### Video Transcript

Determine the intervals on which
the function π of π₯ equals three π₯ cubed minus nine π₯ squared minus four is
increasing and on which it is decreasing.

By definition, a function is
increasing on an interval πΌ if π of π₯ one is less than π of π₯ two whenever π₯
one is less than π₯ two for π₯ one and π₯ two in the interval πΌ. And it is decreasing on πΌ if π of
π₯ one is greater than π of π₯ two whenever π₯ one is less than π₯ two for π₯ one
and π₯ two in πΌ. Thatβs the formal definition. But really the best way to think
about increasing and decreasing functions is using their graphs.

The graph of an increasing function
looks like this: π¦ increases as π₯ increases. Theyβre not always at the same
rate. And the graph of the function which
is decreasing on the interval looks like this. Now letβs turn to the problem we
have. We have to determine the intervals
on which our function is increasing and on which it is decreasing. We could try to graph this function
using a graphing calculator or graphing software. And in fact, weβll do that at the
end of the video.

But the way weβre going to solve
this question is using a test. If the derivative of π, π prime
of π₯, is greater than zero on an interval, πΌ then π is increasing on πΌ. And if the derivative of π, π
prime of π₯, is less than zero on an interval πΌ, then π is decreasing on that
interval πΌ. And this test makes sense given the
graphical interpretation above. A tangent to the increasing
function always has positive slope, so π prime is positive. But the tangents to the decreasing
function have negative slope, so π prime is negative on this interval for this
decreasing function.

Letβs clear away the definition and
graphical interpretation so we have room to apply our test. The test involves π prime, which
is the derivative of our function. So letβs find π prime. We differentiate the polynomial
term by term, using the fact that the derivative of ππ₯ to the π with respect to
π₯ is π times π times π₯ to the π minus one. The derivative of three π₯ to the
power of three comes from multiplying the exponent and coefficient to get nine and
then reducing the exponent by one to get π₯ squared.

Similarly, the derivative of nine
π₯ squared is 18π₯, and the derivative of four is zero. So π prime of π₯ is nine π₯
squared minus 18π₯. π is increasing on intervals for
which this derivative is greater than zero and decreasing on intervals for which the
derivative is less than zero. So we need to find the sign of π
prime. The way we do this is to factor π
prime, writing it as nine π₯ times π₯ minus two.

And from this factored form, we can
easily see that π prime is zero at π₯ equals zero and π₯ equals two. These two numbers are called the
critical numbers of the function π. For π prime to change sign from
positive to negative or negative to positive, it must pass through zero. And so we know that π prime must
have the same sign on the interval π₯ is less than zero. If π prime changed sign below
zero, because itβs continuous, there will be a value of π₯ below zero at which π
prime was zero. And weβve seen that π prime is
only zero at π₯ equals zero and π₯ equals two.

Similarly, π prime canβt change
sign between zero and two. Similarly, π prime canβt change
sign when π₯ is greater than two. If π prime does change sign
anywhere, itβs got to be at zero or two. We want to find the sign of π
prime on these intervals. And we do that by considering the
signs of its factors. So along the top of the table, we
have the two factors of π prime: nine π₯ and π₯ minus two and π prime itself.

Letβs start filling the table
in. What is the sign of nine π₯ when π₯
is less than zero? Nine π₯ is negative when π₯ is less
than zero. You can see this by multiplying
both sides of the inequality, which defines this interval, by nine. What can we say about nine π₯ when
π₯ is between zero and two? In a similar way, we see that nine
π₯ is between zero and 18; in particular, its positive. And finally for the third interval,
we see that is nine π₯ is greater than 18 and so is positive.

We do the same thing for the factor
π₯ minus two. In the first region, it is less
than negative two and so negative. In the second region, it is between
negative two and zero, and so itβs negative here too. And finally in the third region,
itβs positive. We found the sign of the factors of
π prime. Now we just need to find the sign
of π prime itself in each of these regions. In the first region, π₯ is less
than zero, both factors are negative. And so π prime is negative times
negative, which is a positive.

Similarly, in the second region, π
prime is a positive times a negative quantity, which makes it a negative
quantity. And finally, in the third region,
π prime is a positive times a positive, and so it is a positive quantity. Referring back to the increasing
decreasing test, if π prime is positive on πΌ then π is increasing on πΌ; and if
π prime is negative on πΌ, then π is decreasing on πΌ.

So π is increasing on these two
intervals and decreasing on this one. The only thing left to do is to
write down our final answer: π₯ is increasing on the intervals where π₯ is less than
zero and where π₯ is greater than two and is decreasing on the interval where π₯ is
between zero and two. Letβs write these using interval
notation. π₯ is less than zero represents the
open interval between negative infinity and zero. π₯ is greater than two is the open
interval from two to infinity. And zero is less than π₯ is less
than two represents the open interval from zero to two.

So hereβs our answer: π is
increasing on the open interval from negative infinity to zero, and it is also
increasing on the open interval from two to infinity; π is, however, decreasing on
the open interval from zero to two. We solved this problem using the
increasing/decreasing test, which uses the sign of π prime to tell where π is
increasing and where it is decreasing.