Video: Finding the Intervals of Upward and Downward Concavity of a Function Using the Chain Rule

Determine the intervals on which 𝑓(π‘₯) = βˆ’4π‘₯ + (π‘₯ + 3)⁡ βˆ’ 4 is concave up and down.

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Video Transcript

Determine the intervals on which 𝑓 of π‘₯ equals negative four π‘₯ plus π‘₯ plus three to the power of five minus four is concave up and down.

We want to find where this function is concave up and where it is concave down, so we use the concavity test. The concavity test tells us that a function 𝑓 is concave up on an interval 𝐼 if its second derivative, 𝑓 double prime of π‘₯, is greater than zero for values of π‘₯ within that interval 𝐼 and the function 𝑓 is concave down on that interval 𝐼 if its second derivative, 𝑓 double prime of π‘₯, is less than zero for values of π‘₯ in that interval 𝐼.

So in brief, whether a function is concave up or down depends on whether its second derivative is positive or negative. To find the intervals on which the function is concave up and down, we need to solve the two inequalities: 𝑓 double time of π‘₯ is greater than zero and 𝑓 double prime of π‘₯ is less than zero. And before we can solve these inequalities we need to find 𝑓 double prime of π‘₯, and so this is our first step. We have the definition of 𝑓 of π‘₯ from the question, and we can differentiate this term by term to find the first derivative of 𝑓: 𝑓 prime of π‘₯.

So let’s do that. The derivative of negative four π‘₯ is just negative four. And to this, we add the derivative of π‘₯ plus three to the power of five. Now we could expand this out in powers of π‘₯, but there’s an easier way. We can use a special case of the chain rule which gives us the derivative of a power of a function. The derivative of a function 𝑓 of π‘₯ raised to the power of some number 𝑛 is that exponent 𝑛 times the derivative of the function, 𝑓 prime of π‘₯, times 𝑓 of π‘₯ to the power of 𝑛 minus one.

So in our case, the derivative of π‘₯ plus three to the power of five is that exponent five times the derivative of π‘₯ plus three, which is just one, times π‘₯ plus three to the power of five minus one, which is four. Finally, the derivative of four is zero, and so we can subtract zero if we’d like to. It doesn’t do anything because the constant term of a function doesn’t affect its derivative. Rearranging and simplifying, we get five π‘₯ plus three to the power of four minus four.

This is 𝑓 prime of π‘₯, the first derivative of 𝑓, but we want the second derivative, 𝑓 double prime of π‘₯. So we’re going to have to differentiate 𝑓 prime of π‘₯. The derivative of five times π‘₯ plus three to the power of four is five times the derivative of π‘₯ plus three to the four. And the derivative of π‘₯ plus three to the power of four we find just as we did the derivative of π‘₯ plus three to the power of five. It’s the exponent four times the derivative of π‘₯ plus three, which is one, times π‘₯ plus three to the power of four minus one, which is three. And as before, the constant term doesn’t contribute anything to the derivative.

Simplifying then, we find that f double prime of π‘₯ is 20 times π‘₯ plus three cubed. We have found 𝑓 double prime of π‘₯ then, and we can now apply the concavity test. Let’s clear some room so we can do so. Applying the test, we see that 𝑓 is concave up when 20 times π‘₯ plus three cubed is positive and down when 20 times π‘₯ plus three cubed is negative. Our second step therefore is to solve these two inequalities to find the intervals on which 𝑓 is concave up and on which it is concave down.

The first thing we can do is to divide three by 20 on both sides of each inequality. As 20 is positive, we don’t have to flip the sign of the inequality; it stays the same. Now we only have to worry about where π‘₯ plus three cubed is positive and where it is negative. And a number cubed is positive exactly when that number itself is positive. So I claim that the first inequality becomes π‘₯ plus three is greater than zero. As for the second inequality, a number cubed is negative exactly when that number itself is negative. And so I claim the second inequality becomes π‘₯ plus three is less than zero.

You can see this by writing π‘₯ plus three cubed as π‘₯ plus three squared times π‘₯ plus three. π‘₯ plus three squared as the square of a number is always greater than or equal to zero. And so the sign of the product of π‘₯ plus three squared and π‘₯ plus three is just the sign of π‘₯ plus three. The only thing we have to check is what happens when π‘₯ plus three squared is equal to zero. But then π‘₯ is negative three, and so π‘₯ plus three is equal to zero as well. However, you convince yourself, whether it’s with the previous methods or by making a table of the factors of π‘₯ plus three cubed or by looking at the graph of π‘₯ plus three cubed, you should end up with these simple one-step inequalities, which are straight forward to solve.

We subtract three from both sides to find that 𝑓 is concave up when π‘₯ is greater than negative three and concave down when π‘₯ is less than negative three. Using interval notation then, we see that the function 𝑓 is concave down on the open interval from negative infinity to negative three β€” that is where π‘₯ is less than negative three β€” and the function is concave up on the open interval from negative three to infinity β€” that is where π‘₯ is greater than negative three.

We answered this question using the concavity test. We found the second derivative of 𝑓, 𝑓 double prime of π‘₯. And then we solved the inequalities 𝑓 double prime of π‘₯ is greater than zero, which is where 𝑓 is concave up, and 𝑓 double prime of π‘₯ is less than zero, where 𝑓 is concave down. The concavity test doesn’t give the definition of concavity. The definition of concavity is that the function 𝑓 is concave up on that interval 𝐼 if its derivative, 𝑓 prime, is an increasing function on that interval 𝐼 and the function 𝑓 is concave down on 𝐼 if its derivative, 𝑓 prime, is a decreasing function on 𝐼.

Where then does the concavity test come from? Well a test for whether 𝑓 is concave up or down is a test for whether 𝑓 prime is increasing or decreasing. We know a test that helps us find the intervals on which a function is increasing and on which it is decreasing: the increasing/decreasing test. A function 𝑔 is an increasing function on an interval 𝐼 if its derivative, 𝑔 prime of π‘₯, is greater than zero for π‘₯ in that interval, and 𝑔 is a decreasing function on 𝐼 if its derivative, 𝑔 prime of π‘₯, is less than zero for π‘₯ in that interval.

In the definition of concavity, we care about whether the function 𝑓 prime is increasing or decreasing. And so we should apply the increasing/decreasing test to the function 𝑓 prime. Setting 𝑔 to be 𝑓 prime in the increasing/decreasing test, we see the 𝑓 prime is increasing if 𝑓 prime prime of π‘₯ is greater than zero β€” in other words, if 𝑓 double prime of π‘₯ is greater than zero β€” and 𝑓 prime is decreasing if 𝑓 prime prime of π‘₯ is less than zero β€” in other words, if 𝑓 double prime of π‘₯ is less than zero.

In this way, you can see the concavity test is just what you get when you apply the increasing/decreasing test to the definition of concavity.

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