Lesson Video: Convexity and Points of Inflection | Nagwa Lesson Video: Convexity and Points of Inflection | Nagwa

# Lesson Video: Convexity and Points of Inflection Mathematics

In this video, we will learn how to determine the convexity of a function as well as its inflection points using its second derivative.

15:09

### Video Transcript

In this video, we’ll learn how to determine the convexity of a function as well as its inflection points using its second derivative. Before we move on, we must be confident in how to find the first and second derivatives of functions using the standard rules for differentiation. We should also be able to apply the first derivative test to determine the nature of a critical point.

So to understand what we actually mean when we talk about the convexity of a function, we’re going to consider three very common functions. Let’s begin with the graph of the function 𝑓 of 𝑥 equals 𝑥 squared. This is an example of a function that’s convex downward over its entire domain. The critical point appears to point downward. And the value of the slope of the function increases over its entire domain. In other words, before the critical point, the slope is negative. At the critical point, the slope is zero. And after the critical point or turning point of the function, the slope is positive.

And of course, an alternative way to think about this is that if a graph of a function lies above all of its tangents over some interval, the function is said to be convex downward over that interval.

Next, let’s consider the function 𝑔 of 𝑥 equals negative 𝑥 squared. This is an example of a function that’s convex upward over its entire domain. This time, the critical point appears to point upward, whilst the value of the slope decreases over the entire domain of the function. Before the critical point, the slope is positive. And then after, the slope has negative value. Considering the tangents to the curve once again, we see that if a graph of a function lies below all of its tangents over some interval, then it must be convex upwards over that same interval.

So we’ve now considered what it means for the function to be convex down and convex up over some interval. We also saw what this means for the slope of the function over that same interval. Specifically, for the graph to be convex down, its slope must increase, whilst for the graph to be convex up, its slope must decrease. We can generalize this by thinking about the second derivative. If the slope 𝑓 prime of 𝑥 is increasing, then the rate of change of that slope must be positive. But of course, the rate of change of the slope must be 𝑓 double prime of 𝑥. So for our function 𝑓 of 𝑥, it will be convex down over portions of the graph where the second derivative is positive.

In a similar way, for our function 𝑔 of 𝑥, for it to be convex up, the second derivative must be negative. This essentially means that the first derivative 𝑔 prime of 𝑥 will be decreasing over that interval.

Now, with that in mind, let’s think about a slightly different example. For 𝑓 of 𝑥, we were dealing with relative extrema. But now consider the function ℎ of 𝑥 equals 𝑥 cubed. This function does have a critical point at the origin, at zero, zero. But this point is neither a maxima nor a minima. In fact, either side of this critical point, the sign of the slope doesn’t change; it’s positive before the critical point and positive after.

So what does change about that critical point? Well, it changes from being convex up to convex down. We call this a point of inflection. A point of inflection is a point about which the convexity of that function changes. In this case, we see that the graph is changing from convex up to convex down, but the reverse might also be true.

Now, in this case, the point of inflection occurs at the critical point. But this might not always be the case. For example, consider this graph. About some point on this graph, the convexity changes from being down to up. This is not a critical point. The slope is not zero, and it does appear to exist. So we can have a point of inflection occur at a point which is not a critical point.

Now, as with the convex up and convex down definitions, we can define a second derivative definition for a point of inflection. At these points, the second derivative might be equal to zero. This is not always the case, though. So when we have the second derivative equal to zero, we also perform the first derivative test. That is, we check that the slope either side of that critical point has the same sign. Alternatively, we can use the second derivative test and check that the convexity changes over that point of inflection. It changes from convex up to down or vice versa.

So now we have all of these definitions, let’s formalize this somewhat. The second derivative test for a function 𝑓 of 𝑥 tells us that if the second derivative is positive over some interval 𝐼, then 𝑓 is convex down over that interval. And if the second derivative is negative, 𝑓 is convex up over that interval. If the second derivative is equal to zero or is undefined, then a point of inflection might exist.

