Question Video: Finding the Range of an Absolute Value Function | Nagwa Question Video: Finding the Range of an Absolute Value Function | Nagwa

# Question Video: Finding the Range of an Absolute Value Function Mathematics

Which of the following is the range of π(π₯) = β3 β |2 β π₯|? [A] [β3, β) [B] (ββ, 0] [C] (ββ,β 3] [D] β β {2} [E] β

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### Video Transcript

Which of the following is the range of π of π₯ is equal to negative three minus the absolute value of two minus π₯? Option (A) the left-closed, right-open interval from negative three to β. Option (B) the left-open, right-closed interval from negative β to zero. Option (C) the left-open, right-closed interval from negative β to negative three. Option (D) the set of real numbers minus the set including two. Or option (E) the set of real numbers.

In this question, weβre given a function π of π₯, and weβre asked to determine which of five given options is the range of this function. So letβs start by recalling what we mean by the range of a function. Itβs the set of all output values of the function given its domain. And thereβs several different ways we can determine the range of a function. One way of doing this is from a graph of the function. And to determine the range of π of π₯ in this way, weβre going to need to sketch the graph π¦ is equal to π of π₯. To do this, we can note that the equation of π of π₯ is similar to the equation of π of π₯ is the absolute value of π₯. In fact, we can perform a series of transformations of π of π₯ to get the graph of π¦ is equal to π of π₯.

To determine the transformations needed to transform π of π₯ into π of π₯, letβs start by rewriting π of π₯ in terms of π of π₯. First, weβll reorder the two terms of π of π₯. π of π₯ is equal to negative one times the absolute value of two minus π₯ minus three. We could now write this in terms of π of π₯. However, we can simplify this slightly by taking a factor of negative one outside the expression inside the absolute value symbol.

We notice the absolute value of negative one times π₯ minus two will be equal to the absolute value of π₯ minus two, because remember the absolute value of an expression only cares about its size, not its sign. Therefore, π of π₯ is equal to negative one times the absolute value of π₯ minus two minus three. And we can write this as negative one times π of π₯ minus two minus three.

And now that weβve written π of π₯ in this form, we can determine the series of transformations that will transform π¦ is equal to π of π₯ into π¦ is equal to π of π₯. We start with the coefficient of π. This is negative one, which means we need to reflect in the horizontal axis. Next, thereβs no coefficient of π₯, so we move on to subtracting two from the inputs. This means we translate the graph two units right. Finally, weβre subtracting three from the output values. This means we need to translate the graph three units down.

We can now sketch π¦ is equal to π of π₯ by applying these transformations to the graph of π¦ is equal to the absolute value of π₯. First, we need to reflect this in the horizontal axis. Doing this, we get the following. This is the graph π¦ is equal to negative the absolute value of π₯. After this, we need to translate the graph two units to the right. Translating the graph two units to the right gives us the following. Itβs the sketch of π¦ is equal to negative the absolute value of π₯ minus two. And whenever weβre sketching the graph of a function, it can be useful to include useful points on the graph.

For example, in π¦ is equal to negative the absolute value of π₯, the corner of the graph is at the origin. When we translate this two units right, it will be at π₯ is equal to two. This is the π₯-intercept of this graph. Finally, we need to translate this graph three units down. Translating the graph three units down gives us the graph of π¦ is equal to negative the absolute value of π₯ minus two minus three. And we can note that the points at the corner of our curve translate down three units to the coordinates two, negative three.

Weβre now ready to determine the range of our function from its graph. However, letβs start by clearing some space so we can work with a bigger diagram. To determine the range of π¦ is equal to π of π₯ from its graph, we start by recalling the π₯-coordinate of a point on its graph tells us the input value of π₯. And the π¦-coordinate tells us the corresponding output of the function. For example, because the point with coordinates two, negative three lies on the graph, π evaluated at two must be equal to negative three. Negative three is an output of our function. In fact, we can see itβs the largest output. Another way of saying this is itβs the point with largest π¦-coordinate which lies on the graph.

Similarly, we can see that all values of π¦ less than or equal to negative three are possible outputs of the function. Any horizontal line below π¦ is equal to negative three intersects the graph. Therefore, the range of this function is all values less than or equal to negative three. We write this as the left-open, right-closed interval from negative β to negative three, which we can see is option (C). Therefore, the range of π of π₯ is equal to negative three minus the absolute value of two minus π₯ is option (C) the left-open, right-closed interval from negative β to negative three.