Question Video: Using the Graph of a Radical Function to Identify the Domain and Range Mathematics

Consider the function 𝑓(π‘₯) = √(π‘₯ βˆ’ 1). Which of the following graphs could represent 𝑓(π‘₯)? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E. Using the graph of 𝑓(π‘₯), find its domain. Using the graph of 𝑓(π‘₯), find its range.

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Video Transcript

Consider the function 𝑓 of π‘₯ equals the square root of π‘₯ minus one. Which of the following graphs could represent 𝑓 of π‘₯? Using the graph of 𝑓 of π‘₯, find its domain. Using the graph of 𝑓 of π‘₯, find its range.

In order to identify the correct graph of 𝑓 of π‘₯ equals the square root of π‘₯ minus one, let’s remind ourselves what the function the square root of π‘₯ looks like when graphed. It’s the inverse of the function 𝑓 of π‘₯ equals π‘₯ squared, but we restrict it to make sure that it’s one to one. And so it looks a little something like this. In order to use this graph to identify the correct graph of the square root of π‘₯ minus one, let’s recall a function transformation.

Suppose we have the graph of 𝑦 equals 𝑓 of π‘₯. This is mapped onto the graph of 𝑦 equals 𝑓 of π‘₯ minus π‘Ž for some real constant π‘Ž by a transformation by the vector π‘Ž, zero. In other words, the graph is moved π‘Ž units to the right. So in this case, we are subtracting one from the π‘₯ inside the square root symbol, meaning that the graph of 𝑔 of π‘₯ must be translated one unit to the right to map onto 𝑓 of π‘₯. So it will intersect the π‘₯-axis at one and look a little something like this. If we look carefully at all five of our graphs, we can observe that that is graph (B).

The next part of this question asks us to use the graph to find the domain of our function. And so we recall that the domain is a set of possible inputs to the function. In other words, for some function 𝑓 of π‘₯, what π‘₯-values can we substitute into that function to get real outputs? With this definition in mind, it follows that we can use the spread of π‘₯-values on our graph to find the domain of our function. Now, if we look at the spread of π‘₯-values on graph (B), we see that they start at π‘₯ equals one and extend to positive ∞. So π‘₯ can take values greater than or equal to one. In set notation, we say that the domain is the set of values in the left-closed, right-open interval from one to ∞.

And so we’re ready for the final part of this question. It asks us to find the range by using the graph of 𝑓 of π‘₯. The range of the function is the set of possible outputs to the function when the values from the domain are substituted in. In other words, given a function 𝑦 equals 𝑓 of π‘₯, the range is the set of possible 𝑦-values. And so we can look at the spread of possible 𝑦-values in the vertical direction on our graph to establish the range. On graph (B), we see that the 𝑦-values start at zero and they extend towards ∞. Now, whilst it might look like they reach some sort of limit, we know that this is not true since the square root of ∞ minus one is simply ∞.

So the range, the set of possible outputs, is all values greater than or equal to zero. Using set notation, the range of the function is the left-closed, right-open interval from zero to ∞.

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