Video: Concavity and Points of Inflection | Nagwa Video: Concavity and Points of Inflection | Nagwa

# Video: Concavity and Points of Inflection

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

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### Video Transcript

In this video, we will learn how to determine the concavity of a function as well as its inflection points using the second derivative. By this stage, you should feel confident in finding the first and second derivatives of a function using standard rules for differentiation and be well-practiced in using the first derivative test to establish the nature of critical points. Now, weβre going to look at what it means for a function to be concave up or concave down or for it to have a point of inflection. And weβll see how we can use the second derivative as an alternative method to the first derivative test.

Letβs consider the shape of a few well-known graphs. Here, we have the graph of π of π₯ equals π₯ squared, π of π₯ equals negative π₯ squared, and β of π₯ equals π₯ cubed. π of π₯ equals π₯ squared is a good example of a function which is concave upward over its entire domain. The curve bends upwards over its entire domain, and the value of its slope is increasing. Another way to think about this is to say that if a graph of a function lies above all of its tangents over some interval, then it is concave upward on that interval. Similarly, π of π₯ equals negative π₯ squared is a good example of a function which is concave downwards. The curve bends downwards over its entire domain, and the value of its slope is decreasing.

This time, we can say that an alternative way to look at it is to say that if the graph of the function lies below all of its tangents on some interval, then itβs concave downward on that interval. With our function π of π₯ equals π₯ squared, the critical point at zero, zero is a minimum. And in fact, itβs an absolute minimum. Itβs the lowest point of the curve over its entire domain. And for the graph π of π₯ equals negative π₯ squared, the critical point at zero, zero is an absolute maximum. Itβs the highest point of the curve over its entire domain.

β of π₯ equals π₯ cubed offers something a little bit different though. The turning point at zero, zero is known as a point of inflection. This is a critical point at which the behavior of the function changes. It goes from being concave downward to being concave upward, or vice versa. So now that we have a definition, letβs look at how we decide on the nature of the critical point and, therefore, its concavity. Letβs look at the graph of π of π₯ equals π₯ squared again. Its derivative π prime of π₯ equals two π₯ is sometimes called the gradient function because it tells us the gradient or the slope of the tangent of the curve at any point. We can see that the gradient of the tangent of the curve at, say a point π₯ equals negative one, just before the critical point will be negative. And the gradient of the tangent to the curve at a point after the critical point, say π₯ equals one, is positive.

In the past, we would have verified this using the first derivative test. We wouldβve substituted these values into the equation for the first derivative and check that it is indeed negative before the critical point and positive after. But letβs think about what actually is happening to the derivative. Itβs going from a number that is less than zero to a number thatβs greater than zero. In other words, the function π prime of π₯ is increasing. Another way to think about this is to say that the derivative of π prime of π₯ must be greater than zero. In other words, π double prime of π₯, the second derivative of our function, must be greater than zero. And thatβs the second derivative test.

We can evaluate the second derivative at the critical point. And if itβs greater than zero, then we have a local minimum. And this test allows us to test for concavity. If the second derivative of our function is greater than zero for all π₯ in some interval πΌ, then the graph is concave upward on this interval. We can also look at the graph of π of π₯ equals negative π₯ squared. Just before our critical point, the tangent has a positive gradient or a positive slope. And just after our critical point, the tangent has a negative gradient. This means that π prime of π₯ is decreasing. In other words, the derivative of π prime of π₯ must be less than zero. Or π double prime of π₯, the second derivative, must be less than zero.

So if we evaluate the second derivative at the critical point and itβs less than zero, then this tells us that we have a local maximum. And we extend this idea and say that if the second derivative of our function is less than zero for all π₯ in some interval πΌ, then the graph is concave downwards on that interval. But there is something missing. What about if π double prime of π₯ of the second derivative is equal to zero? If the second derivative is equal to zero or in fact undefined, then we could have a point of inflection. But we mustnβt assume that any point where π double prime of π₯ is equal to zero is a point of inflection. Instead, in these situations, we have to check the nature of the second derivative to either side of our point and check that the concavity does indeed change from concave up to concave down, or vice versa. Letβs have a look at an example at how to apply some of these definitions.

Determine the intervals on which the function π of π₯ equals negative four π₯ to the fifth power plus π₯ cubed is concave up and concave down.

Remember, if the second derivative of our function π double prime of π₯ is greater than zero for all π₯ in some interval πΌ, then π is concave upward on this interval. Similarly, if the second derivative is less than zero for all π₯ in some interval πΌ, then π is concave downwards on that interval. So weβre going to need to find the second derivative of our function and use this to determine the intervals on which π double prime of π₯ is greater than zero and less than zero.

