### Video Transcript

For zero is less than π₯ which is
less than two π by five, find the intervals on which π of π₯ equals cos squared
five π₯ plus three cos five π₯ is increasing or decreasing.

We recall, first of all, that a
function is increasing whenever its first derivative π prime of π₯ is greater than
zero. And that same function is
decreasing whenever its first derivative π prime of π₯ is less than zero. We therefore need to find an
expression for the first derivative π prime of π₯ of this trigonometric
function. And we recall, first of all, a
standard result for differentiating cos of ππ₯, which is that its derivative with
respect to π₯ is equal to negative π sin ππ₯. This allows us to differentiate the
second term. The derivative of three cos five π₯
is three multiplied by negative five sin five π₯. But what about the first term?

Well, we can think of it as cos of
five π₯ all squared and then recall the general power rule. This tells us that if we have some
function to a power, then its derivative is equal to that power, so thatβs two,
multiplied by the derivative of the function itself, so that will be negative five
sin five π₯, multiplied by that function to one less power. So, we reduce the power from two to
one.

We therefore have π prime of π₯ is
equal to two multiplied by negative five sin five π₯ cos five π₯ plus three
multiplied by negative five sin five π₯. We can factor by negative five sin
five π₯ to give π prime of π₯ is equal to negative five sin five π₯ multiplied by
two cos five π₯ plus three. Our Function π will, therefore, be
increasing when this first derivative is greater than zero. Now, letβs think about how we can
solve this inequality. And weβll think about that second
bracket first of all.

The graph of cos five π₯, first of
all, is just a horizontal stretch of the graph of cos π₯. And so, it still has negative one
as its minimum value and one as its maximum value. The graph of two cos five π₯ is a
vertical stretch of this graph by a scale factor of two. And so, this will have negative two
as its minimum and two as its maximum. Adding three is a vertical
translation of this graph, which means that the minimum value for two cos five π₯
plus three will be one, and the maximum value will be five.

What this tells us is that two cos
five π₯ plus three itself is always greater than zero, as its minimum value is
one. And therefore, one of the factors
in our product is always positive. In order for the product of two
factors to be positive, they must have the same sign. And therefore, it must also be the
case that π is increasing when the first factor, negative five sin five π₯, is
itself positive. So, our problem has reduced
somewhat. Weβre now just looking for the
region over which negative five sin five π₯ is greater than zero.

We can simplify by dividing both
sides by negative five. And as weβre dividing by a
negative, we must reverse the inequality, to give sin five π₯ is less than zero. Now, remember the domain we were
given for this function was zero is less than π₯ is less than two π by five. If we let π’ equal five π₯, then if
π₯ is between zero and two π by five, π’ will be between zero and two π. So, now, weβre just looking for
where sin π’ is less than zero for values of π’ between zero and two π.

We can answer this by sketching a
graph of π’ against sin π’ for values of π’ between zero and two π. And we see that sin π’ is less than
zero for values of π’ between π and two π. Remember though that π’ is equal to
five π₯, so to convert this back to an inequality in π₯, we need to divide by five,
giving π over five is less than π₯ is less than two π over five. This is the interval on which the
function π is increasing.

By applying the same logic, we see
that π will be decreasing when its first derivative is less than zero, which in
turn leads to sin π’ being greater than zero. Thatβs when π’ is between zero and
π, which leads to π₯ being between zero and π by five. So, weβve completed the
problem. The function π of π₯ is increasing
on the open interval π by five, two π by five and decreasing on the open interval
zero, π by five.