Question Video: Determining the Domain and the Range of a Rational Function given Its Graph | Nagwa Question Video: Determining the Domain and the Range of a Rational Function given Its Graph | Nagwa

# Question Video: Determining the Domain and the Range of a Rational Function given Its Graph Mathematics • Second Year of Secondary School

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Find the domain and range of the function π(π₯) = β(1/(π₯ β 5)).

07:45

### Video Transcript

Find the domain and range of the function π of π₯ equals negative one over π₯ minus five.

Weβre also given a graph of π of π₯, and weβre going to see how the domain and range of a function are linked to its graph and how in some cases we can just read off the domain and range of our function from its graph. We had to find both the domain and the range of the function, but letβs just start with the domain for now and with the definition of the domain.

The domain is a set of inputs to the function π of π₯. Weβre thinking about functions with graphs, so certainly the input to the function has to be a real number. So, the domain is a subset of the real numbers. And because weβre not told anything else about the domain of the function, we assume that we have to make it as big as it can possibly be within the real numbers.

We canβt just set the domain equal to the set of real numbers. The function has to be defined for every element of its domain. So the best we can do is to start with the real numbers and then remove any values of π₯ for which the function π of π₯ is undefined.

Looking at the definition of the functions, can you see how the value of π of π₯ could be undefined? Well, one way for that to happen would be if we were dividing by zero. So certainly, the function π of π₯ is undefined when the denominator of the function π₯ minus five is equal to zero. And this happens when π₯ is equal to five.

And looking at the graph for a moment, we can see that something special happens when π₯ is equal to five. We have a vertical asymptote of the graph. And by looking at this first-class asymptote, even without knowing the equation which defines the function, you would see that the function is undefined for this value of π₯, because any attempt to read off the value of the function would fail.

Are there any more values of π₯ for which the function is undefined? While having a look at the graph, we canβt see any more vertical asymptotes. We also canβt see any hollow circles on the graph which would represent a point for which the function is undefined. And if we imagine moving a vertical line along the π₯-axis from negative β to β, the only value of π₯ for which this line will not intersect with the graph of the function is the value of π₯ that we found already, π₯ equals five. And so, this set of π₯ for which π of π₯ is undefined is just the set of five. And so the domain of our function is just the set of real numbers minus the set of five.

Now that we found the domain of the function, we need to clear some room so we can find the range of the function. The range of a function is the set of outputs of that function. And because weβre looking at a graph, we know itβs going to be a subset of the set of real numbers, perhaps the entire set of real numbers. We donβt know yet. We can use the graph to help us.

Suppose you want to show that four is in the range. Well, that means finding a point on the graph whose π¦-coordinate is four. To find such a point if it exists, we draw the line π¦ equals four and look for intersections with the graph. We see that there is an intersection. And so, there is a value of π₯ just below five which, when you put it into the function π of π₯, you get an output or π¦ value of four. And hence, four is in the range.

We can imagine moving this horizontal line, which currently has the equation π¦ equals four, up and down the π¦-axis. Values of π¦ for which this line intersects the graph are in the range, and values of π¦ for which this line does not intersect the graph are not in the range.

We can see that the π₯-axis with equation π¦ equals zero does not intersect the graph at any point, and so zero is not in the range. But for every other value of π¦, this line does intersect the graph at some point, and so the range is the entirety of the real numbers apart from this value zero. And so we found our final answer that the domain is the set of real numbers minus the set of five, that is, the set of all real numbers apart from five, and the range is the set of real numbers minus the set of zero, that is, the set of all real numbers which arenβt zero.

You may also see this written using a backslash instead of a minus sign. This means exactly the same thing. Itβs just the real numbers minus the set of five or the real numbers minus the set of zero. The backslash symbol is used in the context of sets to avoid confusion with normal subtraction of numbers.

You may have been able to work out the domain and range of this function by thinking of it as a transformation of the reciprocal function π of π₯ equals one over π₯. Recall that we found that the domain of the function is just the set of real numbers minus the set of values of π₯ for which the denominator of the function was zero. This is a general fact about rational functions, which are functions in the form of fractions where both the numerator and the denominator is a polynomial. In that case, the domain of the rational function in question is just the set of real numbers minus the set of zeros of that denominator polynomial. We found the range using graphical methods, imagining a horizontal line moving up and down at the π¦-axis. But we can find it using algebraic methods. And the rest of the video will be dedicated to doing so. If you were happy with the graphical method, then feel free to skip this.

Which values of πΎ are in the range? Well, they are values which are output to the function and hence are equal to negative one over π₯ minus five for some input π₯. So to show that a value of πΎ is in the range, we need to show that thereβs a value of π₯ for which πΎ is equal to negative one over π₯ minus five. And we can show that there is such a value of π₯ by finding a formula for that π₯ in terms of πΎ.

So our goal is to rearrange this equation to make π₯ the subject. We multiply both sides by π₯ minus five and then divide both sides by πΎ to get that π₯ minus five is equal to negative one over πΎ. And adding five to both sides, we can see that π₯ is equal to five minus one over πΎ. So when this value of π₯ is input into the function, we get an output of πΎ.

For example, if we wanted to show that negative 17 is in the range, we just substitute in this value of πΎ. So π₯ is equal to five minus one over negative 17, which is equal to 86 over 17. And if you put this into our function we find that π of 86 over 17 which, substituted in, we see is negative one over 86 over 17 minus five, which simplifies to negative 17. So clearly, negative 17 is an output of the function and hence is in the range. And similarly, we could show that any other number is in the range apart from, of course, πΎ equals zero.

We canβt have πΎ as zero because we have this term of one over πΎ, and so we canβt find an input value of π₯ which will give us an output of πΎ equals zero. And so zero is not in the range. This is the only real value of πΎ for which we have this problem. And so we can see that the range is, as we saw graphically, just the set of real numbers minus the set of zero.

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