### Video Transcript

In this video, weβll will learn how
to find the absolute maximum and minimum values of a function over a given interval
using derivatives. At this stage, you should feel
confident in finding relative or local minima and maxima in evaluating the nature of
these by considering derivatives. Weβll now look to extend these
ideas into finding absolute minima and maxima, in other words, the largest and
smallest values that a function will take.

To begin, weβll recall the extreme
value theorem. This says that if π of π₯ is
continuous over some closed interval π to π, then there are two numbers π and π
which are greater than or equal to π and less than or equal to π. Such that, π of π is an absolute
maximum for the function, and π of π is an absolute minimum for the function over
that closed interval.

In other words, if we have a
continuous function over some closed interval, then weβre guaranteed at some point
in this interval to have both an absolute maximum and an absolute minimum. This theorem is important to
finding absolute extrema on a closed interval. It means weβre never going to be
looking for something that doesnβt exist. We can say that these extreme
values are obtained either at the place or places where critical points occur, or at
the end points of the interval.

This means we have a few steps that
can help us to find absolute extrema for some continuous function π. We begin by finding all critical
points in the closed interval π to π. We then find the values of our
function at these critical points. And then, we evaluate our function
at the end points of our interval. Weβre looking here for values that
might be smaller than any relative minima or larger than any relative maxima. Letβs have a look at how we might
apply this process to an example.

Determine the absolute maximum and
minimum values of the function π of π₯ equals two π₯ to the fourth power minus
eight π₯ squared minus 13 in the closed interval negative one to two.

Remember, to find absolute extrema
for some continuous function π of π₯, we follow three steps. We start by finding all critical
points in the closed interval. We check the values of π of π₯ at
these critical points. And then, we check the end points
for any absolute extrema, values that might be smaller than any relative minima or
larger than any relative maxima. Critical points occur when the
derivative of the function is equal to zero, or does not exist. So, letβs begin by finding the
derivative of our function.

The derivative of π of π₯ is
written as π prime of π₯, and itβs four times two π₯ to the third power minus two
times eight π₯. And, of course, the derivative of
negative 13 is zero. So, we see that the derivative of
our function is eight π₯ cubed minus 16π₯. Weβll set this equal to zero and
solve for π₯. Here, we factor the expression on
the right-hand side to get eight π₯ times π₯ squared minus two.

And we see that for eight π₯ times
π₯ squared minus two to be equal to zero, either eight π₯ must be equal to zero,
which means π₯ is equal to zero. Or we can say that π₯ squared minus
two must be equal to zero. And solving this, we see that π₯ is
equal to both the positive and negative square root of two. So, we have critical points on our
function at π₯ equals zero, π₯ equals negative root two, and π₯ equals root two.

Our second step is to evaluate the
function at these critical points. Thatβs π of zero, π of root two,
and π of negative two. π of zero is two times zero to the
fourth power minus eight times zero squared minus 13, which is negative 13. π of root two is two root two to
the fourth power minus eight root two squared minus 13, which is negative 21. And, actually, π of negative root
two is also negative 21. Now this by itself doesnβt help us
much. It certainly looks like negative 21
and negative 13 might be local extrema, but we need to know whether they are
absolute extrema.

So, weβre going to evaluate our
function at the ends of our interval. Remember, this is because we know
that if we have a continuous function over some closed interval π to π, then weβre
guaranteed at some point in this interval to have both an absolute maximum and an
absolute minimum. And these extreme values are
obtained either at the place or places where local extrema occur, or at the end
points of the interval.

So, weβre going to evaluate π of
negative one, and π of two. π of negative one is two times
negative one to the fourth power minus eight times negative one squared minus 13,
which is negative 19. And π of two is two times two to
the fourth power minus eight times two squared minus 13, which is negative 13. We can quite clearly see then that
the absolute maximum value of our function in the closed interval negative one to
two is negative 13. And the absolute minimum value is
negative 21. This example involves some fairly
simple differentiation, so weβll now look at an example which involves a little more
work.

Determine the absolute maximum and
minimum values of the function π¦ equals π₯ over two π₯ plus eight on the closed
interval two to six.

Remember, to find absolute extrema
for our function π of π₯ over some closed interval, we follow three steps. We begin by finding any critical
points in our closed interval. We then find the values of π of π₯
at these critical points. And then, we check the end points
for absolute extrema, in other words, values that are smaller than the relative
minimum or larger than the relative maximum.

