### Video Transcript

Find the set on which π of π₯ is equal to negative four times π₯ to the power of negative three plus 10 divided by π₯ squared plus nine is continuous. Option (A) the function π is continuous on the set of real numbers. Option (B) the function π is continuous on the set of real numbers excluding zero and three. Option (C) the function π is continuous on the set of real numbers excluding negative three. Option (D) the function π is continuous on the set of real numbers excluding zero. Or option (E) the function π is continuous on the sets of real numbers excluding zero and negative three.

In this question, weβre given an explicit definition of a function π of π₯. And we need to use this to find the set on which this function π of π₯ is continuous. And weβre given five possible options. And thereβs several different ways we could approach this problem. What weβre going to do is weβre going to try and find all of the values of π₯ where our function π is continuous.

To do this, letβs take a closer look at our function π of π₯. First, we can see itβs the sum of two functions. And we recall if two functions are continuous, then their sum is also continuous. So we can look at each of these individually. And this is really helpful because we could look at each of these two parts individually, and we can notice both of these are rational functions.

Remember, a rational function is the quotient of two polynomials. And to see this, it might be easier to rewrite negative four times π₯ to the power of negative three by using our laws of exponents. Itβs equal to negative four over π₯ cubed. Now that weβre trying to find the continuity of rational functions, we can use anything we know about the continuity of rational functions to help us.

The first fact weβre going to want to use is the following. All rational functions are continuous across their entire domain. And remember, the domain of a function is all of the values of π₯ where our function is defined. So all this tells us is our rational functions are continuous everywhere where theyβre defined. But we actually know a very useful result about the domain of rational functions. We also recall rational functions will be defined for all values of π₯ except when their denominator is equal to zero. This is because rational functions are the quotient of two polynomials. And we know all polynomials are defined for all real values of π₯.

So the only way a rational function could not be defined is when weβre dividing by zero. In other words, the denominator would need to be equal to zero. Combining these two facts, we get a very useful result about rational functions. Rational functions are continuous everywhere except when their denominator is equal to zero. We can apply this to both of the rational functions in π of π₯ separately. Letβs start with the first one, which has a denominator of π₯ cubed.

By using our two results, this will be continuous for all values of π₯ except where its denominator is equal to zero. Of course, π₯ cubed being equal to zero means that π₯ must be equal to zero. Therefore, weβve proven that negative four over π₯ cubed is continuous on the set of real numbers excluding zero.

We can do exactly the same thing for our second rational function. Once again, by applying our two rules, we would need to know where the denominator of this rational function is equal to zero. We would need to solve π₯ squared plus nine is equal to zero. However, we can prove that this has no real roots. For example, the discriminant of this quadratic is negative.

However, we could also use the fact that π₯ squared is going to be greater than or equal to zero for all real values of π₯. And then we add nine. This is going to be positive for any value of π₯. So it canβt be equal to zero. Therefore, if the denominator of this rational function is never equal to zero, then itβs defined for all real values of π₯. So 10 divided by π₯ squared plus nine must be continuous for all real values of π₯.

Now we can talk about the continuity of π of π₯ since itβs the sum of these two rational functions. Remember, the sum of two continuous functions is itself continuous. So just by using this, weβve proven π of π₯ is continuous on the set of all real numbers except when π₯ is equal to zero. And just to be absolutely sure, we might be wondering what happens when π₯ is equal to zero in our function π of π₯.

Well, if we were to substitute π₯ is equal to zero into our function π of π₯, we would see weβre dividing by zero. So π evaluated at zero is not defined. And we know a function canβt be continuous at a value of π₯ if itβs not defined at this value of π₯.

Therefore, we were able to show that the function π of π₯ is continuous for all real values except when π₯ is equal to zero. And this was given as option (D).