Question Video: Determining the Domain and the Range of a Piecewise Function | Nagwa Question Video: Determining the Domain and the Range of a Piecewise Function | Nagwa

# Question Video: Determining the Domain and the Range of a Piecewise Function Mathematics • Second Year of Secondary School

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Determine the domain and the range of the function π(π₯) = (π₯Β² β 36)/(π₯ β 6), if π₯ β  6 and π(π₯) = 12, if π₯ = 6.

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### Video Transcript

Determine the domain and the range of the function π of π₯ is equal to π₯ squared minus 36 all divided by π₯ minus six if π₯ is not equal to six and π of π₯ is equal to 12 if π₯ is equal to six.

In this question, weβre asked to determine the domain and the range of a given piecewise-defined function. And we can start by recalling what we mean by the domain and the range of a function. First, the domain of a function is the set of all input values for that function. And the range of a function is the set of all output values of that function given its domain. And we can actually determine the range directly from the piecewise definition of π of π₯, since π of π₯ is a piecewise-defined function.

To input a value of π₯ into this function, we need to determine which subdomain this value lies in. For example, weβre told if we input the value of six into our function, our function outputs 12. So six is an element of the domain of π of π₯. Similarly, if we input a value of π₯ not equal to six, then our function outputs the number squared minus 36 divided by the number minus six. So we can input the value of six and we can input any value of π₯ not equal to six. This is of course any real value of π₯. So we can say the domain of π of π₯ is all real numbers. And itβs worth pointing out we can generalize this result. For any piecewise-defined function, the domain of the piecewise-defined function is the union of its subdomains.

Now that we found the domain of this function, we need to determine its range. And it can be difficult to determine the range of a piecewise-defined function just from its definition. So, instead, letβs sketch our function. Weβll start by sketching our coordinate axes. And itβs usually a good idea to mark any important points of the subdomains of our function. In this case, weβll add π₯ is equal to six onto our sketch. We need to sketch each subfunction over its subdomain separately.

Letβs start with the first subfunction. Thatβs π₯ squared minus 36 all divided by π₯ minus six when π₯ is not equal to six. We can see that this is a rational function, and rational functions are difficult to sketch. So letβs see if we can simplify this first. One way of doing this is to factor our numerator by noticing it is a difference between two squares. We know π₯ squared minus 36 will be equal to π₯ minus six multiplied by π₯ plus six. So by factoring the numerator of our first subfunction, we get π₯ squared minus 36 all over π₯ minus six is equal to π₯ minus six times π₯ plus six all divided by π₯ minus six.

And now we can notice something interesting. In the subdomain of this subfunction, we do not include the value of π₯ is equal to six. And if π₯ is not equal to six, then π₯ minus six is not equal to zero. Therefore, we can cancel the shared factor of π₯ minus six in our numerator and our denominator. Since π₯ minus six is not equal to zero, π₯ minus six divided by π₯ minus six is just equal to one. This just leaves us with π₯ plus six. In fact, we could update our subfunction to just be π₯ plus six, since these two functions output the same values provided π₯ is not equal to six.

We want to sketch this subfunction onto our diagram. We need to sketch π¦ is equal to π₯ plus six, where π₯ is not allowed to be equal to six. Thereβs a few ways of sketching this. First, we note that the π¦-intercept of this line is six. We can then mark this on our diagram. Next, the line has slope one. So for every unit we move right, we move one unit up. Alternatively, we can find the coordinates of the π₯-intercept. We do this by substituting π¦ is equal to zero into our equation and solving. We see the π₯-intercept will be at negative six.

But remember, our subfunction is not sketched for the value of π₯ is equal to six. Itβs not in the subdomain of this subfunction. So at this point, we need to include a hollow circle. This represents that the function is not sketched at this point. And we can find the coordinates of this point by substituting π₯ is equal to six into this linear function. Its π¦-coordinate is going to be 12.

But remember, our input values of π₯ are all values of π₯ except π₯ is equal to six. So our line needs to continue infinitely in both directions. We represent these by using arrows. Therefore, we sketch the first subfunction of π of π₯. However, we still need to sketch the second subfunction of π of π₯. This subfunction is a constant value of 12 at a single value π₯ is equal to six. So the graph of this subfunction is going to be a single point, the point with coordinates six, 12. We need to include this onto our diagram. The point with coordinates six, 12 needs to be included in our sketch of π¦ is equal to π of π₯. So we need a solid dot at this point.

And now that weβve sketched all of the subfunctions of π of π₯ over their subdomain, weβve successfully sketched the graph of π¦ is equal to π of π₯. And now we can notice something interesting at this point. The function π of π₯ is exactly equal to the function π₯ plus six for all real values of π₯. This is because the graph of π of π₯ is just the straight line π¦ is equal to π₯ plus six. And we can use this to determine the range of our function. The range of any nonhorizontal linear function is just all real numbers β. So the range of π of π₯ is all real numbers.

Itβs worth noting we can also see this from our sketch. Since the range is the set of output values of the function, these are given by the π¦-coordinates of any point on our sketch. For example, we can see that 12 is in the range of our function, since thereβs a point on the sketch with π¦-coordinate 12. And this is true for any real value. So the range is all real values.

Therefore, we were able to determine the domain and the range of the function π of π₯. The domain is all real numbers, and the range is also all real numbers.

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