Video: Finding the Intervals of Upward and Downward Concavity of a Polynomial Function

Determine where 𝑓(π‘₯) = π‘₯⁴/2 βˆ’ 3π‘₯Β² + 3 is concave up and where it is concave down.

06:40

Video Transcript

Determine where 𝑓 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 𝑓 is concave upward or concave up on an interval 𝐼 if 𝑓 prime, the first derivative of 𝑓, is an increasing function on 𝐼. A function 𝑓 is concave downward or concave down on an interval 𝐼 if its derivative 𝑓 prime is a decreasing function on 𝐼.

Whether a function 𝑓 is concave up or down depends on whether its derivative 𝑓 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 𝑔 is increasing on an interval 𝐼 if its derivative 𝑔 prime is positive on 𝐼. A function 𝑔 is decreasing on an interval 𝐼 if its derivative 𝑔 prime is negative on 𝐼.

We’re interested in whether 𝑓 prime is increasing or decreasing. So let’s replace 𝑔 by 𝑓 prime. 𝑓 prime is increasing if its derivative, 𝑓 prime prime or 𝑓 double prime, is positive. And 𝑓 prime is decreasing if its derivative, 𝑓 prime prime or 𝑓 double prime, is negative. Applying the increasing and decreasing test to the definition of concavity, we get the concavity test. A function 𝑓 is concave upward on an interval 𝐼 if its second derivative 𝑓 double prime is positive on 𝐼. And the function 𝑓 is concave downward on an interval 𝐼 if its second derivative is negative on 𝐼.

This is a test we’re going to apply to solve our problem. Let’s 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 𝑓 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 that’s 𝑓 prime of π‘₯, the first derivative of 𝑓. But we need the second derivative of 𝑓, so we’re going to have to differentiate 𝑓 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 𝑓 is six π‘₯ squared minus six. 𝑓 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 𝑓 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 let’s 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 it’s 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.

It’s no coincidence that the intervals we’ve 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? It’s 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. Let’s 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 𝑓 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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.