Video Transcript
In this video, we’ll learn how to determine the concavity of a function as well as its inflection points using its second derivative. Before we move on, you must ensure that you’re confident in finding the first and second derivatives of functions using the standard rules for differentiation. And you should also be able to apply the first derivative test to determine the nature of critical points.
To understand what we actually mean by concavity of a function, we’re going to consider three very common functions. Consider the graph of the function 𝑓 of 𝑥 equals 𝑥 squared. This is an example of a function that’s concave upward over its entire domain. It curves upwards, and the value of its slope increases over its entire domain. Before the critical point or turning point of the function, the slope is negative. At the critical point, the slope is zero. And after the critical point, the slope is positive. 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 concave upward over that interval.
Now, consider the function 𝑔 of 𝑥 equals negative 𝑥 squared. This is an example of a function that’s concave downwards over its entire domain. The function curves downwards and the value of the slope decreases over some interval. Before the critical point, the slope is positive. At the critical point, the slope is zero. And after the critical point, the slope has a negative value. Considering the tangents to the curve once again, we can see that if a graph of a function lies below all of its tangents over some interval, then it must be concave downward over that interval.
Now, looking back at both functions 𝑓 and 𝑔, we can also see that the critical point on the graph of 𝑓 of 𝑥 equals 𝑥 squared is an absolute minimum. It’s the lowest point of the curve over its entire domain. The critical point on the graph of 𝑔 of 𝑥 is an absolute maximum. It’s the highest point of the curve over its entire domain. Now consider the graph of the function ℎ of 𝑥 equals 𝑥 cubed. We have something a little bit different going on here. The critical point at the origin zero, zero is called a point of inflection. A point of inflection occurs when the concavity of the function changes. Here it changes from concave downward to concave upward. But the reverse is also true.
Note, though, that a point of inflection doesn’t necessarily need to occur at a critical point. For instance, this graph has portions which are concave downward and portions which are concave upward. There will occur a point on this curve where the concavity changes and so we have a point of inflection. It’s not, however, a critical point. The slope at this point is not zero, and of course we can see it does exist.
So now we know what we mean by concavity and points of inflection. Let’s look at how we can determine the nature of a critical point and, therefore, its concavity. Let’s begin by going back to the function 𝑓 of 𝑥. We need to find its gradient function, which is given by its first derivative. The first derivative of 𝑥 squared with respect to 𝑥 is two 𝑥. We can find the location of any critical points by setting 𝑓 prime of 𝑥 equal to zero, so zero is equal to two 𝑥, which means 𝑥 is equal to zero. So we have a critical point at zero. Then the first derivative test tells us that the nature of the critical point can be established by finding the slope of the tangent to the curve either side of this point.
We know the slope at zero is equal to zero. When 𝑥 is equal to negative one, 𝑓 prime of 𝑥 is two times negative one, which is negative two. And when 𝑥 is equal to one, the gradient function is two times one, which is equal to two. The slope changes from negative to positive about the critical point, giving us that concave-upward shape. What this means is that over that interval, the slope 𝑓 prime of 𝑥 is increasing. In other words, the rate of change of 𝑓 prime of 𝑥 must be positive. But of course, the rate of change of 𝑓 prime of 𝑥 is its derivative, so it’s 𝑓 double prime of 𝑥. And 𝑓 double prime of 𝑥 must be greater than zero.
And so this gives us the first definition. If 𝑓 double prime of 𝑥 is greater than zero for all values of 𝑥 in some interval 𝐼, then the graph must be concave upward over that interval. And this is known as the second derivative test since evaluating the second derivative gives us information about the nature of the extremer and thus the concavity of the curve. The reverse is true for a graph which is concave downwards. The value of its slope is decreasing. It moves from positive to negative. This means the rate of change of the slope must be negative. So 𝑓 double prime of 𝑥 is less than zero. And we can therefore say that if 𝑓 double prime of 𝑥 is less than zero for all 𝑥 in some interval 𝐼, then 𝑓 of 𝑥 is concave downward over that interval.
