Question Video: Finding the Domain and Range of a Function from Its Graph | Nagwa Question Video: Finding the Domain and Range of a Function from Its Graph | Nagwa

# Question Video: Finding the Domain and Range of a Function from Its Graph Mathematics

Which of the following are the range and domain of the function π(π₯) = |βπ₯ + 3| β 5? [A] The domain is β, and the range is [β5, β). [B] The domain is β, and the range is (ββ, β5]. [C] The domain is [β5, β), and the range is β. [D] The domain is β, and the range is [0, β). [E] The domain is β, and the range is (3, β).

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### Video Transcript

Which of the following are the range and domain of the function π of π₯ is equal to the absolute value of negative π₯ plus three minus five? Is it option (A) the domain is the set of real numbers and the range is the left-closed, right-open interval from negative five to β? Is it option (B) the domain is the set of real numbers and the range is the left-open, right-closed interval from negative β to negative five? Is it option (C) the domain is the left-closed, right-open interval from negative five to β and the range is the set of real numbers? Is it option (D) the domain is the set of real numbers and the range is the left-closed, right-open interval from zero to β? Or is it option (E) the domain is the set of real numbers and the range is the open interval from three to β?

In this question, weβre given the graph of the function π¦ is equal to π of π₯. And weβre given the definition of π of π₯: π of π₯ is the absolute value of negative π₯ plus three minus five. We need to use these to determine which of five possible options give us the range and domain of this function. To do this, letβs start by recalling what we mean by the range and domain of a function. First, we recall that the domain of a function is the set of all possible input values for that function. Second, we recall that the range of a function is the set of all output values of the function given its domain.

And we can determine the domain and range of this function in two different ways. We can either do this by using the definition of π of π₯ or we can do this from its graph. Weβll go through both of these methods. Letβs start by doing this from the definition of π of π₯. First, we need to recall that the absolute value of π₯ has the domain of all real numbers and a range of the left-closed, right-open interval from zero to β. We can then see that π of π₯ is actually a transformation of the absolute value function. And we can even determine the exact series of transformations which take us from π¦ is equal to the absolute value of π₯ to π¦ is equal to π of π₯.

One way of doing this would be to write π of π₯ as the absolute value of negative one times π₯ minus three minus five. This then allows us to determine the series of transformations. Itβs a reflection in the vertical axis, a translation of three units right, and then a translation of five units down. However, we can use our rules of absolute values to make this even simpler. The absolute value of negative one times π₯ minus three will just be the absolute value of π₯ minus three. So π of π₯ is equal to the absolute value of π₯ minus three minus five. This is then just a translation of the absolute value graph three units right and five units down.

However, we donβt need to use this graph to answer this question. Instead, we know translating vertically wonβt affect the domain of the function since these are the π₯ input values and translating horizontally wonβt affect the input either since this is all real numbers. So we can already conclude the domain of π of π₯ is the same as the domain of the absolute value of π₯. The domain is all real numbers β.

Next, we can determine the range. Horizontal translations will not affect the range. However, vertical translations will. Weβre translating the entire graph five units down. Or in other words, weβre subtracting five from every possible output value. So the range will change from the left-closed, right-open interval from zero to β to the left-closed, right-open interval from negative five to β. And we can see that this is the answer given in option (A).

And we could stop here. However, letβs also go over how to determine the domain and range of this function from its graph. In the graph of a function, the π₯-coordinate of any point on the curve tells us the input value and the π¦-coordinate tells us the corresponding output value of the function. For example, we can see the point with coordinates three, negative five lies on the curve. Since the π₯-coordinate tells us the input value of the function and the π¦-coordinate tells us the corresponding output, we can conclude that π evaluated at three must be equal to negative five, which we can confirm by substituting three into our function π of π₯.

This tells us two pieces of information. First, three is in the domain of our function since we can substitute three into the function. And negative five is an element of the range since itβs a possible output value. Hence, the π₯-coordinates of points on the graph tell us the elements of the domain of the function and the π¦-coordinates of points on the graph tell us the elements of the range of the function. This will allow us to determine the domain and range of the function from its graph.

First, we know that the graph of the function continues indefinitely in both directions. Next, we note that any possible vertical line intersects the graph, which means we can input any real value we want into the function. The domain of the function is the set of all real numbers. However, the same is not true for horizontal lines, we can see that there is a lowest possible horizontal line which still intersects the graph. The lowest π¦-coordinate of a point on the graph is negative five. So the lowest possible output value of the function is negative five. And we can see every vertical line above negative five will intersect the curve in at least one place. Therefore, the range of this function will be all values greater than or equal to negative five. Thatβs the left-closed, right-open interval from negative five to β.

Therefore, we were able to show two different ways of determining the range and domain of the function π of π₯ is equal to the absolute value of negative π₯ plus three minus five. In both cases, we showed the domain is the set of real numbers and the range is the left-closed, right-open interval from negative five to β, which was option (A).