Question Video: Finding the Domain and Range of Absolute Value Functions Mathematics

Find the domain and range of the function 𝑓(π‘₯) = |3π‘₯ βˆ’ 1| + 3π‘₯ + 5.

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Video Transcript

Find the domain and range of the function 𝑓 of π‘₯ is equal to the absolute value of three π‘₯ minus one plus three π‘₯ plus five.

In this question, we’re given a function 𝑓 of π‘₯ and we’re asked to determine the domain and the range of this function. So to answer this question, let’s start by recalling what we mean by the domain and range of a function. First, the domain of a function is the set of all input values for that function. Second, the range of a function is the set of all output values for the function given its domain. And there’s many different ways of determining the domain and range of a function. One way of doing this is to sketch its graph. To help us sketch the graph of 𝑦 is equal to 𝑓 of π‘₯, let’s call the function the absolute value of three π‘₯ minus one 𝑔 of π‘₯. Then, since this contains an absolute value, we can rewrite this as a piecewise function.

To do this, we recall the absolute value of a function changes definition depending on whether it’s taking the absolute value of a positive or a negative number. Therefore, if three π‘₯ minus one is greater than or equal to zero, the absolute value of three π‘₯ minus one is just three π‘₯ minus one. However, if three π‘₯ minus one is less than zero, when we take its absolute value, we need to multiply this by negative one. And negative one times three π‘₯ minus one is one minus three π‘₯. This then gives us a piecewise definition of the function 𝑔 of π‘₯. And it’s worth noting we can simplify this definition by simplifying its subdomains. We can add one to both inequalities and then divide both inequalities through by three. We get π‘₯ is greater than or equal to one-third and π‘₯ is less than one-third, giving us the following expression for 𝑔 of π‘₯.

Now, we can use this to find a piecewise definition of the function 𝑓 of π‘₯. First, we see that our function 𝑓 of π‘₯ is equal to 𝑔 of π‘₯ plus three π‘₯ plus five. This means the output of our function 𝑓 of π‘₯ is going to depend on the output of our function 𝑔 of π‘₯. If π‘₯ is greater than or equal to one-third, 𝑔 of π‘₯ is going to output three π‘₯ minus one. We can then substitute this into our function 𝑓 of π‘₯. If π‘₯ is greater than or equal to one-third, 𝑓 of π‘₯ is equal to three π‘₯ minus one plus three π‘₯ plus five, which we can simplify as six π‘₯ plus four. However, if our input value of π‘₯ is less than one-third, then 𝑔 of π‘₯ is one minus three π‘₯.

We can then substitute this into our function 𝑓 of π‘₯. If π‘₯ is less than one-third, 𝑓 of π‘₯ is equal to one minus three π‘₯ plus three π‘₯ plus five, which we can simplify. Negative three π‘₯ plus three π‘₯ is zero, so 𝑓 of π‘₯ is just equal to six when π‘₯ is less than one-third. And if 𝑓 of π‘₯ is equal to the constant value of six when π‘₯ is less than one-third and 𝑓 of π‘₯ is equal to six π‘₯ plus four when π‘₯ is greater than or equal to one-third, 𝑓 of π‘₯ is a piecewise-defined function. And now we can sketch the graph 𝑦 is equal to 𝑓 of π‘₯ to determine its domain and range.

However, before we do this, recall that the domain of a piecewise-defined function is the union of its subdomains. So in fact, we can determine the domain of our function 𝑓 of π‘₯ from its piecewise definition. It’s all values of π‘₯ less than one-third or values of π‘₯ greater than or equal to one-third. In other words, it’s all real values of π‘₯.

Now we need to determine the range of this function. We’ll do this by sketching its graph. First, the function is a constant value of six when π‘₯ is less than one-third. And since the output values are a constant value of six, this is the horizontal line 𝑦 is equal to six, where our input values of π‘₯ must be less than one-third. And we represent this by a hollow circle when π‘₯ is one-third. Next, when π‘₯ is greater than or equal to one-third, our function 𝑓 of π‘₯ is a linear function: six π‘₯ plus four. And to sketch this linear function, let’s start by finding its endpoint, the value when π‘₯ is one-third. We substitute π‘₯ is equal to one-third into the linear function to get six times one-third plus four. And if we evaluate this, we see it’s equal to six.

Therefore, the endpoint of this linear function is the point with coordinates one-third, six. And this is the hollow dot we already had on our graph, which means we can now fill this in because this is now a point on the graph of our function. It’s the endpoint of the linear portion. We could now sketch the rest of this graph accurately. However, we’ll see it’s not necessary to determine its range. All we need to know is that this line has positive slope. Its slope is six. And so as π‘₯ approaches ∞, the line is going to approach ∞.

We’re now ready to determine the range of this function from its graph. Remember, the range of the function is the set of all possible output values of the function. And when we sketch the graph of a function, the π‘₯-coordinate of the point on the curve tells us the input value and the 𝑦-coordinate tells us the corresponding output value. Therefore, the 𝑦-coordinates are points on the curve that are telling us the range of our function. For example, from the graph, we can see that six is an output value of our function. In fact, it’s the lowest output. It’s the lowest 𝑦-coordinate of a point on the graph. And we also know that our line extends to ∞. Therefore, any value greater than or equal to six is a possible output of the function. So the range of this function contains all values greater than or equal to six.

Remember, we need to write this as a set. This is the left-closed, right-open interval from six to ∞. Therefore, we were able to show the domain of the function 𝑓 of π‘₯ given to us in the question is the set of real numbers, and its range is the left-closed, right-open interval from six to ∞.

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