Weβll begin by finding the first derivative of our function. Itβs five multiplied by negative four π₯ to the fourth power plus three times π₯ squared, which is negative 20π₯ to the fourth power plus three π₯ squared. Weβll differentiate again to find the second derivative. This time, itβs four times negative 20π₯ cubed plus two times three squared, which is negative 80π₯ cubed plus six π₯. Our job is now to determine the interval in which this derivative is greater than zero and on which itβs less than zero. Weβll begin by setting it equal to zero and solving for π₯. We can factor to get two π₯ times negative 40π₯ squared plus three. And then we know that for the product of two π₯ and negative 40π₯ squared plus three to be equal to zero, either to π₯ itself is equal to zero, which means π₯ is equal to zero, or negative 40π₯ squared plus three is equal to zero.

Weβll solve by adding 40π₯ squared to both sides, dividing through by 40, and then finding the square root, remembering to find both the positive and negative square root of three over 40. We can actually rationalize the denominator here, and we get π₯ equals positive or negative root 30 over 20. Next, weβll sketch the curve of π double prime of π₯ out to help us decide where itβs less than zero and equal to zero. Itβs a cubic graph with a negative coefficient of π₯ cubed, which has roots of negative root 30 over 20, positive root 30 over 20, and zero. So itβll look something like this. We can see that π double prime of π₯ is less than zero here and here. And itβs greater than zero here and here. Since the second derivative is greater than zero on the open interval negative infinity to negative root 30 over 20 and the open interval zero to root 30 over 20, π of π₯ is concave upward on these intervals. Similarly, itβs concave downward on the open interval negative root 30 over 20, zero, and root 30 over 20 infinity.

In our next example will look at how to find the inflection points of a curve.

Determine the inflection points of the curve π¦ equals π₯ squared plus two π₯ minus five.

Remember, we say that we could have an inflection point on our curve π¦ equals π of π₯ if the second derivative is equal to zero or the second derivative is undefined at some point. Letβs formalize this a little though. Instead, weβll say that if π is an inflection point on a continuous function π, then π double prime of π₯, the second derivative, is equal to zero or is undefined and the curve changes from concave upward to downward or vice versa at π. We will begin, as usual, by finding the location of any critical points, then determining their nature. Weβll differentiate our function π¦ with respect to π₯. The first derivative is two π₯ plus two.

To find the location of any critical points, weβre going to set this equal to zero. So two π₯ plus two equals zero. To solve for π₯, we subtract two from both sides. And we divide through by two. So there is one critical point at π₯ equals negative one. And what about its nature? This time will find the second derivative. The second derivative of our function is simply two. Interestingly, the second derivative here is a constant. And itβs greater than zero. In other words, the second derivative is greater than zero across the entire domain. π¦ is concave upwards and the concavity never changes. So we can say that this curve has no inflection points.

In our next example, weβll look at how to use the second derivative to find the point of inflection on a curve.

Find the inflection point on the graph of π of π₯ equals π₯ cubed minus nine π₯ squared plus six π₯.

If π is an inflection point on a continuous function π, then the second derivative of π of π₯ is equal to zero or is undefined at this point. And the concavity of the curve changes at this point. To answer this question then, weβll begin by finding the second derivative of our function. The first derivative is three π₯ squared minus two times nine π₯ plus six, which simplifies to three π₯ squared minus 18π₯ plus six. The second derivative is six π₯ minus 18. We know that there could be a point of inflection when the second derivative is equal to zero. So weβll set this equal to zero and solve for π₯. We add 18 to both sides of our equation and then we divide through by six. And we see the π₯ is equal to three.

But just because the second derivative of π of three is equal to zero, that doesnβt guarantee itβs a point of inflection. Weβre going to double-check the concavity of the curve either side of this point. Weβll check π double prime of two and π double prime of four. π double prime of two is six times two minus 18, which is negative six. And π double prime of four is six times four minus 18, which is six. The second derivative of π at two is less than zero and at four is greater than zero. The curve is going from concave downwards to concave upwards. So π₯ equals three is indeed a point of inflection. Now, we know this; we can substitute π₯ equals three into the equation π of π₯ to find π of three. Thatβs three cubed minus nine times three squared plus six times three, which is negative 36. The inflection point of our function is at three negative 36.

In our final two examples, weβll look at how the standard rules of differentiation can also be applied to help us test for concavity in points of inflection, in particular, by looking at trigonometric and logarithmic functions.

Given that π of π₯ equals sin of four π₯ plus cos of four π₯, where π₯ is greater than or equal to zero and less than or equal to π by two, determine the inflection points of π.

