### Video Transcript

Find the domain and range of the function π of π₯ equals negative four times the absolute value of π₯ minus five minus one.

Weβll begin by recalling what we actually mean by the domain and range of a function. We say that the domain of a function is the complete set of possible values of our independent variable. In other words, the set of all possible values for π₯ which make the function work and it will ensure it outputs real π¦-values.

Formally, we also say that the range of a function is the complete set of possible resulting values for the independent variable after weβve substituted the domain. In other words, the resulting π¦-values we get after substituting in possible π₯-values.

So letβs think about what our function is telling us. We take a value for π₯, and we substitute it into the expression π₯ minus five. No matter our result, we find the absolute value of that. In other words, we make it a positive number. We then multiply that by negative four and subtract one. So are there any values of π₯ which will make the function not work?

Well, no, all real values of π₯ will give us a real value for our function. And so we can say that the domain of our function is the set of real numbers. And in fact, the domain of a function which is of the form the absolute value of some polynomial plus or minus or multiplied by constants will always be the set of all real numbers. But what about the range of this?

Well, we have two ways we can check this. One way is to consider it algebraically. Well, firstly, we know that if π₯ is equal to five, the absolute value of π₯ minus five is the absolute value of zero, which is just zero. But if π₯ is not equal to five, we substitute it in. We get π₯ minus five, and we make that positive. So the absolute value of π₯ minus five will always be greater than zero as long as π₯ is not equal to five. And therefore, we can say that the absolute value of π₯ minus five will always be greater than or equal to zero.

Remember, our function is negative four times the absolute value of π₯ minus five minus one. So letβs multiply both sides of our inequality by negative four. When we do, the left-hand side of our inequality becomes negative four times the absolute value of π₯ minus five. And the right-hand side is still zero.

But remember, when multiplying an inequality by a negative number, we have to reverse the inequality symbol. So we get negative four times the absolute value of π₯ minus five is less than or equal to zero.

Next, weβll subtract one from both sides. And we get negative four times the absolute value of π₯ minus five minus one is less than or equal to negative one. So we can see that the output of this function will always be less than or equal to negative one. And using set notation, the range is as shown. Our output will be greater than negative infinity but less than or equal to negative one.

But there was another way we could find the range. And thatβs to consider the graph of our function. Letβs begin by considering the graph of π¦ equals π₯ minus five. Itβs a single straight line that passes through the π¦-axis at negative five and the π₯-axis at five. π¦ is equal to the absolute value of π₯ minus five is as shown.

Essentially, we make all π¦-values positive. And what this ends up looking like is a reflection of the part of the graph that lies below the π₯-axis in the π₯-axis. Notice that this does include the point of intersection of the graph with the π₯-axis.

Multiplying our function by negative four reflects in the π₯-axis and makes it four times steeper. Then we subtract one, which results in a vertical translation by one unit. The range is the possible π¦-values of this graph. We can see that the highest point of the graph is at five, negative one. So our π¦-values will always be less than or equal to negative one. And the range is the same as before.