### Video Transcript

Determine the intervals on which
the function π of π₯ equals negative one plus one over π₯ minus four over π₯
squared is increasing and decreasing.

We begin by recalling how we
determine whether a function is increasing or decreasing at a given point. We say that a function π of π₯ is
increasing at a given point π₯ if its first derivative at this point is greater than
zero; thatβs π prime of π₯. Similarly, the function will be
decreasing at this point if the value of its first derivative here is less than
zero. If itβs negative. And so we see weβre going to need
to begin by finding the first derivative of our function. Thatβs π prime of π₯. Our function π of π₯ consists of
three terms, so weβre simply going to differentiate term by term.

Before we do, though, we begin by
rewriting one over π₯ as π₯ to the power of negative one and negative four over π₯
squared as negative four times π₯ to the power of negative two. And this is really useful because
we know, to differentiate a power term of the form ππ₯ to the πth power where π
and π are real constants, we multiply the entire term by the exponent, then reduce
that exponent by one. And of course, the first derivative
of any constant is zero. So differentiating negative one
with respect to π₯, and we get zero. Then when we multiply the entire
term of π₯ to the power of negative one by the exponent and reduce that exponent by
one, we get negative one times π₯ the power of negative two.

Similarly, letβs multiply four π₯
to the power of negative two by negative two but then reduce the exponent by
one. So our third term is negative
negative two times four π₯ to the power of negative three. Weβll forget about the zero, and
weβll rewrite negative one π₯ to the power of negative two as negative one over π₯
squared. Then our third term is negative
negative two times four. So thatβs eight. And similarly, we can write π₯ to
the power of negative three as one over π₯ cubed. And so π prime of π₯ is negative
one over π₯ squared plus eight of π₯ cubed. Before we can find the values of π₯
such that this is greater than or less than zero, weβre going to begin by adding our
fractions. And weβll do so by creating a
common denominator.

To find the common denominator, we
multiply the numerator and denominator of our first fraction by π₯. So we now have negative π₯ over π₯
cubed plus eight over π₯ cubed. And since the denominators are
equal, we can simply add the numerators. And so we see that π prime of π₯
is negative π₯ plus eight all over π₯ cubed. Now remember, our job is to find
the values of π₯ such that this is either positive or negative. And here, we can use a little bit
of logic to do so. Letβs begin by looking for
intervals of increase of our function, in other words, values of π₯ such that the
first derivative is greater than zero. Now of course, our first
derivative, π prime of π₯, is a fraction. Itβs negative π₯ plus eight all
over π₯ cubed.

So how do we ensure this fraction
is positive? Well, either we need to be dividing
a positive by a positive, so we need the numerator and denominator of our fraction
to both be positive. Or, alternatively, to achieve a
positive number, we divide a negative by a negative. So we need negative π₯ plus eight
to be less than zero and π₯ cubed to be less than zero.

Letβs begin with the first
scenario. We have negative π₯ plus eight is
greater than zero and π₯ cubed is greater than zero. The interval of increase here will
be the intersection of the solutions to these two inequalities. Letβs solve this first inequality
by adding π₯ to both sides. So π₯ is less than eight. We simply take the cube root of
both sides of this inequality, and we find π₯ is greater than zero. The intersection of the solutions
to these two inequalities are values of π₯ greater than zero and less than
eight. And so weβre not using interval
notation just yet. But we can see our function is
increasing for these values of π₯.

Now, of course, we did say that a
negative number divided by a negative number will yield a positive result. So our first derivative will also
be greater than zero if negative π₯ plus eight is less than zero and π₯ cubed is
also less than zero. Itβs negative. This time, though, when we solve
these two inequalities, we find π₯ is greater than eight and less than zero. There are no values of π₯ which
satisfy both of these inequalities at the same time. In other words, there are no values
of π₯ in the intersection of the solutions to these inequalities. And we, therefore, say that the
function is increasing when π₯ is greater than zero and less than eight.

Weβll now consider the values of π₯
such that π prime of π₯ is less than zero. These will tell us the values of π₯
on which our function is decreasing. Once again, we have negative π₯
plus eight over π₯ cubed. And we want that to be less than
zero, to be negative. Well, we know a positive number
divided by a negative number or vice versa will give us a negative number. So either our numerator has to be
negative and our denominator has to be positive or our numerator needs to be
positive and our denominator needs to be negative.

Weβll consider first the situation
where negative π₯ plus eight is positive and the denominator, π₯ cubed, is
negative. Solving these two inequalities and
we find π₯ is less than eight and π₯ is less than zero. Well, the intersection of solutions
to these inequalities, in other words, the values of π₯ that satisfy both, are
values of π₯ less than zero. But of course, we said we needed to
consider the scenario where our numerator is negative and our denominator is
positive. In this case, we get π₯ is greater
than eight and π₯ is greater than zero. Weβll consider again the
intersection of solutions. Well, the values of π₯ which
satisfy both of these inequalities are π₯ is greater than eight.

And of course, these are the values
of π₯ where our function is decreasing. Now, we do need to be a little bit
careful. If π₯ is equal to zero, our
derivative becomes eight divided by zero, which we know to be undefined. And weβre therefore interested in
the existence of any critical points. However, the value π₯ is equal to
zero is not included in any of our intervals for increase or decrease.

So in this case actually, we donβt
need to worry about it. We use the curly brackets to
represent our intervals. We say that itβs increasing on the
open interval from zero to eight. Those are values of π₯ between zero
and eight, but not including zero and eight. And itβs decreasing on the open
interval from negative β to zero and eight to β.