Video: Finding the Intervals of Increasing and Decreasing of a Polynomial Function

Video Transcript

Determine the intervals on which the function 𝑓 of 𝑥 equals three 𝑥 cubed minus nine 𝑥 squared minus four is increasing and on which it is decreasing.

By definition, a function is increasing on an interval 𝐼 if 𝑓 of 𝑥 one is less than 𝑓 of 𝑥 two whenever 𝑥 one is less than 𝑥 two for 𝑥 one and 𝑥 two in the interval 𝐼. And it is decreasing on 𝐼 if 𝑓 of 𝑥 one is greater than 𝑓 of 𝑥 two whenever 𝑥 one is less than 𝑥 two for 𝑥 one and 𝑥 two in 𝐼. That’s the formal definition. But really the best way to think about increasing and decreasing functions is using their graphs.

The graph of an increasing function looks like this: 𝑦 increases as 𝑥 increases. They’re not always at the same rate. And the graph of the function which is decreasing on the interval looks like this. Now let’s turn to the problem we have. We have to determine the intervals on which our function is increasing and on which it is decreasing. We could try to graph this function using a graphing calculator or graphing software. And in fact, we’ll do that at the end of the video.

But the way we’re going to solve this question is using a test. If the derivative of 𝑓, 𝑓 prime of 𝑥, is greater than zero on an interval, 𝐼 then 𝑓 is increasing on 𝐼. And if the derivative of 𝑓, 𝑓 prime of 𝑥, is less than zero on an interval 𝐼, then 𝑓 is decreasing on that interval 𝐼. And this test makes sense given the graphical interpretation above. A tangent to the increasing function always has positive slope, so 𝑓 prime is positive. But the tangents to the decreasing function have negative slope, so 𝑓 prime is negative on this interval for this decreasing function.

Let’s clear away the definition and graphical interpretation so we have room to apply our test. The test involves 𝑓 prime, which is the derivative of our function. So let’s find 𝑓 prime. We differentiate the polynomial term by term, using the fact that the derivative of 𝑎𝑥 to the 𝑛 with respect to 𝑥 is 𝑎 times 𝑛 times 𝑥 to the 𝑛 minus one. The derivative of three 𝑥 to the power of three comes from multiplying the exponent and coefficient to get nine and then reducing the exponent by one to get 𝑥 squared.

Similarly, the derivative of nine 𝑥 squared is 18𝑥, and the derivative of four is zero. So 𝑓 prime of 𝑥 is nine 𝑥 squared minus 18𝑥. 𝑓 is increasing on intervals for which this derivative is greater than zero and decreasing on intervals for which the derivative is less than zero. So we need to find the sign of 𝑓 prime. The way we do this is to factor 𝑓 prime, writing it as nine 𝑥 times 𝑥 minus two.

And from this factored form, we can easily see that 𝑓 prime is zero at 𝑥 equals zero and 𝑥 equals two. These two numbers are called the critical numbers of the function 𝑓. For 𝑓 prime to change sign from positive to negative or negative to positive, it must pass through zero. And so we know that 𝑓 prime must have the same sign on the interval 𝑥 is less than zero. If 𝑓 prime changed sign below zero, because it’s continuous, there will be a value of 𝑥 below zero at which 𝑓 prime was zero. And we’ve seen that 𝑓 prime is only zero at 𝑥 equals zero and 𝑥 equals two.

Similarly, 𝑓 prime can’t change sign between zero and two. Similarly, 𝑓 prime can’t change sign when 𝑥 is greater than two. If 𝑓 prime does change sign anywhere, it’s got to be at zero or two. We want to find the sign of 𝑓 prime on these intervals. And we do that by considering the signs of its factors. So along the top of the table, we have the two factors of 𝑓 prime: nine 𝑥 and 𝑥 minus two and 𝑓 prime itself.

Let’s start filling the table in. What is the sign of nine 𝑥 when 𝑥 is less than zero? Nine 𝑥 is negative when 𝑥 is less than zero. You can see this by multiplying both sides of the inequality, which defines this interval, by nine. What can we say about nine 𝑥 when 𝑥 is between zero and two? In a similar way, we see that nine 𝑥 is between zero and 18; in particular, its positive. And finally for the third interval, we see that is nine 𝑥 is greater than 18 and so is positive.

We do the same thing for the factor 𝑥 minus two. In the first region, it is less than negative two and so negative. In the second region, it is between negative two and zero, and so it’s negative here too. And finally in the third region, it’s positive. We found the sign of the factors of 𝑓 prime. Now we just need to find the sign of 𝑓 prime itself in each of these regions. In the first region, 𝑥 is less than zero, both factors are negative. And so 𝑓 prime is negative times negative, which is a positive.

Similarly, in the second region, 𝑓 prime is a positive times a negative quantity, which makes it a negative quantity. And finally, in the third region, 𝑓 prime is a positive times a positive, and so it is a positive quantity. Referring back to the increasing decreasing test, if 𝑓 prime is positive on 𝐼 then 𝑓 is increasing on 𝐼; and if 𝑓 prime is negative on 𝐼, then 𝑓 is decreasing on 𝐼.

So 𝑓 is increasing on these two intervals and decreasing on this one. The only thing left to do is to write down our final answer: 𝑥 is increasing on the intervals where 𝑥 is less than zero and where 𝑥 is greater than two and is decreasing on the interval where 𝑥 is between zero and two. Let’s write these using interval notation. 𝑥 is less than zero represents the open interval between negative infinity and zero. 𝑥 is greater than two is the open interval from two to infinity. And zero is less than 𝑥 is less than two represents the open interval from zero to two.

So here’s our answer: 𝑓 is increasing on the open interval from negative infinity to zero, and it is also increasing on the open interval from two to infinity; 𝑓 is, however, decreasing on the open interval from zero to two. We solved this problem using the increasing/decreasing test, which uses the sign of 𝑓 prime to tell where 𝑓 is increasing and where it is decreasing.