### Video Transcript

In this video, we’ll learn what it
means for a function to be either increasing or decreasing on a given interval. And we’ll see how to determine
whether a function is increasing or decreasing on a particular interval using
derivatives. You should be familiar with how to
differentiate polynomial, trigonometric, exponential, and logarithmic functions, and
also how to differentiate combinations of these using the product, quotient, and
chain rules.

Firstly then, let’s look at what
these terms, increasing and decreasing, mean in relation to functions. The formal definition for a
function to be decreasing on an interval is as follows. A function is decreasing on an
interval 𝐼 if 𝑓 of 𝑥 one is greater than 𝑓 of 𝑥 two whenever 𝑥 one is less
than 𝑥 two for any two points 𝑥 one and 𝑥 two in the interval 𝐼. If we consider the left-hand
portion of the quadratic graph I’ve drawn, we can consider two points 𝑥 one and 𝑥
two where 𝑥 one is less than 𝑥 two. We see that 𝑓 of 𝑥 one is greater
than 𝑓 of 𝑥 two. And therefore, our function would
be considered to be decreasing on this interval.

In practical terms though, what
this means is that the slope of the graph of our function is negative, which makes
sense. If the value of the function is
decreasing, getting smaller, then the function must be sloping downwards. If we recall that the slope of a
function is given by its first derivative, then we can form an alternative
definition. A function is decreasing on an
interval 𝐼 if its first derivative 𝑓 prime of 𝑥 is less than zero, that is
negative, for all 𝑥-values in the interval 𝐼.

Here, we see the link to
derivatives. If we can find the first derivative
of a function 𝑓 prime of 𝑥, we can then consider where this derivative is negative
in order to determine the intervals over which the function is decreasing. We can consider the definition of
an increasing function in the same way. Formally, first of all, a function
is increasing on an interval 𝐼 if 𝑓 of 𝑥 one is less than 𝑓 of 𝑥 two whenever
𝑥 one is less than 𝑥 two for any values 𝑥 one and 𝑥 two in the interval 𝐼. This time, we see that larger
values of 𝑥 are associated with larger values of the function itself. So, our function is increasing as
the 𝑥-values increase.

In practical terms, this means that
the slope of the graph of our function will be positive. The graph will be sloping
upwards. Again, recalling that the first
derivative of a function gives its slope, we see that a function will be increasing
on an interval 𝐼 if the first derivative 𝑓 prime of 𝑥 is greater than zero for
all 𝑥-values in that interval 𝐼. Let’s now see how we can apply our
definition of increasing and decreasing functions in terms of their first
derivatives to some problems.

Given that 𝑓 of 𝑥 equals five 𝑥
squared minus three 𝑥 minus the natural log of 𝑥, find the intervals on which 𝑓
is increasing or decreasing.

First, we recall that a function is
increasing when its first derivative 𝑓 prime of 𝑥 is greater than zero, and a
function is decreasing when its first derivative 𝑓 prime of 𝑥 is less than
zero. We’re therefore going to need an
expression for the first derivative of this function. We can differentiate
term-by-term. The derivative of five 𝑥 squared
is five multiplied by two 𝑥; that’s 10𝑥. The derivative of negative three 𝑥
is negative three. And the derivative of negative the
natural log of 𝑥 is negative one over 𝑥. So, we have our first
derivative. 𝑓 prime of 𝑥 equals 10𝑥 minus
three minus one over 𝑥.

By our definition of an increasing
function, first of all, 𝑓 will be increasing when its first derivative 10𝑥 minus
three minus one over 𝑥 is greater than zero. And therefore, we have an
inequality in 𝑥 that we need to solve. Now, we know that there is an 𝑥 in
the denominator of this fraction here, so the step we’d like to take first of all is
to multiply by 𝑥 in order to eliminate this fraction. But we need to be a little bit
careful because we have an inequality and no guarantee that 𝑥 is positive. If we multiply by a negative
𝑥-value, then we would need to reverse the direction of our inequality.

