### Video Transcript

Consider a function whose first
derivative is π prime of π₯ equals π₯ plus 11 times π₯ minus seven over π₯ minus
three. On which interval or intervals is
π increasing?

Letβs consider what it actually
means for a function to be increasing. A function is increasing when its
rate of change is positive. In terms of its derivatives, we say
that a function is increasing when its first derivative is greater than zero.

Now, weβve actually been given the
first derivative of our function π. Weβre told that π prime of π₯ is
π₯ plus 11 times π₯ minus seven over π₯ minus three. Our job then is to find the value
or values of π₯ such that this is greater than zero.

Now, weβll need to be really
careful. This is a rational function. So for the quotient of two
functions to be greater than zero, both functions themselves must be greater than
zero. Or both functions must be less than
zero since we know a negative divided by a negative is a positive.

Letβs look at the first
scenario. Weβre interested in the values of
π₯ such that π₯ plus 11 times π₯ minus seven is greater than zero and π₯ minus three
is greater than zero. In other words, weβre looking for
the intersection of the solutions. Now, one way we can find the values
of π₯ such that π₯ plus 11 times π₯ minus seven is greater than zero is to first
consider where this is equal to zero. Well, for the product of two values
to be zero, either one of those values must themselves be zero. So we can say that π₯ plus 11 times
π₯ minus seven will be equal to zero when π₯ plus 11 is equal to zero or π₯ minus
seven is equal to zero. Solving each equation, and we see
that this is when π₯ is equal to negative 11 and π₯ is equal to seven.

We know that the graph of π¦ equals
π₯ plus 11 times π₯ minus seven is a quadratic curve. And we now know it passes through
the π₯-axis at negative 11 and seven. The graph is greater than zero. And therefore, π₯ plus 11 times π₯
minus seven is greater than zero in these two locations. So, in other words, when π₯ is less
than negative 11 or greater than seven.

But if we go back to our other
inequality β π₯ minus three is greater than zero β and we add three to both sides,
we see that π₯ must be greater than three. Remember, weβre looking for the
intersection of these solutions. Well, π₯ cannot be greater than
three and less than negative 11. It also canβt purely be greater
than three and greater than seven. So the intersection is in fact π₯
is greater than seven.

In interval notation then, we say
that π is increasing on the open interval from seven to positive β. But remember, we also said we
needed to look for the values of π₯ such that both the numerator and denominator of
our fraction are negative. Well, this time, itβs the other
part of the graph weβre interested in that will tell us the values of π₯ such that
π₯ plus 11 times π₯ minus seven is less than zero. In this case, itβs values of π₯
greater than negative 11 and less than seven.

But if we go to our second
inequality, we see that π₯ is less than three. So whatβs the intersection? Whatβs the overlap of our solutions
here? Well, the intersection the values
of π₯ which satisfy both sets of inequalities is π₯ is greater than negative 11 and
less than three. In interval notation, thatβs the
open interval from negative 11 to three.

And so the function π is
increasing on the open interval from seven to positive β and negative 11 to
three.