# Video: Finding the Intervals of Increasing and Decreasing of a Function Involving Using the Power Rule for Differentiation

Determine the intervals on which the function 𝑓(𝑥) = −1 + (1/𝑥) − (4/𝑥²) is increasing and decreasing.

06:41

### Video Transcript

Determine the intervals on which the function 𝑓 of 𝑥 equals negative one plus one over 𝑥 minus four over 𝑥 squared is increasing and decreasing.

We begin by recalling how we determine whether a function is increasing or decreasing at a given point. We say that a function 𝑓 of 𝑥 is increasing at a given point 𝑥 if its first derivative at this point is greater than zero; that’s 𝑓 prime of 𝑥. Similarly, the function will be decreasing at this point if the value of its first derivative here is less than zero. If it’s negative. And so we see we’re going to need to begin by finding the first derivative of our function. That’s 𝑓 prime of 𝑥. Our function 𝑓 of 𝑥 consists of three terms, so we’re simply going to differentiate term by term.

Before we do, though, we begin by rewriting one over 𝑥 as 𝑥 to the power of negative one and negative four over 𝑥 squared as negative four times 𝑥 to the power of negative two. And this is really useful because we know, to differentiate a power term of the form 𝑎𝑥 to the 𝑛th power where 𝑎 and 𝑛 are real constants, we multiply the entire term by the exponent, then reduce that exponent by one. And of course, the first derivative of any constant is zero. So differentiating negative one with respect to 𝑥, and we get zero. Then when we multiply the entire term of 𝑥 to the power of negative one by the exponent and reduce that exponent by one, we get negative one times 𝑥 the power of negative two.

Similarly, let’s multiply four 𝑥 to the power of negative two by negative two but then reduce the exponent by one. So our third term is negative negative two times four 𝑥 to the power of negative three. We’ll forget about the zero, and we’ll rewrite negative one 𝑥 to the power of negative two as negative one over 𝑥 squared. Then our third term is negative negative two times four. So that’s eight. And similarly, we can write 𝑥 to the power of negative three as one over 𝑥 cubed. And so 𝑓 prime of 𝑥 is negative one over 𝑥 squared plus eight of 𝑥 cubed. Before we can find the values of 𝑥 such that this is greater than or less than zero, we’re going to begin by adding our fractions. And we’ll do so by creating a common denominator.

To find the common denominator, we multiply the numerator and denominator of our first fraction by 𝑥. So we now have negative 𝑥 over 𝑥 cubed plus eight over 𝑥 cubed. And since the denominators are equal, we can simply add the numerators. And so we see that 𝑓 prime of 𝑥 is negative 𝑥 plus eight all over 𝑥 cubed. Now remember, our job is to find the values of 𝑥 such that this is either positive or negative. And here, we can use a little bit of logic to do so. Let’s begin by looking for intervals of increase of our function, in other words, values of 𝑥 such that the first derivative is greater than zero. Now of course, our first derivative, 𝑓 prime of 𝑥, is a fraction. It’s negative 𝑥 plus eight all over 𝑥 cubed.

So how do we ensure this fraction is positive? Well, either we need to be dividing a positive by a positive, so we need the numerator and denominator of our fraction to both be positive. Or, alternatively, to achieve a positive number, we divide a negative by a negative. So we need negative 𝑥 plus eight to be less than zero and 𝑥 cubed to be less than zero.

Let’s begin with the first scenario. We have negative 𝑥 plus eight is greater than zero and 𝑥 cubed is greater than zero. The interval of increase here will be the intersection of the solutions to these two inequalities. Let’s solve this first inequality by adding 𝑥 to both sides. So 𝑥 is less than eight. We simply take the cube root of both sides of this inequality, and we find 𝑥 is greater than zero. The intersection of the solutions to these two inequalities are values of 𝑥 greater than zero and less than eight. And so we’re not using interval notation just yet. But we can see our function is increasing for these values of 𝑥.

Now, of course, we did say that a negative number divided by a negative number will yield a positive result. So our first derivative will also be greater than zero if negative 𝑥 plus eight is less than zero and 𝑥 cubed is also less than zero. It’s negative. This time, though, when we solve these two inequalities, we find 𝑥 is greater than eight and less than zero. There are no values of 𝑥 which satisfy both of these inequalities at the same time. In other words, there are no values of 𝑥 in the intersection of the solutions to these inequalities. And we, therefore, say that the function is increasing when 𝑥 is greater than zero and less than eight.

We’ll now consider the values of 𝑥 such that 𝑓 prime of 𝑥 is less than zero. These will tell us the values of 𝑥 on which our function is decreasing. Once again, we have negative 𝑥 plus eight over 𝑥 cubed. And we want that to be less than zero, to be negative. Well, we know a positive number divided by a negative number or vice versa will give us a negative number. So either our numerator has to be negative and our denominator has to be positive or our numerator needs to be positive and our denominator needs to be negative.

We’ll consider first the situation where negative 𝑥 plus eight is positive and the denominator, 𝑥 cubed, is negative. Solving these two inequalities and we find 𝑥 is less than eight and 𝑥 is less than zero. Well, the intersection of solutions to these inequalities, in other words, the values of 𝑥 that satisfy both, are values of 𝑥 less than zero. But of course, we said we needed to consider the scenario where our numerator is negative and our denominator is positive. In this case, we get 𝑥 is greater than eight and 𝑥 is greater than zero. We’ll consider again the intersection of solutions. Well, the values of 𝑥 which satisfy both of these inequalities are 𝑥 is greater than eight.

And of course, these are the values of 𝑥 where our function is decreasing. Now, we do need to be a little bit careful. If 𝑥 is equal to zero, our derivative becomes eight divided by zero, which we know to be undefined. And we’re therefore interested in the existence of any critical points. However, the value 𝑥 is equal to zero is not included in any of our intervals for increase or decrease.

So in this case actually, we don’t need to worry about it. We use the curly brackets to represent our intervals. We say that it’s increasing on the open interval from zero to eight. Those are values of 𝑥 between zero and eight, but not including zero and eight. And it’s decreasing on the open interval from negative ∞ to zero and eight to ∞.