Video: Finding Intervals Where a Rational Function Increases and Decreases

Find all possible intervals on which the function 𝑓(π‘₯) = (π‘₯Β² βˆ’ 4)/(π‘₯Β² + 1) is increasing and decreasing.

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Video Transcript

Find all possible intervals on which the function 𝑓 of π‘₯ is equal to π‘₯ squared minus four all divided by π‘₯ squared plus one is increasing and decreasing.

We’re given a function 𝑓 of π‘₯, and we can see that 𝑓 of π‘₯ is a rational function. It’s the quotient of two polynomials. We need to determine on which intervals is this function increasing and on which intervals is this function decreasing. And the first thing we should always check when we’re asked a question like this is, what is the domain of our function 𝑓 of π‘₯? And the reason for this is, a function can only be increasing or decreasing on an interval if it’s defined across that entire interval.

So finding the domain of our function 𝑓 of π‘₯ will help us determine which intervals our function is increasing or decreasing. In our case, our function 𝑓 of π‘₯ is a rational function. And rational functions are defined everywhere except where the denominator is equal to zero. So to find the domain of 𝑓 of π‘₯, we need to check which values of π‘₯ our denominator is equal to zero. In this case, we need to check when π‘₯ squared plus one is equal to zero.

However, we can see that this has no solutions. For example, subtracting one from both sides of the equation gives us that π‘₯ squared should be equal to negative one. Therefore, our function 𝑓 of π‘₯ is defined for all real values of π‘₯. So its domain is all real values.

Now, we want to use this to help us determine the intervals on which 𝑓 of π‘₯ is increasing and decreasing. And to do this, we can recall the following useful bit of information. For a function with a continuous derivative, if 𝑓 prime is greater than zero on some interval 𝐼, we can conclude that 𝑓 must be increasing on the interval 𝐼. However, if 𝑓 prime of π‘₯ is less than zero on this interval 𝐼, then 𝑓 must be decreasing on this interval 𝐼. In other words, we can find information about the intervals on which 𝑓 of π‘₯ is increasing or decreasing by looking at the sign of its derivative.

So we want to differentiate 𝑓 of π‘₯. Since 𝑓 of π‘₯ is the quotient of two polynomials, we’ll do this by using the quotient rule. So let’s start by recalling the quotient rule. The quotient rule tells us if we have two differentiable functions 𝑒 of π‘₯ and 𝑣 of π‘₯, then the derivative of 𝑒 of π‘₯ divided by 𝑣 of π‘₯ with respect to π‘₯ is equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ minus 𝑣 prime of π‘₯ times 𝑒 of π‘₯ all divided by 𝑣 of π‘₯ all squared.

And it’s worth pointing out this won’t be valid if 𝑣 of π‘₯ is equal to zero. We want to apply this to our function 𝑓 of π‘₯. So we need to set 𝑒 of π‘₯ to be the polynomial in our numerator, that’s π‘₯ squared minus four, and 𝑣 of π‘₯ to be the polynomial in our denominator, that’s π‘₯ squared plus one. And something worth pointing out here is π‘₯ squared is greater than or equal to zero. This means that 𝑣 of π‘₯ is greater than or equal to one. Therefore, 𝑣 of π‘₯ is never equal to zero. So in this case, our derivative will be defined for all real values of π‘₯. In fact, it will be a continuous function.

Now, to use the quotient rule, we see we need to find expressions for 𝑒 prime of π‘₯ and 𝑣 prime of π‘₯. We can do both of these term by term by using the power rule for differentiation. We want to multiply by our exponent of π‘₯ and reduce this exponent by one. We get 𝑒 prime of π‘₯ and 𝑣 prime of π‘₯ are both equal to two π‘₯.

Now, we’re ready to find our expression for 𝑓 prime of π‘₯ by using the quotient rule. We just need to substitute our expressions for 𝑒 of π‘₯, 𝑣 of π‘₯, 𝑒 prime of π‘₯, and 𝑣 prime of π‘₯ into the quotient rule. This gives us 𝑓 prime of π‘₯ is equal to two π‘₯ times π‘₯ squared plus one minus two π‘₯ times π‘₯ squared minus four all divided by π‘₯ squared plus one all squared.

And we can simplify this expression. In our numerator, we can take out the shared factor of two π‘₯. Doing this, we get two π‘₯ multiplied by π‘₯ squared plus one minus π‘₯ squared minus four all divided by π‘₯ squared plus one all squared. And we can simplify this even further. In our numerator, we can distribute negative one over our innermost parentheses. And if we were to do this, we would get negative π‘₯ squared plus four. And then we could just simplify our numerator. We have π‘₯ squared minus π‘₯ squared, which is equal to zero, and one plus four, which is equal to five. So, in fact, our numerator simplifies to give us two π‘₯ times five, which we know is equal to 10π‘₯. Therefore, we’ve shown 𝑓 prime of π‘₯ is equal to 10π‘₯ divided by π‘₯ squared plus one all squared.

We’re now ready to start finding the intervals on which 𝑓 of π‘₯ is increasing and decreasing. We need to find where 𝑓 prime of π‘₯ is greater than zero and when 𝑓 prime of π‘₯ is less than zero. To do this, let’s take a closer look at our expression for 𝑓 prime of π‘₯. First, as we’ve already discussed, but it’s worth repeating, this is a rational function where its denominator is never equal to zero. So, in fact, 𝑓 prime of π‘₯ is continuous. And it’s defined for all real values of π‘₯. And we want to determine when this is positive and when this is negative.

Let’s take a look at our denominator. We can see our denominator is π‘₯ squared plus one all squared. And because we’re taking the square of our value, we know this will be greater than or equal to zero. However, we already know it’s not equal to zero. So, in fact, our denominator is always positive. And if our denominator is always positive, that means the sign of 𝑓 prime of π‘₯ is entirely defined by its numerator. And in our numerator, of course 10 is positive. So the only thing which will determine the sign of 𝑓 prime of π‘₯ is just π‘₯ itself.

Therefore, if π‘₯ is greater than zero, we can see that 𝑓 prime of π‘₯ is the quotient of two positive numbers. So 𝑓 prime of π‘₯ is positive. However, if π‘₯ is less than zero, then 𝑓 prime of π‘₯ will be the quotient of a negative number and a positive number. So 𝑓 prime of π‘₯ will be negative. And we know this is exactly the same as saying 𝑓 is increasing when π‘₯ is greater than zero and 𝑓 is decreasing when π‘₯ is less than zero.

And of course we’re told to write this in interval notation. Saying that π‘₯ is greater than zero is the same as saying it’s on the open interval from zero to ∞. And saying that π‘₯ is less than zero is the same as saying it’s on the open interval from negative ∞ to zero.

And there is one small thing worth pointing out here. Why don’t we include the value of zero in our interval? If the derivative of 𝑓 at the end point of our interval is equal to zero, then we could say that our function is increasing or decreasing on this interval. It’s actually personal preference if you want to include it or not. However, in our definitions, we use 𝑓 prime of π‘₯ is greater than zero and prime of π‘₯ is less than zero to determine whether 𝑓 is increasing or decreasing.

We’ll just leave our intervals open, giving us our final answer. For the function 𝑓 of π‘₯ is equal to π‘₯ squared minus four all divided by π‘₯ squared plus one, this function will be decreasing on the open interval from negative ∞ to zero and will be increasing on the interval from zero to ∞.

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