### 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.