### 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, weβre 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 weβll see how we can use the
second derivative as an alternative method to the first derivative test.

Letβs 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 itβs 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, itβs an absolute
minimum. Itβs 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. Itβs 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,
letβs look at how we decide on the nature of the critical point and, therefore, its
concavity. Letβs 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 wouldβve 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 letβs think about what actually
is happening to the derivative. Itβs going from a number that is
less than zero to a number thatβs 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 thatβs the second derivative
test.

We can evaluate the second
derivative at the critical point. And if itβs 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 πΌ, 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 itβs 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 πΌ, 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 mustnβt 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. Letβs 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 πΌ, then π is concave upward on this interval. Similarly, if the second derivative
is less than zero for all π₯ in some interval πΌ, then π is concave downwards on
that interval. So weβre 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.

Weβll begin by finding the first
derivative of our function. Itβs 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. Weβll differentiate again to find
the second derivative. This time, itβs 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 itβs less than
zero. Weβll 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.

Weβll 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, weβll sketch the curve of π
double prime of π₯ out to help us decide where itβs less than zero and equal to
zero. Itβs 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 itβll look something like
this. We can see that π double prime of
π₯ is less than zero here and here. And itβs 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, itβs 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. Letβs formalize this a little
though. Instead, weβll 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. Weβll differentiate our function π¦
with respect to π₯. The first derivative is two π₯ plus
two.

To find the location of any
critical points, weβre 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 itβs 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, weβll 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, weβll
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 weβll 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 doesnβt guarantee itβs a point of
inflection. Weβre going to double-check the
concavity of the curve either side of this point. Weβll 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. Thatβs 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, weβll
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 doesnβt guarantee we have a point of infection. So we always perform a second test
to check. Letβs 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₯. Weβre looking to find a point of
inflection. So letβs set this equal to zero and
solve for π₯, noticing that weβre 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 weβll 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. Weβre 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 doesnβt 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 weβll 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,
weβll 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 weβll 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₯. Weβre 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₯. Letβs 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, thereβs 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 letβs 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, weβve seen that if
the second derivative of π₯ is greater than zero for all π₯ in some interval πΌ,
then the graph of π is concave upward on πΌ. We also saw that if the reverse is
true, then π is concave downward on πΌ. 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 doesnβt actually guarantee an
inflection point. And so we perform the second
derivative test and check for the concavity either side.