### Video Transcript

Find the points π₯, π¦ where π¦
equals nine π₯ plus nine over π₯ has a local maximum or a local minimum.

Local maxima and local minima are
examples of critical points. And we recall that at the critical
points of a function, the first derivative dπ¦ by dπ₯ is equal to zero. Before differentiating, we may find
it helpful to rewrite the second term in our function as nine π₯ to the negative
one. We can then use the power rule of
differentiation to find the first derivative dπ¦ by dπ₯. Remember that when we
differentiate, we decrease the power by one. So when we decrease that power of
negative one, it will become negative two not zero. Watch out for that! Thatβs a common mistake. We can rewrite this derivative as
nine minus nine over π₯ squared and weβll then set this derivative equal to
zero.

Weβll now solve the resulting
equation in order to find the π₯-values at the critical points. We begin by multiplying every term
in the equation by π₯ squared. We can then divide through by nine
to give π₯ squared minus one equals zero. Add one to both sides and then
finally take the square root, remembering that we have both positive and negative
solutions. We find that π₯ is equal to
positive or negative one. So this function has two critical
points.

Next, we need to find the π¦-values
at each critical point by evaluating the function itself. When π₯ is equal to positive one,
π¦ is equal to nine multiplied by one plus nine over one, which is equal to 18,
giving a critical point of one, 18. When π₯ is equal to negative one,
π¦ is equal to negative 18. So our second critical point has
coordinates negative one, negative 18. We now need to determine whether
these critical points are local minima or local maxima, which weβll do using the
second derivative test. Weβll clear some space in order to
do this.

To find the second derivative d two
π¦ by dπ₯ squared, we need to differentiate the first derivative, which was nine
minus nine π₯ to the power of negative two with respect to π₯. Doing so, we obtain negative nine
multiplied by negative two π₯ to the power of negative three, which we can write as
18 over π₯ cubed. Next, we need to evaluate this
second derivative at each of our critical points. When π₯ is equal to negative one,
the second derivative is 18 over negative one cubed, which is equal to negative
18. This is less than zero. And we recall that if the second
derivative of a function is negative at a critical point, then the critical point is
a local maximum. Evaluating the second derivative
when π₯ is equal to positive one gives 18 over one cubed, which is 18. And as this is greater than zero,
we conclude that the critical point when π₯ is equal to one is a local minimum.

So weβve completed the problem. We answered that the point one, 18
is a local minimum and the point negative one, negative 18 is a local maximum.