Video Transcript
In this video, we will see how to
apply the second derivative test to classify a critical point as either a local
minimum, a local maximum, or a point of inflection.
We should already be familiar with
the definition of critical points as points of a function where the slope of the
tangent to the curve is equal to zero or is undefined and how to find the critical
points for function using differentiation. You may also be familiar with the
first derivative test for classifying critical points, which is also called
determining their nature. Critical points can be either local
minima, local maxima, or points of inflection. And they are classified according
to the shape of the curve at that point. This is determined by the behavior
of the slope of the curve around the point.
Recall that the first derivative of
a function 𝑓 prime of 𝑥 or d𝑦 by d𝑥, if we’re using Leibniz’s notation, tells us
the slope of a curve. That’s the rate of change of the
curve itself. And at critical points, the first
derivative is equal to zero. Therefore, the second derivative of
a function, which is the derivative of the first derivative, tells us the slope of
the slope. Or more usefully, it tells us about
the rate of change of the slope of a curve. Let’s consider how the slope of a
curve changes around a critical point, starting with a local minima.
By drawing in tangents to the curve
either side of the critical point, we see that the slope of the curve and therefore
the first derivative of the function is negative to the left of our critical point
and is positive to the right of the critical point. The slope and therefore the value
of 𝑓 prime of 𝑥 changes from negative to zero to positive. And therefore, the value of the
slope is increasing. Recall that if a function is
increasing, it has a positive derivative. So this tells us that as the slope
is increasing, the derivative of the slope is positive.
The derivative of the slope is the
second derivative of the original function. So we can conclude that at a local
minimum, the second derivative of the function will be positive: 𝑓 double prime of
𝑥 is greater than zero. We can apply the same reasoning to
a local maximum. This time, the slope 𝑓 prime of 𝑥
changes from positive to zero to negative and therefore the value of 𝑓 prime of 𝑥
is decreasing. If a function is decreasing, then
its derivative is negative. So we can conclude that at a local
maximum, the derivative of 𝑓 prime of 𝑥, that’s 𝑓 double prime of 𝑥, the second
derivative of the original function, will be negative.
Unfortunately, the second
derivative test is not particularly useful for identifying points of inflection. At a point of inflection, the slope
either changes from positive to zero to positive or from negative to zero to
negative. And so, the sign of 𝑓 prime of 𝑥
is the same either side of a point of inflection. Consequently, we can’t use results
about increasing or decreasing functions to use the second derivative test to
classify a point of inflection. In fact, it turns out that at
points of inflection, the second derivative of a function is equal to zero. But this can also be true at some
local minima or local maxima. So it isn’t enough in order to
conclude that at a critical point must be a point of inflection.
For example, consider the function
𝑓 of 𝑥 equals 𝑥 to the power of four. We know from its graph that it has
a local minimum point at the origin. If we find the first derivative 𝑓
prime of 𝑥, this is equal to four 𝑥 cubed. And setting this equal to zero, we
confirm that there is indeed a critical point when 𝑥 is equal to zero. The second derivative of the
function 𝑓 double prime of 𝑥 is 12𝑥 squared. And if we were to substitute 𝑥
equals zero into the second derivative, we would get zero. But as we’ve seen, this critical
point is a local minimum, not a point of inflection. What this tells us is that if the
second derivative at a critical point is equal to zero, we must instead use the
first derivative test to determine the nature of the critical point because it could
be a point of inflection, but it may also be a local minimum or a local maximum. Let’s now consider some
examples.
Determine the local maximum and
minimum values of the function 𝑦 equals negative three 𝑥 squared minus six 𝑥
minus four.
First, we recall that at critical
points, the first derivative of the function — in this case d𝑦 by d𝑥 — is equal to
zero. So our first step is going to be to
find the first derivative of this function. By applying the power rule of
differentiation, we find that d𝑦 by d𝑥 is equal to negative six 𝑥 minus six. We then set our expression for d𝑦
by d𝑥 equal to zero and solve for 𝑥, giving 𝑥 is equal to negative one. Our function, therefore, has one
critical point, which occurs when 𝑥 is equal to negative one.
Next, we need to evaluate the
function at the critical point, which we do by substituting 𝑥 equals negative one
into the equation we’ve been given. We obtain 𝑦 equals negative three
multiplied by negative one squared minus six multiplied by negative one minus four
which simplifies to negative one. This tells us then that the only
critical point of this function is the point with coordinates negative one, negative
one. But we need to determine whether
this is a local minimum or local maximum, which we’ll do by applying the second
derivative test.
