### Video Transcript

Determine the intervals of increase
and decrease of the piecewise-defined function π of π₯ is equal to negative nine π₯
if π₯ is less than zero and π of π₯ is equal to nine π₯ if π₯ is greater than or
equal to zero.

In this question, weβre given a
piecewise-defined function π of π₯. And weβre asked to determine the
intervals in increase and decrease of this function. And we can recall we say a function
is increasing on an interval if taking a larger input value on the interval
increases its output. And we say a function is decreasing
on an interval if taking a larger input decreases its output on the interval. And this in fact gives us two ways
of determining the intervals in increase and decrease of this function.

The first way we can do this is to
sketch a graph of π¦ is equal to π of π₯. We can do this by sketching each
subfunction separately. Letβs start by sketching π¦ is
equal to negative nine π₯ where π₯ must be less than zero. First, this is a straight line
written in slopeβintercept form. We can see its π¦-intercept is at
zero and its slope is negative nine. But remember, weβre only sketching
this subfunction for values of π₯ less than zero. So we need to have a hollow dot at
this π¦-intercept. This gives us the following
sketch.

We now need to sketch the graph of
the second subfunction. Thatβs π¦ is equal to nine π₯ for
values of π₯ greater than or equal to zero. Once again, this is a straight line
in slopeβintercept form. Its π¦-intercept is zero and its
slope is nine. Since π₯ is equal to zero is
included in this subfunction, we need to include the π¦-intercept which is at the
origin. This gives us a sketch which looks
like the following.

We can use this to determine the
intervals of increase and decrease of our function. Letβs start with the intervals of
increase. The intervals of increase of our
function are the intervals where taking a larger input value increases the
output. In the diagram, this will be any
interval where our function is sloping upwards. Since this is a piecewise function
made up of two straight lines, we can see this will only be true on the interval
where π₯ is positive. So our function is only increasing
when π₯ is positive. We want to write this as an
interval. Thatβs the open interval from zero
to β.

We can do exactly the same to
determine the interval of decrease from the sketch. Once again, an interval of decrease
is an interval where taking a larger input value of π₯ decreases the output value of
the function. In the graph, this will be anywhere
where our function is sloping downwards. And we can see on our diagram this
is true when π₯ is negative. Therefore, from our sketch, we can
see the interval of decrease of this function is the open interval from negative β
to zero. This is then enough to answer our
question. We can say that our function π of
π₯ is increasing on the open interval from zero to β and decreasing on the open
interval from negative β to zero.

We could stop here. However, there is a second method
we can use to answer this question. We can notice that π of π₯ is a
piecewise linear function. In particular, since linear
functions are differentiable, this means that π of π₯ is a piecewise differentiable
function. And we know that we can determine
the intervals of increase and decrease of a differentiable function by using the
sign of its derivative. In particular, if π prime of π₯ is
positive on an interval πΌ, then we can say that π is increasing on this interval
πΌ. And if π prime of π₯ is negative
on an interval πΌ, then π is decreasing on πΌ.

And itβs worth noting we can see
this on the graph we sketched. When the function has a negative
slope, its outputs are decreasing. And when our function has a
positive slope, its outputs are increasing. So we want to find an expression
for π prime of π₯. We can do this by differentiating
each of the subfunctions separately since this is a piecewise differentiable
function.

However, there is one small thing
we need to note. We do need to consider what happens
at the endpoints of our subdomains. In particular, in this case, we can
know our function wonβt be differentiable when π₯ is zero. And this is because as π₯
approaches zero from the left, its derivative is negative nine. However, as π₯ approaches zero from
the right, its derivative is positive nine. And since the limit as π₯
approaches zero from the left and the limit as π₯ approaches zero from the right of
this function are not equal, its limit as π₯ approaches zero does not exist. So we donβt include the endpoint of
this interval.

π prime of π₯ is negative nine if
π₯ is less than zero. And π prime of π₯ is nine if π₯ is
greater than zero. And we can then answer the question
directly from this function. π prime of π₯ is positive when π₯
is greater than zero, so itβs increasing when π₯ is greater than zero. And π prime of π₯ is negative when
π₯ is less than zero, so π prime of π₯ is decreasing when π₯ is less than zero. And this gives us a second method
of answering this question.

Therefore, we were able to show π
of π₯ is increasing on the open interval from zero to β and is decreasing on the
open interval from negative β to zero.