However, if we look back at our
original function, we see that it includes this term the natural log of 𝑥. And the natural logarithm of 𝑥 is
undefined for 𝑥-values less than or equal to zero. This means that the domain of our
function 𝑓 of 𝑥 is 𝑥 greater than zero. We’re only working with positive
values of 𝑥. And therefore, we can multiply our
inequality by 𝑥 without worrying about needing to change the direction of the
inequality sign.

Multiplying by 𝑥 gives 10𝑥
squared minus three 𝑥 minus one is greater than zero. And we see that we have a quadratic
inequality which we need to solve. There are a number of different
methods that we can use, but almost certainly we need to factorise to begin
with. By following the formal method of
factoring by grouping, all with a bit of trial and error, we see that this quadratic
factors as five 𝑥 plus one multiplied by two 𝑥 minus one.

We then need to find the critical
values for this quadratic, which we do by setting each of our two brackets equal to
zero, not greater than zero. We then solve each linear equation
to give 𝑥 equals negative one-fifth and 𝑥 equals one-half. So, these are the two critical
values for this quadratic. Now, there are two ways that we can
proceed from here. One is to use a table of values to
check the sign of our quadratic either side and in-between our critical values. The other is to sketch a graph. And that’s the one that I’m going
to choose to demonstrate.

We know that we have a quadratic
with a positive leading coefficient. So, its graph will be a
parabola. And we know the critical values,
which are the values at which the graph crosses the 𝑥-axis, are negative one-fifth
and one-half. So, the graph looks like this. Remember, this is the graph of our
first derivative, 10𝑥 minus three minus one over 𝑥. We said that our function 𝑓 will
be increasing when its first derivative 𝑓 prime of 𝑥 is greater than zero. That is when the graph of its
derivative 𝑓 prime of 𝑥 is above the 𝑥-axis.

This will correspond to two
sections of our graph, the section where 𝑥-values are less than negative one-fifth
and the part where 𝑥-values are greater than one-half. But remember, we said that the
domain of our function 𝑓 of 𝑥 was just 𝑥 greater than zero. And therefore, we can actually
ignore one-half of our graph completely. We can say then that our function
𝑓 is increasing on the open interval one-half, infinity. That’s all 𝑥-values greater than
one-half.

To see where our function is
decreasing, we’re looking at where the first derivative 𝑓 prime of 𝑥 is less than
zero, which means we’re looking at where its graph is below the 𝑥-axis. Now, on our original graph, this
would’ve been everywhere between the two critical values. But as we’ve reduced the graph to
be only 𝑥-values greater than zero, this is for all 𝑥-values greater than zero but
less than one-half. We therefore say that 𝑓 is
decreasing on the open interval zero, one-half.

So, we’ve completed the
problem. We had to differentiate the
function 𝑓 of 𝑥 to find its first derivative 𝑓 prime of 𝑥, and then use our
knowledge of quadratic inequalities to find where 𝑓 prime of 𝑥 was greater than
zero and where 𝑓 prime of 𝑥 was less than zero.

Now, we may also need to apply key
rules of differentiation such as the chain rule, product rule, or quotient rule in
order to answer questions involving more complex functions. Let’s see an example of this.

Determine the intervals on which
the function 𝑓 of 𝑥 equals seven 𝑥 over 𝑥 squared plus nine is increasing and
where it is decreasing.

We recall, first of all, that
whether a function is increasing or decreasing can be determined by considering its
derivative. A function will be increasing when
its first derivative is positive, and it will be decreasing when its first
derivative is negative. We therefore need to find an
expression for 𝑓 prime of 𝑥. We note, first of all, that 𝑓 is a
quotient. So, in order to find this
derivative, we’re going to need to apply the quotient rule.

The quotient rule tells us that for
two differentiable functions 𝑢 and 𝑣, the derivative with respect to 𝑥 of their
quotient, 𝑢 over 𝑣, is equal to 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all
over 𝑣 squared. We therefore let 𝑢 equal the
numerator of our quotient, that’s seven 𝑥, and 𝑣 equal the denominator, that’s 𝑥
squared plus nine. d𝑢 by d𝑥 and d𝑣 by d𝑥 can each be found using the power rule
of differentiation. d𝑢 by d𝑥 is seven, and d𝑣 by d𝑥 is two 𝑥.

