### Video Transcript

Which of the following statements
is true for the function π of π₯ is equal to negative π₯ minus six all cubed? Is it option (A) π of π₯ is
increasing on the set of real numbers? Option (B) π of π₯ is decreasing
on the set of real numbers. Option (C) π of π₯ is increasing
on the open interval from negative β to negative six and decreasing on the open
interval from negative six to β. Or is it option (D) π of π₯ is
increasing on the open interval from negative six to β and decreasing on the open
interval from negative β to negative six?

In this question, weβre given four
statements involving the monotonicity of a given function π of π₯. π of π₯ is equal to negative π₯
minus six all cubed. We need to determine which of these
statements is true. To do this, we can start by
recalling for a differentiable function π of π₯, we can determine the intervals
which are decreasing and increasing by considering its derivative.

In particular, we recall if π is a
differentiable function and if π prime of π₯ is greater than zero on an open
interval from π to π, then we can conclude that π of π₯ is increasing on the open
interval from π to π. Similarly, if π prime of π₯ is
less than zero on an open interval from π to π, then π of π₯ is decreasing on the
open interval from π to π. This is particularly useful because
we can see that our function π of π₯ is differentiable for all real numbers. This is because weβre taking the
cube of a linear function. So π of π₯ is a cubic
polynomial. We can then use these results to
determine the intervals on which π of π₯ is increasing and decreasing. We just need to determine an
expression for π prime of π₯.

And thereβs a few different ways we
can find π prime of π₯. For example, we could use the
binomial formula to distribute our exponent over our parentheses. Or we could expand by using the
FOIL method. Either case, we would end up with a
cubic polynomial. We could differentiate this term by
term by using the power rule for differentiation. This is not the only way we can
differentiate this function. We can also notice that π of π₯ is
the composition of two functions. We take a linear function and then
we cube this value. And we can differentiate the
composition of two functions by using the chain rule.

And in particular, because our
outer function is a power function, we can differentiate this by using the general
power rule, which is an application of the chain rule. This tells us the derivative of π
of π₯ raised to the πth power is equal to π times π prime of π₯ multiplied by π
of π₯ to the power of π minus one, where our value of π is a real constant and π
of π₯ is a differentiable function. Weβll use this method to
differentiate the function. However, itβs personal preference
which one we want to use.

So weβll set our value of π equal
to three and our inner function π of π₯ to be negative π₯ minus six. And itβs worth noting π of π₯ is a
linear function. So π prime of π₯ will be the
coefficient of π₯, which is negative one. We can now evaluate π prime of π₯
by substituting these values into the general power rule. π prime of π₯ is equal to three
times negative one multiplied by negative π₯ minus six raised to the power of three
minus one. We can then simplify this
expression. π prime of π₯ is equal to negative
three times negative π₯ minus six squared.

And now, we can notice something
interesting. Negative π₯ minus six all squared
is a square. This means itβs greater than or
equal to zero for any value of π₯. In particular, thereβs only one
input value of π₯ where this derivative will be equal to zero. Thatβs when π₯ is equal to negative
six. We then multiply this by a negative
constant. And remember, a negative multiplied
by a positive is a negative. Therefore, we have shown that π
prime of π₯ is less than or equal to zero for all values of π₯. And in particular, π prime of π₯
is equal to zero only when π₯ is equal to negative six.

We can then use our property. The derivative of π of π₯ is
negative for all values of π₯ except when π₯ is negative six. This then tells us the function is
decreasing on the open interval from negative β to negative six and the open
interval from negative six to β. However, since our function is
defined at negative six and its derivative is zero only at negative six, we can also
say itβs decreasing at this point. In other words, weβve shown the
function π of π₯ is equal to negative π₯ minus six all cubed is decreasing for all
real values. This is option (B).