So what happens if the second derivative is equal to zero? Well, in fact, when the second derivative is equal to zero or does not exist, we might have a point of inflection. But we can’t assume any point where the second derivative is equal to zero or does not exist is necessarily a point of inflection. Instead, we evaluate the second derivative either side of our critical point, and we check that the concavity changes from concave up to down or vice versa. Now that we have our definitions, we’re going to look at an example of how to calculate the intervals over which a polynomial function is concave up or concave down.
Determine the intervals on which the function 𝑓 of 𝑥 equals negative four 𝑥 to the fifth power plus 𝑥 cubed is concave up and down.
We have a function which is a fifth-degree polynomial, and we’re asked to determine the concavity of the function. Specifically, we need to look for the intervals over which it’s concave upward and concave downward. And so we recall that given a function 𝑓 of 𝑥, if its second derivative is positive, it’s greater than zero for all 𝑥 in some interval 𝐼, then 𝑓 of 𝑥 is concave upward over that interval. Similarly, if the second derivative is less than zero, then 𝑓 of 𝑥 is concave downwards over that interval. If it’s equal to zero, then we might have a point of inflection. But we’d need to perform further testing to be sure. In fact, we’re only interested in the concavity, so these two definitions are enough.
We are going to then need to find the second derivative of our function, so we’ll differentiate it twice. And of course, to differentiate a power term, we multiply the entire term by the exponent and then reduce that exponent by one. So the first derivative of negative four 𝑥 to the fifth power is five times negative four 𝑥 to the fourth power. And when we differentiate 𝑥 cubed, we get three 𝑥 squared. So our first derivative is negative 20𝑥 to the fourth power plus three 𝑥 squared. To find the second derivative, we repeat this process one more time. We differentiate 𝑓 prime of 𝑥 and we get four times negative 20𝑥 cubed plus two times three 𝑥, giving us that the second derivative is negative 80𝑥 cubed plus six 𝑥.
To find the interval on which our function is concave up, we need to solve the inequality negative 80𝑥 cubed plus six 𝑥 is greater than zero. Remember, we’re finding the locations where the second derivative is positive. To do that, we’re going to begin by solving negative 80𝑥 cubed plus six 𝑥 is equal to zero. We’ll then sketch the graph of the function to determine the intervals over which it’s positive and negative. We factor our two 𝑥 from the left-hand side of our equation, and we get two 𝑥 times negative 40𝑥 squared plus three. For this expression to be equal to zero, we can say that either two 𝑥 is equal to zero or negative 40𝑥 squared plus three is equal to zero. The solution to our first equation is 𝑥 is equal to zero. And we solve our second by adding 40𝑥 squared to both sides.
We then divide through by 40, so 𝑥 squared is equal to three over 40. And then we find the positive and negative square root of three over 40, so those are our second lot of solutions to 𝑥. In fact, if we distribute the root over our fraction and then rationalize the denominator, we get plus or minus the square root of 30 over 20. Next, we’ll sketch the graph of the function negative 80𝑥 cubed plus six 𝑥. It’s a cubic function with a negative leading coefficient, so it will look a little something like this. By solving negative 80𝑥 cubed plus six 𝑥 equals zero, we found the roots of that equation, the places where it intersects the 𝑥-axis, and of course we’re looking to find the locations where this is positive. And if we look at our graph, we see it’s positive here and it’s positive over here.
In other words, it’s positive when 𝑥 is less than negative root 30 over 20 and when 𝑥 is greater than zero and less than root 30 over 20. And so those are the intervals using inequality notation for which 𝑓 of 𝑥 is concave up. When it’s concave down, the second derivative is less than zero. And if we look at our graph, we see that’s here and over here. So, that’s when 𝑥 is greater than negative root 30 over 20 and less than zero and when 𝑥 is greater than root 30 over 20. Using interval notation, 𝑓 of 𝑥 is concave up on the open intervals from negative ∞ to negative root 30 over 20 and zero to root 30 over 20 and concave down on the open interval from negative root 30 over 20 to zero and root 30 over 20 to ∞.
Now that we’ve established how to determine the concavity of a function, we’ll look at how to determine whether a graph 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.