Remember, the critical points are
the points on our curve where the derivative is equal to zero or possibly does not
exist. So, we need to find the derivative
of our function and set it equal to zero. But how do we differentiate π₯ over
two π₯ plus eight? In fact, we have a number of
methods we could use. But since itβs the quotient of two
differentiable functions, we can use the quotient rule.

This says that the derivative of
the quotient of two differentiable functions π’ and π£ is π£ times dπ’ by dπ₯ minus
π’ times dπ£ by dπ₯ all over π£ squared. The numerator of our fraction is
π₯, so weβre going to let π’ be equal to π₯ and π£ be equal to two π₯ plus
eight. Then, dπ’ by dπ₯ is equal to
one. And dπ£ by dπ₯ is equal to two.

So, the derivative of π¦ with
respect to π₯ is two π₯ plus eight times dπ’ by dπ₯, which is one, minus π’ times
dπ£ by dπ₯, thatβs π₯ times two, all over π£ squared, thatβs two π₯ plus eight all
squared. So, we can say that dπ¦ by dπ₯ is
equal to eight over two π₯ plus eight all squared. Next, weβre going to set this equal
to zero and solve for π₯. But look at what happens when we
do.

For an algebraic fraction to be
equal to zero, the numerator must itself be equal to zero. In this case, we end up with the
statement zero equals eight, which we know to be rubbish. This means dπ¦ by dπ₯, in this
case, cannot be equal to zero. There are no turning points for us
to evaluate. And we, therefore, jump straight to
step three and evaluate the function at the end points of the interval. Thatβs π of two and π of six.

π of two is two over two times two
plus eight, thatβs two twelfths, which simplifies to one-sixth. π of six is six over two times six
plus eight. Thatβs six twentieths, which
simplifies to three-tenths. Three-tenths is greater than
one-sixth, so we can say the absolute maximum value of our function is three-tenths
and the absolute minimum value is one-sixth. Now letβs backtrack a little
though.

We said that critical points occur
at places on the function where the derivative does not exist, and there is one
point on our function where the derivative doesnβt exist. Thatβs the point where two π₯ plus
eight is equal to zero, or π₯ is equal to negative four. Now since this was outside the
closed interval two to six, we actually didnβt need to worry about this critical
point. And we could focus solely on the
end points of the intervals π of two and π of six.

In our next example, weβll look at
how to apply the process to finding absolute extrema for piecewise functions.

Determine the absolute maximum and
minimum values of the function π of π₯ is equal to six π₯ minus three squared, if
π₯ is less than or equal to two and two minus nine π₯, if π₯ is greater than two in
the closed interval one to six.

Remember, to find absolute extrema
for continuous functions, we follow three steps. We begin by finding any critical
points in the closed interval weβre looking at. We find the values of our function
π of π₯ at these critical points. And then, we check the end points
for absolute extrema, values that are smaller than the relative minimum or larger
than the relative maximum.

Now we have a slight problem
here. This is a piecewise function. And we donβt yet know if this
piecewise function is itself continuous. We do have a note that the function
six π₯ minus three all squared is continuous, and the function two minus nine π₯ is
continuous. So, what weβll do is consider that
π₯ equals two might be a critical point. And to test this, weβre going to
evaluate the right-hand and left-hand derivative of our function at π₯ equals
two.

Weβll begin by evaluating the
right-hand derivative of π at π₯ equals two. This is given by the limit as β
approaches zero from the right of π of two plus β minus π of two all over β. Weβre looking at the right-hand
derivative, so weβre interested in the function that applies when π₯ is greater than
two. Thatβs π of π₯ equals two minus
nine π₯.

So, weβre looking for the limit as
β approaches zero from the right of two minus nine times two plus β minus two minus
nine times two over β. Thatβs two minus 18 minus nine β
minus two plus 18 all over β. This quite quickly simplifies to
negative nine β over β. And then, we simplify further, and
we see that weβre looking for the limit as β tends to zero from the right of
negative nine. But this is independent of β, so we
know that this is just going to be equal to negative nine. And so, our right-hand derivative
is negative nine.

Weβll now repeat this process for
the left-hand derivative. This time, weβre evaluating π of
two plus β minus π of two over β as β approaches zero from the left. And so, weβre interested in the
part of π of π₯ where π₯ is less than or equal to two. So, we have the limit as β
approaches zero from the left of six times two plus β minus three all squared minus
six times two minus three all squared all over β. This simplifies to nine plus six β
all squared minus nine squared over β.

