### Video Transcript

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

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

By our definition of an increasing
function, first of all, π will be increasing when its first derivative 10π₯ minus
three minus one over π₯ is greater than zero. And therefore, we have an
inequality in π₯ that we need to solve. Now, we know that there is an π₯ in
the denominator of this fraction here, so the step weβd like to take first of all is
to multiply by π₯ in order to eliminate this fraction. But we need to be a little bit
careful because we have an inequality and no guarantee that π₯ is positive. If we multiply by a negative
π₯-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 π₯. And the natural logarithm of π₯ is
undefined for π₯-values less than or equal to zero. This means that the domain of our
function π of π₯ is π₯ greater than zero. Weβre only working with positive
values of π₯. And therefore, we can multiply our
inequality by π₯ without worrying about needing to change the direction of the
inequality sign.

Multiplying by π₯ gives 10π₯
squared minus three π₯ 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 π₯ plus one multiplied by two π₯ 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 π₯ equals negative one-fifth and π₯ 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 thatβs the one that Iβm 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 π₯-axis, are negative one-fifth
and one-half. So, the graph looks like this. Remember, this is the graph of our
first derivative, 10π₯ minus three minus one over π₯. We said that our function π will
be increasing when its first derivative π prime of π₯ is greater than zero. That is when the graph of its
derivative π prime of π₯ is above the π₯-axis.

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

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

So, weβve completed the
problem. We had to differentiate the
function π of π₯ to find its first derivative π prime of π₯, and then use our
knowledge of quadratic inequalities to find where π prime of π₯ was greater than
zero and where π prime of π₯ was less than zero.