We recall that an inflection point on a curve is the point at which the concavity of the function changes. An inflection point might not appear at a critical point, and so we know that an inflection point can be determined by using the second derivative. If the second derivative is equal to zero or does not exist, then we might have an inflection point. To be sure, we check that the concavity either side of that point changes from either concave up to downwards or vice versa. And so we’re going to begin by finding the first derivative of our function. Differentiating term by term and remembering that when we differentiate a power term, we multiply by the exponent and then reduce this exponent by one, and we find 𝑓 prime of 𝑥 is three 𝑥 squared minus 18𝑥 plus six.
We then repeat this process again to find the second derivative, and that’s just six 𝑥 minus 18. Of course, a constant differentiates to zero. Note that we said that an inflection point might occur when the second derivative is equal to zero or does not exist. But of course, 𝑓 of 𝑥 was a polynomial, so continuous and differentiable over its entire domain. Similarly, 𝑓 prime of 𝑥 was also a polynomial, so it’s also differentiable and continuous over its entire domain. This tells us that the second derivative does indeed exist for all values of 𝑥 in the domain of the function. So we’re only going to solve six 𝑥 minus 18 equals zero. We add 18 to both sides and then we divide by six. And we see that the second derivative is equal to zero at the point 𝑥 equals three.
Now that we’ve established that a point of inflection might occur at the point 𝑥 equals three, we’re going to check the concavity either side by evaluating the second derivative at 𝑥 equals two and 𝑥 equals four. We choose points which are local to where the second derivative is equal to zero. In this case, because there’s only one point where this is the case and the second derivative is a linear function, it’s not hard to do that. If our second derivative function had been a higher-order polynomial though and there have been multiple points where the second derivative was equal to zero, we’d need to take care to pick points that are not beyond the adjacent points where the second derivative is equal to zero.
When 𝑥 is equal to two, the second derivative is six times two minus 18, which is negative six. Then when 𝑥 is equal to four, our expression for the second derivative is six times four minus 18, and this time that’s positive six. And so about the point 𝑥 equals three, the second derivative changes from being negative to positive. This means it changes from being concave down to concave up, and that does indeed confirm to us that the point 𝑥 equals three is a point of inflection. To find the coordinates of that point of inflection then, we substitute 𝑥 equals three into the original function. That’s three cubed minus nine times three squared plus six times three, which is negative 36. The inflection point on the graph of our function then is at three, negative 36.
Throughout this video, we’ve made reference to the fact that, firstly, a point of inflection might occur at a critical point, but it won’t always. We’ve also said that the second derivative being equal to zero doesn’t guarantee the existence of a point of inflection. A prime example here is the function 𝑓 of 𝑥 equals 𝑥 to the fourth power. The first derivative 𝑓 prime of 𝑥 of this function is four 𝑥 cubed. Then the second derivative is given by 12𝑥 squared. Now, if we were to set the second derivative equal to zero, we get the equation 12𝑥 squared equals zero, and solving gives us a value of 𝑥 equals zero. So the second derivative is equal to zero at the point where 𝑥 is equal to zero.
But what happens on the graph of the function at the point 𝑥 equals zero? We do, in fact, appear to have a turning point of some description. But the concavity of the function doesn’t change. It’s concave up before the point 𝑥 equals zero and concave up after the point 𝑥 equals zero. And so this is why it’s really important to check the value of the second derivative when you think you’ve got a point of inflection. Remember, if the second derivative is less than zero, the curve will be concave down in that region. If it’s greater than zero, it will be concave up. But if it’s equal to zero, it might be a point of inflection. But we need to do a little bit more work by checking the value of the second derivative just to the left and just to the right to see whether it’s concave up, concave down, or a point of inflection.
We’re now going to recap the key points from this lesson. In this video, we learned that if for some function 𝑓 of 𝑥, 𝑓 double prime of 𝑥 is greater than zero for all 𝑥 in some interval, then 𝑓 of 𝑥 is concave up over that interval. Similarly, if 𝑓 double prime of 𝑥 is less than zero for all 𝑥 in some interval, then 𝑓 of 𝑥 is concave down over that interval. Finally, we learned that a point of inflection is a point about which the concavity of the function changes. We can find possible locations of points of inflection by finding points at which the second derivative is equal to zero or possibly doesn’t exist. But we do need to check that the concavity of the function will change to either side of this point.