And then, if we distribute the
parentheses, we see that weβre left with the limit as β approaches zero from the
left of 108β plus 36β squared. We can do a little bit of
simplification, and this becomes 108 plus 36β. And then, we see that as β
approaches zero from the left, weβre left with 108.

We can see that our left-hand and
right-hand derivatives are not equal. And so, π prime of π₯, our
derivative, doesnβt actually exist at π₯ equals two. And so, we know that we have a
critical point at π₯ equals two. And we know weβre going to need to
find π of π₯ at this point. We should also though check for any
critical points on each part of our piecewise function. So, weβll differentiate each part
with respect to π₯ and set that equal to zero.

We can use the general power rule
to differentiate six π₯ minus three all squared with respect to π₯. Itβs two times six π₯ minus
three. And then, we reduce the power by
one. And then, we multiply that by the
derivative of six π₯ minus three, which is just six. So, the derivative of this bit is
12 times six π₯ minus three. And the derivative of two minus
nine π₯ is negative nine.

It should be quite clear that
thereβs no way for negative nine to be equal to zero. The derivative of this part of our
function is always negative nine. But we can set 12 times six π₯
minus three equal to zero. And when we solve for π₯, we get π₯
equals 0.5. So, we have one more critical point
at π₯ equals 0.5. Weβre going to evaluate our
function at the points π₯ equals two and π₯ equals 0.5 then. Letβs clear some space.

0.5 is less than two, so we
evaluate the function at this point by using six π₯ minus three all squared. And when we substitute 0.5 in, we
get zero. So, π of 0.5 is zero. We use the same part of our
piecewise function to evaluate π of two. And when we do, we get 81. So, weβve found π of π₯ at the
critical points on our function. Next, we need to check the end
points.

These are π of one and π of
six. We use six π₯ minus three all
squared once again to evaluate π of one. And that gives us nine. But six is greater than two. So, to evaluate π of six, we use
two minus nine π₯ and we get negative 52. We can see that the absolute
maximum value of our function is 81 and the absolute minimum value is negative
52. The key point to remember here is
that if weβre dealing with a piecewise function, we must check the behaviour at the
end of each piece of our function. In our final example, weβll
consider how we can apply ideas about finding absolute extrema to exponential
functions.

Find the absolute maximum and
minimum values rounded to two decimal places of the function π of π₯ equals five
π₯π to the negative π₯, given that π₯ is a part of the closed interval zero to
four.

Remember, to find absolute extrema
for our function π of π₯, we follow three steps. We find all critical points in our
closed interval. We then find the values of π of π₯
at these critical points. And then, we check the end points
for absolute extrema. The critical points are the points
where the derivative is equal to zero or does not exist. So, weβre going to find the
derivative of our function and set that equal to zero.

And here, we notice that this
itself is the product of two differentiable functions. So, weβre going to use the product
rule. This says that the derivative of
the product of two differentiable functions π’ and π£ is π’ times dπ£ by dπ₯ plus π£
times dπ’ by dπ₯. So, we let π’ be equal to five π₯
and π£ be equal to π to the negative π₯. Then, dπ’ by dπ₯ is equal to five
and dπ£ by dπ₯ π₯ is equal to negative π to the negative π₯. This means the derivative of our
function is five π₯ times negative π to the negative π₯ plus π to the negative π₯
times five, which we can simplify to five π to the negative π₯ times one minus
six

Letβs set this equal to zero. Now thereβs no way for five π to
the negative π₯ to be equal to zero. So, for the statement five π to
the negative π₯ times one minus π₯ equals zero to be true, we know that one minus π₯
itself must be equal to zero, which means that π₯ equals one is a critical
point. Weβre, therefore, going to evaluate
our function at this critical point and at the end points of the function, so π of
one, π of zero, and π of four.

π of one is five times one times
π to the negative one, which is 1.8393 and so on, or correct to two decimal places
as required 1.84. π of zero is zero. And π of four is five times four
times π to the negative four, which is 0.37 correct to two decimal places. And we can, therefore, say that the
absolute maximum value of our function is 1.84 and the absolute minimum value is
zero.

In this video, weβve learned that
if we have a continuous function over some closed interval, then weβre guaranteed at
some point in this interval to have both an absolute maximum and an absolute
minimum. We also saw that these extreme
values are obtained either at the place or places where local extrema occur or at
the end points of the interval. We saw that we can apply these
ideas to exponential functions and functions which are both products and quotients
of other differentiable functions, but that we need to be careful with piecewise
functions to evaluate the ends of each piece of the function.