Substituting into the formula for
the quotient rule, we have then that 𝑓 prime of 𝑥 is equal to 𝑥 squared plus nine
multiplied by seven minus seven 𝑥 multiplied by two 𝑥 all over 𝑥 squared plus
nine all squared. Distributing the parentheses in the
numerator gives seven 𝑥 squared plus 63 minus 14𝑥 squared all over 𝑥 squared plus
nine all squared. Which then simplifies to 63 minus
seven 𝑥 squared over 𝑥 squared plus nine all squared. And so, we have our expression for
the first derivative.

Our function 𝑓 will be increasing
when its first derivative is greater than zero. So, we have an inequality in 𝑥
that we need to solve. Now, we can actually simplify this
somewhat. Note that the denominator of this
fraction is something squared, 𝑥 squared plus nine all squared. And therefore, the denominator
itself will always be greater than zero. In order for the whole fraction to
be greater than zero then, we only need to ensure that its numerator is greater than
zero because a positive divided by a positive will give something which is
positive.

The inequality, therefore,
simplifies to 63 minus seven 𝑥 squared is greater than zero. We can divide through by seven and
then add 𝑥 squared to each side to give nine is greater than 𝑥 squared. Or written the other way around, 𝑥
squared is less than nine. So, we have a relatively
straightforward quadratic inequality to solve. If 𝑥 squared must be less than
nine, that’s strictly less than nine, then we can have any 𝑥-value between negative
three and three, although not including these values themselves. The solution to this quadratic
inequality then is negative three is less than 𝑥 is less than three, or the open
interval negative three to three.

So, we found the only interval on
which the function 𝑓 of 𝑥 is increasing. To determine where the function is
decreasing, we require its first derivative to be less than zero, which in turn
leads to 63 minus seven 𝑥 squared is less than zero. We therefore reverse the direction
of all the inequality signs in our previous working out leading to 𝑥 squared is
greater than nine. This is only the case for 𝑥-values
strictly less than negative three and 𝑥-values strictly greater than positive
three. So, we find that there were two
intervals on which our function is decreasing. The open intervals negative
infinity to negative three and three, infinity.

So, by applying the quotient rule
to find the first derivative of our function 𝑓 of 𝑥, and then solving a relatively
straightforward quadratic inequality. We find that the function 𝑓 of 𝑥
is increasing on the open interval negative three to three and decreasing on the
open intervals negative infinity to negative three and three, infinity.

In our next example, we’ll consider
a problem involving trigonometric functions.

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

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

Well, we can think of it as cos of
five 𝑥 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 that’s two,
multiplied by the derivative of the function itself, so that will be negative five
sin five 𝑥, multiplied by that function to one less power. So, we reduce the power from two to
one.

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

The graph of cos five 𝑥, first of
all, is just a horizontal stretch of the graph of cos 𝑥. And so, it still has negative one
as its minimum value and one as its maximum value. The graph of two cos five 𝑥 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 𝑥
plus three will be one, and the maximum value will be five.

What this tells us is that two cos
five 𝑥 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 𝑓 is increasing when the first factor, negative five sin five 𝑥, is
itself positive. So, our problem has reduced
somewhat. We’re now just looking for the
region over which negative five sin five 𝑥 is greater than zero.

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

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

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

In summary then, we’ve seen that a
function 𝑓 is increasing whenever its first derivative 𝑓 prime of 𝑥 is greater
than zero. And the function 𝑓 is decreasing
whenever its first derivative 𝑓 prime of 𝑥 is less than zero. We can use differentiation and the
rules of differentiation such as the quotient rule, product rule, and chain rule in
order to find the first derivatives of functions. And then, solve the resulting
inequalities to determine the intervals on which those functions are increasing or
decreasing.