There are two ways we can check whether this is the case. We can check for a change of concavity. Does the graph change from being convex up to down, or vice versa, about this point? Alternatively, we can consider the sign of the first derivative. If the sign of the first derivative either side of this point does not change, then that’s a good indication to us that we have a point of inflection. With all this in mind, let’s now look at an example of how we could calculate the intervals over which a polynomial function is convex up or convex down.

Determine the intervals on which the function 𝑓 of 𝑥 equals negative four 𝑥 to the fifth power plus 𝑥 cubed is convex up and down.

Our function 𝑓 of 𝑥 is a fifth-degree polynomial, and we’re looking to determine its convexity. We know that we can use the second derivative to do so. And in fact, since 𝑓 of 𝑥 is a polynomial, this means it’s differentiable and continuous over its entire domain. This means we can be confident that we can evaluate both the first and second derivative of this function and ensure that it exists.

So how does the second derivative help us identify the convexity of a function? Well, we know that the function is convex up over any intervals where the second derivative is less than zero, where it’s negative. And it’s convex down over intervals where the second derivative is positive, in other words, where the first derivative, the slope, is increasing. So let’s begin by evaluating the first then the second derivative of our function.

To begin, we’ll find the first derivative. And we know we can simply differentiate term by term. The power rule for differentiation says that we can multiply by the exponent and then reduce that exponent by one. So when we differentiate negative four 𝑥 to the fifth power with respect to 𝑥, we get five times negative four 𝑥 to the fourth power. Similarly, the derivative of 𝑥 cubed is three 𝑥 squared. And so our first derivative is negative 20𝑥 to the fourth power plus three 𝑥 squared.

We’ll repeat this process for 𝑓 double prime of 𝑥. This time, differentiating term by term and simplifying gives us negative 80𝑥 cubed plus six 𝑥. Now this expression for the second derivative is a cubic polynomial. It has a negative leading coefficient, which tells us information about the shape of the curve. This means, if we think about the curve in particular, we’ll be able to identify the regions where the second derivative is negative and where it’s positive.

We’ll begin by finding where it’s actually equal to zero. This will tell us the points where it intersects the 𝑥-axis, in other words, for what values of 𝑥 is negative 80𝑥 cubed plus six 𝑥 equal to zero. Factoring the left-hand side, and we get two 𝑥 times negative 40𝑥 squared plus three. And for the product of these two expressions to be zero, we know one or other of the expressions must itself be equal to zero. In other words, either two 𝑥 is equal to zero or negative 40𝑥 squared plus three equals zero.

Dividing our first equation by two, and we get 𝑥 equals zero. With our second equation, we’ll add 40𝑥 squared then divide through by 40. So 𝑥 squared is three over 40. To solve for 𝑥, we take the positive and negative square root of three over 40. And 𝑥 simplifies to positive or negative root 30 over 20. So this tells us that our cubic curve negative 80𝑥 cubed plus six 𝑥 passes through the 𝑥-axis at zero and positive and negative root 30 over 20. Since it has a negative leading coefficient, we also know that it should look like this.

So with that in mind, let’s identify the regions over which this second derivative is negative. This will tell us where the original function 𝑓 of 𝑥 is convex up. Specifically, we see that the second derivative is less than zero. In other words, the graph of this function lies below the 𝑥-axis over the open interval negative root 30 over 20 to zero and the open interval root 30 over 20 to ∞. So these are the regions over which it’s convex up.

With that in mind, let’s repeat this process and identify the portions where 𝑓 double prime of 𝑥 is positive. On the graph of 𝑦 equals 𝑓 double prime of 𝑥, these are the regions where it lies above the 𝑥-axis. So we see that 𝑓 double prime of 𝑥 is greater than zero over the open interval from negative ∞ to negative root 30 over 20 and the open interval from zero to root 30 over 20. This means these are the regions over which the original function is convex down.

