### Video Transcript

Consider the function π of π₯
equals π₯ minus one squared times π₯ plus two. Find π prime of π₯. Find and classify the critical
points of π. Find the intervals of increase and
decrease for π. Find the limit as π₯ approaches
infinity of π of π₯.

And then thereβs one more part of
this question, which asks us to identify the correct graph for our function. So weβll look at our options when
weβve completed the first four parts. So how are we going to evaluate π
prime of π₯?

Thatβs the first derivative of our
function with respect to π₯. Well, taking a look at our
function, there are a number of ways we can do this. We could distribute our parentheses
and differentiate term by term. Alternatively, notice that we have
a product of two functions, one of which is a composite function. So we could use the product rule
alongside the general power rule.

The product rule says that the
derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£
by dπ₯ plus π£ times dπ’ by dπ₯. So letβs let π’ be equal to π₯
minus one all squared and π£ be equal to π₯ plus two. Weβll need to use the general power
rule to find the derivative of π₯ minus one all squared. This is a special case of the chain
rule. And it says we can multiply the
entire function by the exponent, reduce the exponent by one, and then multiply all
of this by the derivative of the inner function.

Well, the derivative of π₯ minus
one is just one. So dπ’ by dπ₯ is two times π₯ minus
one to the power of two minus one times one. And that simplifies to two times π₯
minus one. dπ£ by dπ₯ is much more straightforward. Itβs simply one. So we substitute what we have into
the product rule. Itβs π’ times dπ£ by dπ₯ β thatβs
π₯ minus one all squared times one β plus π£ times dπ’ by dπ₯. Distributing our parentheses, and
the first derivative of our function simplifies really nicely to three π₯ squared
minus three.

In the second part of this
question, we need to find and classify the critical points of our function. Remember, these occur when our
first derivative is either equal to zero or does not exist. Well, we actually know that
polynomial functions are all differentiable. So weβre just going to consider the
case for which our first derivative is equal to zero. That is, three π₯ squared minus
three is equal to zero.

Letβs solve for π₯. We divide through by one. And then we use the difference of
two squares to factor π₯ squared minus one. Itβs π₯ plus one times π₯ minus
one. Now for the product of these two
terms to be equal to zero, either π₯ plus one has to be equal to zero or π₯ minus
one has to be equal to zero. And solving each equation for π₯,
we obtain π₯ must be either negative one or π₯ equals one. So we have critical points where π₯
is either equal to positive or negative one.

Our next job is to classify
these. We can perform the first derivative
test and evaluate the first derivative just before and just after each critical
point. Letβs add a table. We know that the first derivative
is equal to zero when π₯ is equal to negative one or positive one. The first derivative evaluated at
π₯ equals negative two is three times negative two squared minus three, which is
nine. The first derivative evaluated when
π₯ is equal to zero is three times zero squared minus three, which is negative
three. And the first derivative evaluated
at π₯ equals two is also nine. The function is increasing before
π₯ equals negative one and decreasing after. So the critical point π₯ equals
negative one is a local maximum. The opposite is true when π₯ equals
one. So that must be our local
minimum.

In the third part of this question,
weβre looking to find the intervals of increase and decrease for our function. Letβs clear some space. Now this might seem awfully similar
to what we just did. But at that point, we were simply
considering the nature of the function at specific points. We now want to know over what
intervals the function is increasing or decreasing. So weβre going to work out when the
first derivative is less than zero, decreasing, or greater than zero,
increasing.

We know the graph of π¦ equals
three π₯ squared minus three looks a little something like this. We can see that the graph of the
first derivative is greater than zero here and here. In other words, when π₯ is less
than negative one or greater than one. The first derivative is less than
zero here. Thatβs when π₯ is greater than
negative one or less than one. So the intervals of increase are
the open intervals from negative infinity to negative one and one to infinity. And the interval of decrease is the
open interval from negative one to one.

And finally, we need to find the
limit as π₯ approaches infinity of the function. We can do this by direct
substitution. We see that as π₯ approaches
infinity, the function itself also grows larger and larger. So the function also approaches
infinity. Letβs now combine everything weβve
done to identify the graph of our function.

Which of the following is the graph
of π? If we were to distribute the
parentheses of our function, weβd see we have a cubic graph with a positive leading
coefficient. In other words, the coefficient of
π₯ cubed is positive. That tells us the shape of the
graph will be a little something like this. We know it has critical points π₯
equals negative one and one and a maximum and minimum in those locations,
respectively.

We can work out the value of our
function at these points by substituting negative one and one in. And when we do, we see that π of
negative one is four and π of one is zero. So we know the graph has a maximum
at negative one, four and a minimum at one, zero. We also saw that it had intervals
of increase over the open interval negative infinity to negative one and one to
infinity and decrease over the open interval negative one to one. The only graph that satisfies all
of these conditions is A. In fact, we could have very quickly
spotted they could not have been D or E as these are graphs of quadratic
functions. C is a cubic graph which has a
negative coefficient of π₯ cubed. So we really only had two
options.

Notice that we didnβt worry about
evaluating the limits of our functions to look for asymptotes. As we know, a cubic graph doesnβt
have any.