To find the second derivative, we
need to differentiate our first derivative with respect to 𝑥. So we’re finding the derivative of
negative six 𝑥 minus six with respect to 𝑥. Applying the power rule, we see
that this derivative is just equal to negative six. Now, this second derivative is
actually just a constant because we’ve differentiated a quadratic expression
twice. So we don’t need to substitute the
𝑥-coordinate at our critical point in in order to evaluate because the second
derivative is constant for all values of 𝑥. We note that negative six is less
than zero. We recall that if the second
derivative of a function is negative at the critical point, then the critical point
is a local maximum. So the point negative one, negative
one is indeed a local maximum of this function.
So we can conclude that this
function has no local minimum value but has a local maximum value of negative
one. Notice that is the value of the
function itself that we are giving here, not the 𝑥-value, although they are both
the same in this instance. We can also confirm this result
using our knowledge of the graphs of quadratic functions. As the coefficient of 𝑥 squared in
this curve is negative, the graph of this quadratic will be a negative parabola. We know that parabolas have only a
single critical point. And if the coefficient of 𝑥
squared is negative, then their critical point will be a local maximum.
Let’s now consider another
example.
Find the points 𝑥, 𝑦 where 𝑦
equals nine 𝑥 plus nine over 𝑥 has a local maximum or a local minimum.
Local maxima and local minima are
examples of critical points. And we recall that at the critical
points of a function, the first derivative d𝑦 by d𝑥 is equal to zero. Before differentiating, we may find
it helpful to rewrite the second term in our function as nine 𝑥 to the negative
one. We can then use the power rule of
differentiation to find the first derivative d𝑦 by d𝑥. Remember that when we
differentiate, we decrease the power by one. So when we decrease that power of
negative one, it will become negative two not zero. Watch out for that! That’s a common mistake. We can rewrite this derivative as
nine minus nine over 𝑥 squared and we’ll then set this derivative equal to
zero.
We’ll now solve the resulting
equation in order to find the 𝑥-values at the critical points. We begin by multiplying every term
in the equation by 𝑥 squared. We can then divide through by nine
to give 𝑥 squared minus one equals zero. Add one to both sides and then
finally take the square root, remembering that we have both positive and negative
solutions. We find that 𝑥 is equal to
positive or negative one. So this function has two critical
points.
Next, we need to find the 𝑦-values
at each critical point by evaluating the function itself. When 𝑥 is equal to positive one,
𝑦 is equal to nine multiplied by one plus nine over one, which is equal to 18,
giving a critical point of one, 18. When 𝑥 is equal to negative one,
𝑦 is equal to negative 18. So our second critical point has
coordinates negative one, negative 18. We now need to determine whether
these critical points are local minima or local maxima, which we’ll do using the
second derivative test. We’ll clear some space in order to
do this.
To find the second derivative d two
𝑦 by d𝑥 squared, we need to differentiate the first derivative, which was nine
minus nine 𝑥 to the power of negative two with respect to 𝑥. Doing so, we obtain negative nine
multiplied by negative two 𝑥 to the power of negative three, which we can write as
18 over 𝑥 cubed. Next, we need to evaluate this
second derivative at each of our critical points. When 𝑥 is equal to negative one,
the second derivative is 18 over negative one cubed, which is equal to negative
18. This is less than zero. And we recall that if the second
derivative of a function is negative at a critical point, then the critical point is
a local maximum. Evaluating the second derivative
when 𝑥 is equal to positive one gives 18 over one cubed, which is 18. And as this is greater than zero,
we conclude that the critical point when 𝑥 is equal to one is a local minimum.
So we’ve completed the problem. We answered that the point one, 18
is a local minimum and the point negative one, negative 18 is a local maximum.
In our next example, we’ll apply
our knowledge of the second derivative test for local extrema to a problem involving
differentiation of trigonometric functions.
Find, if any, the local maximum and
minimum values of 𝑓 of 𝑥 equals 19 sin 𝑥 plus 15 cos 𝑥, together with their
type.
We recall first of all that at the
critical points of a function, the first derivative 𝑓 prime of 𝑥 is equal to
zero. We’ll also need to recall the cycle
that we can use for differentiating sine and cosine. 𝑓 prime of 𝑥 is, therefore, equal
to 19 cos 𝑥 minus 15 sin 𝑥 and we set this equal to zero. To solve, we can first separate the
two terms onto opposite sides of the equation and then divide both sides of the
equation by both cos 𝑥 and 15 to give sin 𝑥 over cos 𝑥 is equal to 19 over
15. At this point, we recall one of our
trigonometric identities: tan 𝜃 is equal to sin 𝜃 over cos 𝜃. So we have tan 𝑥 equals 19 over
15.
To solve, we apply the inverse tan
function. And we must recall at this point
that in order to differentiate trigonometric functions, we must be working with the
angle measured in radians as the key limits used when we first derive the
derivatives from first principles are only true in radians. So when we evaluate 𝑥 on our
calculators, we must make sure we’re working in radians. We find then that 𝑥 is equal to
0.9025 radians. However, tan 𝑥 is a periodic
function with a period of 𝜋. So there are other solutions to
this equation which will correspond to other critical points of the function 𝑓. This means that critical points
will occur at this 𝑥-value that we’ve just found plus or minus integer multiples
of 𝜋. Adding 𝜋 to our value of 0.9025
gives 4.0441 radians. So this will be the second 𝑥-value
at which a local maximum or minimum occurs.
