### Video Transcript

Determine where π of π₯ equals π₯
of the four over two minus three π₯ squared plus three is concave up and where it is
concave down.

This question is about on which
intervals our function is concave up and on which intervals it is concave down. To answer this question, it is
helpful to know the definitions of concave up and concave down. A function π is concave upward or
concave up on an interval πΌ if π prime, the first derivative of π, is an
increasing function on πΌ. A function π is concave downward
or concave down on an interval πΌ if its derivative π prime is a decreasing
function on πΌ.

Whether a function π is concave up
or down depends on whether its derivative π prime is increasing or decreasing. Now we know a test which tells us
whether a given function is increasing or decreasing on a given interval. A function π is increasing on an
interval πΌ if its derivative π prime is positive on πΌ. A function π is decreasing on an
interval πΌ if its derivative π prime is negative on πΌ.

Weβre interested in whether π
prime is increasing or decreasing. So letβs replace π by π
prime. π prime is increasing if its
derivative, π prime prime or π double prime, is positive. And π prime is decreasing if its
derivative, π prime prime or π double prime, is negative. Applying the increasing and
decreasing test to the definition of concavity, we get the concavity test. A function π is concave upward on
an interval πΌ if its second derivative π double prime is positive on πΌ. And the function π is concave
downward on an interval πΌ if its second derivative is negative on πΌ.

This is a test weβre going to apply
to solve our problem. Letβs clear some space so we have
room to apply it. We need to find the second
derivative of our function. We differentiate once to find π
prime of π₯. We differentiate term by term using
the known derivative of a monomial function. The derivative of π₯ to the four
over two is two π₯ to the three. The derivative of the three π₯
squared is six π₯. And the derivative of three is
zero.

So thatβs π prime of π₯, the first
derivative of π. But we need the second derivative
of π, so weβre going to have to differentiate π prime. The derivative of two π₯ cubed is
six π₯ squared. And the derivative of six π₯ is
six. So we find that the second
derivative of π is six π₯ squared minus six. π is concave upward when six π₯
squared minus six is greater than zero and concave downward when six π₯ squared
minus six is less than zero.

This is just applying the concavity
test with our newly found expression for π double prime of π₯. Now we have two quadratic
inequalities to solve. And we can solve them in a variety
of ways. We can start by dividing both sides
by six, leaving π₯ squared minus one is greater than zero, and then factoring the
left-hand side. The first inequality becomes π₯
plus one times π₯ minus one is greater than zero.

We can do exactly the same thing to
the other inequality. The only difference is that the
greater than sign becomes a less than sign. Now letβs solve these inequalities
using a table. We find the intervals on which the
factors π₯ plus one and π₯ minus one are positive or negative and use this
information to find where the product π₯ plus one times π₯ minus one is positive or
negative.

For example, for π₯ less than
negative one, π₯ plus one is less than zero, and so negative. And π₯ minus one is less than two,
and so certainly negative. When π₯ is between negative one and
one, π₯ plus one is between zero and two. So itβs positive, whereas π₯ minus
one is between negative two and zero. And so it is negative. Finally, when π₯ is greater than
one, both π₯ plus one and π₯ minus one are positive.

Itβs no coincidence that the
intervals weβve ended up with split the real line at the roots of π₯ plus one times
π₯ minus one. π₯ plus one and π₯ minus one are
both negative on the interval π₯ is less than negative one. And so their product is
positive. A negative times a negative is a
positive. On this interval, we have the
product of a positive and a negative quantity, which makes a negative quantity. And finally, when π₯ is greater
than one, π₯ plus one times π₯ minus one is a positive times a positive, which is a
positive.

So when is π₯ plus one times π₯
minus one positive or greater than zero? Itβs positive on two intervals: the
intervals where π₯ is less than negative one and where π₯ is greater than one. And when is it negative? Well, we can see from the table
that this happens when π₯ is between negative one and one. Letβs switch from using
inequalities like π₯ is less than negative one to using interval notation.

π₯ is less than negative one on the
open interval from a negative infinity to negative one and greater than one on the
open interval from a one to infinity. And π₯ is between negative one and
one on the open interval from negative one to one. Using these intervals in our answer
then, we see that the function π of π₯ equals π₯ to the four over two minus three
π₯ squared plus three is concave up on the open intervals from negative infinity to
negative one and from one to infinity and concave down on the open interval from a
negative one to one.

We found this answer using the
concavity test, which is just the application of the increasing/decreasing function
test to the definition of concavity. Applying this test to our function
gave a couple of quadratic inequalities which we solved it in the normal way.