And so we’ve answered the question. The function 𝑓 of 𝑥 is convex up over the open interval negative root 30 over 20 to zero and the open interval root 30 over 20 to ∞. And it’s convex down over the open interval negative ∞ to negative root 30 over 20 and zero to root 30 over 20.

Now that we’ve demonstrated how to find the convexity of a function, we’ll look at how to determine whether it has a point of inflection.

Find the inflection point on the graph of 𝑓 of 𝑥 equals 𝑥 cubed minus nine 𝑥 squared plus six 𝑥.

Remember, in order to find the inflection point on the graph of a function, we need to identify the point at which the convexity of that function changes. And whilst an inflection point might appear at a critical point, this is not necessarily the case. We can find the possible locations of one, though, by using the second derivative. Specifically, if the second derivative is equal to zero or does not exist, then inflection point might occur. To be sure, we can check whether the convexity either side of that point changes. In other words, does the graph change from being convex up to down or vice versa?

Now 𝑓 of 𝑥 is a polynomial. And this is great because it means it’s continuous and differentiable over its entire domain. In fact, when we differentiate 𝑓 of 𝑥, we get another polynomial, which is still continuous and differentiable. This means we can begin by finding the first derivative and then differentiate one more time to get the second derivative.

We differentiate term by term. The derivative of 𝑥 cubed with respect to 𝑥 is three 𝑥 squared. Remember, we multiply by the exponent and then reduce that the exponent by one. Similarly, differentiating negative nine 𝑥 squared with respect to 𝑥, and we get negative 18𝑥. Finally, the derivative of six 𝑥 is simply six. So the first derivative 𝑓 prime of 𝑥 is three 𝑥 squared minus 18𝑥 plus six.

We’ll repeat this process one more time to find the expression for the second derivative. This time, we get six 𝑥 minus 18. And to establish where an inflection point might occur, we’re going to set this equal to zero. In other words, six 𝑥 minus 18 is equal to zero. To solve for 𝑥, we begin by adding 18 to both sides, so 18 equals six 𝑥 or six 𝑥 equals 18. Then, we divide through by six. So we get 𝑥 equals 18 over six, or simply three. So we can deduce that an inflection point might occur at the point where 𝑥 equals three. This is the point where the second derivative is equal to zero.

To confirm that this is indeed an inflection point, we’ll look at the second derivative either side of 𝑥 equals three and check that the convexity does indeed change. Specifically, let’s check the points where 𝑥 equals two and 𝑥 equals four. Substituting 𝑥 equals two into our expression for the second derivative, and we get six times two minus 18, which is negative six. We know that if the second derivative is negative, the slope of the function is decreasing. This means over the interval where the second derivative is negative, the function is convex up. So our function is convex up at the point where 𝑥 equals two.

Repeating this for 𝑥 equals four, and we get that the second derivative is six times four minus 18, which is positive six. This time, 𝑓 of 𝑥 is convex down since the second derivative is positive at 𝑥 equals four. So we observe that the graph of the function changes from being convex up to convex down about the point 𝑥 equals three.

We can determine the full coordinates of this point by substituting 𝑥 equals three into the original function. 𝑓 of three is three cubed minus nine times three squared plus six times three, which is equal to negative 36. So the inflection point on the graph of our function is at three, negative 36.

Let’s now recap the key points from this lesson. In this lesson, we learned that for some function 𝑓 of 𝑥, if its second derivative is positive over some interval, then the function itself is convex down over that same interval. And the reverse is true. If the second derivative is negative over that interval, then 𝑓 of 𝑥 is convex up. Finally, we saw that an inflection point occurs when the convexity of the graph changes. This will be when 𝑓 double prime of 𝑥 is equal to zero or does not exist. But alone, this criteria is not sufficient. So we must check either that the convexity of the graph changes or that the sign of the first derivative does not.