Next, we need to evaluate the
function 𝑓 of 𝑥 at each of the critical points. For our first critical point, when
𝑥 is equal to 0.9025, we get 24.21 to two decimal places. And at our second critical point,
where 𝑥 is equal to 4.0441, we get negative 24.21 to two decimal places. Now, this makes sense because the
sine and cosine functions each have a horizontal line of symmetry on the 𝑥-axis and
therefore so will a sum or difference of the sine and cosine functions, meaning that
the absolute value of the local maximum will be the same as the absolute value of
the local minimum.
Now, finally, we need to apply the
second derivative test to classify these critical points. So let’s make a little bit of
room. We differentiate 𝑓 prime of 𝑥 to
give negative 19 sin 𝑥 minus 15 cos 𝑥. Now we need to evaluate this
function at each of our critical points, but there’s a trick that we can use
here. As we’ve differentiated twice,
we’ve been halfway around our cycle of differentiation, which means that the second
derivative is actually almost identical to the original function. What’s different is that both terms
are negative instead of positive. But if we factor this negative one
out of our expression, we see that in this instance, 𝑓 double prime of 𝑥 is
actually equal to negative 𝑓 of 𝑥.
The reason this is useful is
because we’ve already evaluated 𝑓 of 𝑥 at each of our critical points. So we can use the values we’ve
already found in order to determine the second derivative at our critical
points. At our first critical point, when
𝑥 is equal to 0.9025, 𝑓 of 𝑥 was equal to 24.21. So the second derivative 𝑓 double
prime of 𝑥 will be equal to negative 24.21. As this is less than zero, it tells
us that this critical point will be a local maximum. At our second critical point, the
value of 𝑓 of 𝑥 was negative 24.21. So the value of 𝑓 double prime of
𝑥 will be positive 24.21 and as this is greater than nought, our second critical
point is a local minimum.
Now these are local minimum and
maximum values. But due to the shape of the graph
of 𝑓 of 𝑥, they’re also the absolute minimum and maximum values of the
function. So we can conclude that the local
and absolute minimum value of the function is negative 24.21 and the local and
absolute maximum value of the function is 24.21.
Let’s consider our final
example.
Suppose 𝑓 prime of four equals
zero and 𝑓 double prime of four equals negative four. What can you say about 𝑓 at the
point 𝑥 equals four? 𝑓 has a local minimum at 𝑥 equals
four. 𝑓 has a local maximum at 𝑥 equals
four. 𝑓 has a point of inflection at 𝑥
equals four. It is not possible to state the
nature of the turning point of 𝑓 at 𝑥 equals four. Or 𝑓 has a vertical tangent at 𝑥
equals four.
Let’s take each of the pieces of
information we’ve been given in turn. Firstly, we’re told that 𝑓 prime
of four is equal to zero. And if the first derivative of a
function is equal to zero at a given point, then the function has a critical point
at that point. So we know that 𝑓 has a critical
point when 𝑥 is equal to four. Next, we’re told that 𝑓 double
prime of four is equal to negative four. So the second derivative of our
function 𝑓 is negative when 𝑥 is equal to four. The second derivative will be
negative at a local maximum. So we can conclude that 𝑓 has a
local maximum at 𝑥 equals four.
That’s the second option in the
list we’ve been given. The first, third, and fourth
options are therefore false. If a point is a local maximum, it
can’t also be a local minimum or a point of inflection. And it has been possible for us to
determine the nature of this turning point. Let’s consider the fifth
option. We know that the first derivative
of our function 𝑓 is zero when 𝑥 is equal to four, which means that the slope of
the curve and the slope of the tangent will be zero. Therefore, 𝑓 will have a
horizontal, not a vertical tangent at 𝑥 equals four. So we’ve completed the problem. 𝑓 has a local maximum at 𝑥 equals
four.
Let’s summarize what we’ve seen in
this video.
If 𝑓 is a differentiable function
such that the first derivative 𝑓 prime of 𝑎 is equal to zero, then 𝑓 has a
critical point at 𝑥 equals 𝑎. If the second derivative 𝑓 double
prime of 𝑎 is positive, then the critical point is a local minimum. But if the second derivative 𝑓
double prime of 𝑎 is negative, the critical point is a local maximum. If the second derivative 𝑓 double
prime of 𝑎 is equal to zero, then the critical point could be a point of
inflection. But it’s possible that it could
also be a local minimum or a local maximum. So in this instance, we’d need to
use the first derivative test in order to classify the critical point.
