### Video Transcript

Find the absolute maximum and
minimum values of the function π¦ equals π₯ over four plus one over π₯ minus four on
the interval from one to three inclusive.

We have a closed interval that is
an interval which includes both of its endpoints, one and three. And we want to find the absolute
maximum and minimum values of the function given on this interval. So for values of π₯ between one and
three inclusive, whatβs the greatest and smallest that π¦ can be.

To solve this problem, weβre going
to use the closed interval method which tells us how to find the absolute maximum
and minimum values of a continuous function π on a closed interval π, π. We evaluate π at its critical
numbers in the open interval from π to π and also at the endpoints π and π. The largest value obtained in this
way is the absolute maximum and the smallest is the absolute minimum.

Before applying this method, we
should check that it is applicable. This method finds the absolute
maximum and minimum values of a continuous function. Is our functions continuous? As a rational function, it is
continuous wherever it is defined. However, itβs not defined when this
denominator polynomial is zero. So itβs not defined at π₯ equals
four which is the only value of π₯ for which π₯ minus four is zero. And so itβs not continuous at π₯
equals four. So what do we do?

Well, for this problem, itβs not an
issue because weβre restricting our attention to the closed interval from one to
three. And π₯ equals four does not lie in
this interval. And so our function is continuous
on the smaller domain, the closed interval from one to three. This allows us to apply the closed
interval method. So letβs apply the method.

We first have to evaluate π at its
critical numbers in the open version of the interval on which it is defined. Our function is π₯ plus four plus
one over π₯ minus four. And weβre looking for the critical
numbers in the open interval from one to three. But what are the critical numbers
of this function in this interval? In fact, more generally, what are
critical numbers of a function? Period.

A number π in the domain of a
function f is a critical number of π if the derivative π prime evaluated at π is
zero or if π prime evaluated at π does not exist. In other words, π is not in the
domain of π prime. Weβre looking for a critical number
of our function and this depends on the derivative of the function. So letβs differentiate. We can do this term by term.

The derivative of π₯ over four or a
quarter π₯ is straightforward. Itβs a quarter. The derivative of one over π₯ minus
four is slightly more complicated. So letβs give it a bit of
thought. If we let π§ equal π₯ minus four,
then this is the derivative with respect to π₯ of one over π§. And we can apply the chain rule π
by ππ₯ is ππ§ by ππ₯ times π by ππ§. So now we can consider ππ§ by ππ₯
and π by ππ§ of one over π§ separately. What is ππ§ by ππ₯? π§ is π₯ minus four by
definition. And so differentiating this with
respect to π₯, we see that ππ§ by ππ₯ is one.

How about π by ππ§ of one over
π§. We can write one over π§ as π§ to
the negative one. And then we can apply the fact that
π by π whatever of whatever to the power of π is π times whatever to the power
of π minus one. With π equal to negative one, we
get that π by ππ§ of π§ to the negative one is negative one times π§ to the
negative two. Letβs clear some room. We can write π§ to the negative two
as one over π§ squared. Here, weβve used the fact that
adding one times negative one times something is just the same as subtracting that
something.

Now, we use the fact that π§, the
auxiliary variable, is π₯ minus four to get our derivative ππ¦ by ππ₯ in terms of
π₯. Letβs clear away some working to
make room for some more. Letβs remind ourselves why we
differentiated our function. It was to find the critical numbers
of our function, where the derivative is zero or does not exist. Letβs start with where the
derivative does not exist.

Because of the denominator π₯ minus
four squared, we can see that the derivative doesnβt exist when π₯ is four. But this isnβt an issue because
four wasnβt in the domain of our original function. And it certainly isnβt in the open
interval from one to three. The derivative does in fact exist
everywhere in this open interval. Therefore, the only critical
numbers come from when the derivative is zero. Letβs find out when this happens by
solving ππ¦ by ππ₯ equals zero.

We solve this by writing the
derivative as one fraction. Itβs simple to write the two
fractions we have with a common denominator, and so combine them. Now, this fraction is going to be
equal to zero exactly when the numerator is equal to zero, assuming that the
denominator is not also equal to zero at the same time. We expand and simplify the
numerator getting π₯ squared minus eight π₯ plus 12, which we can factor by
inspection.

Now that the numerator is factored,
we can see that the derivative is zero when π₯ is six or π₯ is two. These are therefore our critical
numbers. But remember, we are only
interested in the critical numbers in the open interval from one to three. Six is not in the interval. And so the only critical number in
the open interval from one to three is two. Our first line of working becomes
evaluate π₯ over four plus one over π₯ minus two at two. Letβs clear some room so we can do
this.

To find the value of π¦ when π₯ is
two, we just substitute two for π₯. So π¦ is two over four plus one
over two minus four. And this simplifies to zero. We have completed the first step of
the closed interval method. We have evaluated our function at
its critical numbers in the open interval from π to π. We also need to evaluate our
function π at the end points π and π of the closed interval.

π and π are one and three,
respectively. Letβs evaluate then, starting with
π₯ equals one. π¦ is then one over four plus one
over one minus four which is negative a twelfth. How about for the other endpoint
when π₯ is three? There, π¦ is three over four plus
one over three minus four which simplifies to negative a quarter. What do we do now? Well, we look at the closed
interval method. And it says that the largest value
obtained is the absolute maximum and the smallest value obtained is the absolute
minimum. The largest of our three values β
zero, negative a twelfth, and negative a quarter β is zero. And the smallest value we obtained
is negative a quarter.

What do we conclude then? The absolute maximum value of our
function π¦ equals π₯ over four plus one over π₯ minus four on the closed interval
from one to three is zero. And the absolute minimum of our
function on this interval is negative a quarter. This was a relatively
straightforward application of the closed interval method. While our function wasnβt
continuous on the set of real numbers, it was continuous on the interval that we
cared about, the closed interval from one to three. And so we could apply the
method.

We found the critical members of
our function by differentiating and finding which numbers made the derivative
zero. There were no numbers in the
interval for which the derivative does not exist. We found that only one of the
critical numbers, two, was in the open interval from one to three. And we evaluated our function at
this critical number as well as the two endpoints one and three of the closed
interval on which the function was defined.

The closed interval method told us
that the largest value we obtained, zero, is the absolute maximum of our function
over our interval and the smallest, negative a quarter, is the absolute minimum.