ by dπ₯ squared are both positive. Letβs start by considering dπ¦ by dπ₯. We need this to be positive. Is it positive at point π΄? If we sketch or even just imagine the tangents to the graph at the point π΄, we can see that its slope is negative. And the value of this slope is dπ¦ by dπ₯ at π΄. So we see that dπ¦ by dπ₯ is negative at π΄. dπ¦ by dπ₯ must be positive at the point weβre looking for. So we can eliminate π΄. Thatβs not our answer.

How about at point π΅? Well here, the value of dπ¦ by dπ₯, the slope of the tangent to the graph, at π΅ is positive. We can see this somehow even without sketching in the tangent at π΅. The function is increasing in a suitably small interval around π΅. And so the instantaneous rate of change at π΅, dπ¦ by dπ₯ at π΅, is positive. With the tangent shown, itβs clear that this slope is positive. But of course in order to sketch the tangent, we had to know this. As dπ¦ by dπ₯ is positive at π΅, it could be our answer.

Letβs look at πΆ now. Much like at π΅, we can see that the dπ¦ by dπ₯ is positive at πΆ. As the tangent at πΆ has a positive slope, πΆ is also a possibility for our answer then. π· lies at the local maximum where the tangent is horizontal, and hence dπ¦ by dπ₯ is zero at π·. Zero is not a positive number, and so dπ¦ by dπ₯ is not positive at π· and we can eliminate π·. This is not our answer. And finally at πΈ, much like at π΄, the derivative dπ¦ by dπ₯ is negative. As the tangent at πΈ has negative slope, πΈ therefore is also not our answer.

So weβve eliminated three options our answer is either π΅ or πΆ. And where weβre going to choose between options π΅ and πΆ is by considering the second derivative, d squared π¦ by dπ₯ squared. Remember that we require this to be positive too. Letβs clear our diagram so we have some space to work. We could tell where dπ¦ by dπ₯ was positive or negative or indeed zero just by looking at the graph. This is because we know that dπ¦ by dπ₯ is the slope function, and weβre good at seeing where the slope is positive negative or zero. Can we do something similar for d squared π¦ by dπ₯ squared? Can we link it to some concept that we can recognize on a graph just in the same way that we linked dπ¦ by dπ₯ to the concept of slope?

The answer is yes. d squared π¦ by dπ₯ squared is to concavity as dπ¦ by dπ₯ is to slope. Understanding what concavity looks like will help us determine where d squared π¦ by dπ₯ squared is positive and where it is negative in the same way that understanding what slope looked like helped us to determine where dπ¦ by dπ₯ was positive and negative. By definition, a function π is concave upward on an interval πΌ if its derivative π prime is an increasing function on πΌ. And π is concave downwards on an interval πΌ if its derivative π prime is a decreasing function on that interval πΌ. How does this help us determine where d squared π¦ by dπ₯ squared is positive or negative?

Well instead we look for where the function is concave upward or concave downward. Imagine a point on the graph of π to the left of the point π΅. What happens to the derivative π prime as this point as the point moves through π΅ to the right of π΅? Well π prime or dπ¦ by dπ₯ is the slope at this point or more precisely the slope of the tangent to this point. Can you see that the slope has increased as we move from the orange point to the left of π΅ to the purple point to the right?

The slope at the orange point is more gradual than the slope at the purple point which is steeper. The slope function π prime has therefore increased on this interval. And so itβs natural to believe that the slope π prime is an increasing function on this small interval. π prime is an increasing function and so π is concave upward. Remember that we said that d squared π¦ by dπ₯ squared is to concavity as dπ¦ by dπ₯ is to slope. Letβs make that slightly more precise. If d squared π¦ by dπ₯ squared is positive on an interval πΌ, then π¦ equals π of π₯ is concave upward on πΌ. And if d squared π¦ by dπ₯ squared is negative on an interval πΌ, then π¦ equals π of π₯ is concave downward on πΌ.

Weβve already seen that our function is concave upward at π΅. And this strongly suggests that d squared π¦ by dπ₯ squared is positive at π΅. But we have to be slightly careful here. While itβs true that if d squared π¦ by dπ₯ squared is positive and the function is concave upward, the converse is not necessarily true. It doesnβt follow that if the function is concave upward that d squared π¦ by dπ₯ squared has to be positive. We have to see whatβs happening at πΆ therefore. Imagine a point on the graph of the function just to the left of πΆ, moving along the graph through πΆ to the right of πΆ.

Is π prime increasing or decreasing on the interval that the point covers? Itβs not too hard to see that the derivative π prime is decreasing on this interval from a maximum at the orange point to a minimum at the purple point. And as π prime is a decreasing function on this interval, π is concave downward on this interval. Now if d squared π¦ by dπ₯ squared is positive as it is at the point that weβre looking for, then our function should be concave upward. As the function is concave downward at πΆ, we can eliminate πΆ. πΆ is not our answer. Our answer is therefore the only point remaining, the point π΅, where the function is concave upward. This is the point at which both dπ¦ by dπ₯ and d squared π¦ by dπ₯ squared are positive.

Notice that the tangents near π΅ where the function is concave upward lie below the graph of the function and the tangents near πΆ where the function is concave downward lie above the graph of the function. This is no coincidence. It can be shown that if the tangents to a graph at a point lies below the graph nearby that point then the function is concave upward at that point and if the tangent to the graph at a point lies above the graph nearby then the function is concave downward at that point.

This fact allows us to tell just by looking at the graph where the function is concave upward and where it is concave downward, and hence, to determine whether d squared π¦ by dπ₯ squared is positive or negative at a given point on a graph.