Remember, a point of infection on a curve is a point where the curve changes from being concave downwards to concave upwards, or vice versa. This occurs when π double prime of π₯ is equal to zero or is undefined. But remember, π double prime of π₯ being equal to zero doesnβt guarantee we have a point of infection. So we always perform a second test to check. Letβs begin by finding the first derivative of our function. We can quote the standard derivatives of sin π of π₯ and cos of π of π₯. And we see the π prime of π₯ is equal to four cos of four π₯ minus four sin of four π₯.

Next, we find the second derivative. And we see the π double prime of π₯ is equal to negative 16 sin of four π₯ minus 16 cos of four π₯. Weβre looking to find a point of inflection. So letβs set this equal to zero and solve for π₯, noticing that weβre looking to the closed interval zero π by two radiance. We divide through by negative 16 and then subtract cos of four π₯ from both sides. We then recall the fact that tan is equal to sin π₯ over cos π₯. So sin four π₯ divided by cos four π₯ is tan four π₯. And we see that tan of four π₯ is equal to negative one. We find the arctan of negative one which we know to be negative π by four radiance.

Remember though, tan of π₯ is a periodic with a period of π radiance. So this tells us that there might be more than one solution. We amend our interval by multiplying by four. And we see that four π₯ must be greater than or equal to zero and less than or equal to two π. And weβll find all values of four π₯ in this interval by adding multiples of π to our solution. Adding π to negative π by four and we get three π by four. Weβre adding π again and we get seven π by four. Finally, we can divide through by four and we see that π₯ is equal to three π by 16 and seven π by 16 radiance. Remember, just because the second derivative is equal to zero, that doesnβt guarantee we have a point of inflection. So we check for concavity either side of these values.

We can choose π₯ to be equal to 0.5 and 0.6. These are values either side of three π by 16. And weβll also check with π₯ equals 1.3 and π₯ equals 1.4, which are values either side of seven π by 16. π double prime of 0.5 is a negative value and π double prime of 0.6 is a positive value. Similarly, π double prime of 1.3 is greater than zero and π double prime of 1.4 is less than zero. And we see that about the point π₯ equals three π by 16, the graph goes from being concave down to being concave up. And about π₯ equals seven π by 16, it goes from being concave upwards to being concave downwards. And these are indeed both points of inflection. We can substitute each value into π of π₯ to find the matching π¦-coordinates. The points of inflection lie at three π by 16 zero and seven π by 16 zero.

Find, if any, the inflection points of π of π₯ equals three π₯ squared times the natural log of two π₯.

To find the points of inflection, weβll evaluate the second derivative of our function and set it equal to zero. Notice that our function is itself the product of two functions. So weβll need to use the product rule to differentiate it. This says that for two differentiable functions, π’ and π£, the derivative of their product is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. We let π’ be equal to three π₯ squared and π£ be equal to the natural log of two π₯. Then dπ’ by dπ₯ is six π₯ and dπ£ by dπ₯ is one over π₯. So π prime the first derivative of our function is three π₯ squared times one over π₯ plus six π₯ times the natural log of two π₯ or three π₯ plus six π₯ times the natural log of two π₯. Weβre going to differentiate this again.

We use the product rule to find that the derivative of six π₯ times the natural log of two π₯ is six plus six times the natural log of two π₯. And the second derivative is nine plus six times the natural log of two π₯. Letβs set this equal to zero. To solve for π₯, we subtract nine and then divide through by six. We raised both sides as the power of π. And then we divide through by two. So potentially, thereβs an inflection point at π₯ equals one-half π to the negative three over two. But we ought to check whether this is actually an inflection point by checking the values of π double prime or the second derivative to either side of this.

A half π to the negative three over two is approximately 0.112. So letβs try π₯ equals 0.1 and π₯ equals 0.12. The second derivative π double prime of 0.1 is less than zero. And π double prime of 0.12 is greater than zero. The curve goes from being concave down to being concave up. And we can say we do indeed have a point of inflection at π₯ equals one-half π to the negative three over two. Substituting this value of π₯ into original function, we get negative nine over eight π cubed. π has an inflection point at π to the negative three over two over two, negative nine over eight π cubed.

In this video, weβve seen that if the second derivative of π₯ is greater than zero for all π₯ in some interval πΌ, then the graph of π is concave upward on πΌ. We also saw that if the reverse is true, then π is concave downward on πΌ. We also saw that an inflection point occurs when the concavity of the graph changes. And this occurs when the second derivative is equal to zero or undefined at this point. And we saw that alone π double prime of π₯ being equal to zero or undefined doesnβt actually guarantee an inflection point. And so we perform the second derivative test and check for the concavity either side.

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