Lesson Video: The Domain and the Range of a Radical Function | Nagwa Lesson Video: The Domain and the Range of a Radical Function | Nagwa

# Lesson Video: The Domain and the Range of a Radical Function Mathematics • Second Year of Secondary School

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In this video, we will learn how to find the domain and the range of a radical function either from its graph or from its defining rule.

17:35

### Video Transcript

In this video, we will learn how to find the domain and the range of a radical function, either from its graph or from its defining rule. In particular, we will focus on the domain and range of functions involving the square and cube roots.

Letβs begin by recalling the definitions of the domain and range of a function. The domain of a function π of π₯ is the set of all possible values of π₯ such that the expression π of π₯ is defined. In formulae, we can think of the domain of a function as the set of all values on which the function acts, or its input values. The range of a function π of π₯ is the set of all possible values the expression π of π₯ can take when π₯ is any number from the functionβs domain. We can therefore think of the range of a function as a set of values the function produces, or its output values.

Letβs begin by considering the most straightforward radical function: π of π₯ is equal to the square root of π₯. If there is a real number π¦ satisfying π¦ is equal to the square root of π₯, then it follows that π₯ is equal to π¦ squared. We know that the square of any real number is nonnegative, so π₯ must also be nonnegative. This means that any input value to the function π of π₯ must be nonnegative, which tells us that the domain of the square root function is π₯ is greater than or equal to zero. We can also express this in interval notation as π₯ belongs in the left-closed, right-open interval from zero to β.

Now, letβs consider the range of the square root function. Here is a sketch of its graph. We can see that the graph appears to become flatter as π₯ increases. But its value still continues to increase without bound. This tells us that any positive real value is a possible output value from the function π of π₯ equals the square root of π₯. And the value of zero is also included because the square root of zero is zero. The range of the square root function is therefore the nonnegative real numbers. So we can say π of π₯ is greater than or equal to zero. Again, we can express this in interval notation as π of π₯ belongs in the left-closed, right-open interval from zero to β.

Notice that we use π₯ to describe the domain of the function because it is the input values, whereas we use π of π₯ to describe the range because weβre discussing the output values. We also note that, in this instance, the values in the domain and range of the square root function are in fact the same. But that certainly wonβt always be the case. We also observe at this point that, on the graph of a function, the functionβs domain corresponds to the part of the horizontal axis where the graph exists, whereas the range of the function corresponds to the part of the vertical axis where the graph exists. This gives us a useful method for finding the domain and range of a function if weβre given its graph.

Letβs now consider a more general square root function: π of π₯ is equal to the square root of π of π₯. This time, weβre not finding the square root simply of π₯, but of a function of π₯, which weβre calling π of π₯. We therefore have a composite function or a function of a function. It follows from what weβve already seen that the input values for the function π of π₯ must be nonnegative. And so π of π₯ must be nonnegative. The domain of composite square root function as shown then is the set of all π₯-values such that the function π of π₯ is greater than or equal to zero. The range of the function π of π₯ will depend on what the function π of π₯ is. But we can determine the range by considering the greatest and smallest values the function can take, as weβll see in our examples.

Letβs begin then with an example in which weβll find the domain of a composite square root function.

Find the domain of the function π of π₯ equals the square root of seven π₯ minus seven.

We have here a composite square root function. If we allow π of π₯ to be the function under the square root, so π of π₯ is equal to seven π₯ minus seven, then π of π₯ is equal to the square root of π of π₯. We recall that the domain of a composite square root function of the form π of π₯ equals the square root of π of π₯ can be found by finding the set of all values of π₯ such that π of π₯ is nonnegative. We therefore need to solve the inequality seven π₯ minus seven is greater than or equal to zero.

We can do this by first adding seven to each side, giving seven π₯ is greater than or equal to seven. And then we can divide each side of the inequality by seven to give π₯ is greater than or equal to one. We can express this in interval notation as the left-closed, right-open interval from one to β.

Letβs now consider an example in which weβll identify the graph of a composite square root function by considering both its domain and range.

Which of the following is the graph of π of π₯ equals the square root of one minus two π₯?

Weβre going to approach this problem by considering the domain and the range of the function π of π₯. This is a composite square root function because we can think of it as π of π₯ equals the square root of π of π₯, where π of π₯ is the function one minus two π₯. We recall that the domain of a composite square root function of this form is the set of all values of π₯ for which π of π₯ is greater than or equal to zero.

We can therefore find the domain of this function by solving the inequality one minus two π₯ is greater than or equal to zero. Subtracting one from each side gives negative two π₯ is greater than or equal to negative one. And then we divide both sides by negative two, remembering that when we divide an inequality by a negative value, we need to reverse the direction of the sign. So we have π₯ is less than or equal to negative one over negative two, which is one over two, or one-half. Weβve therefore found that the domain of the function is the set of values from negative β to one-half.

Now, recall that, on a graph of a function, its domain corresponds to the part of the horizontal axis where the graph exists. Considering the five graphs weβve been given, we can see that this rules out options (A), (C), and (E) as they are defined for values of π₯ not in the interval from negative β to one-half, whereas options (B) and (D) are only defined on the correct interval.

Letβs now consider the range of the function π of π₯. We recall that the range of the square root function is the set of all nonnegative real numbers, which we can express in interval notation as the left-closed, right-open interval from zero to β. As the range of the function π of π₯ under the square root is all real numbers, the square root of π of π₯ has the same range as the square root of π₯. Hence, the range of π of π₯ is all nonnegative real numbers.

Remember that, on the graph of a function, its range corresponds to the part of the vertical axis where the graph exists. In graph (B), we can see that the function is entirely below the π₯-axis. And so the range of this function is the set of all negative real numbers and zero. This therefore rules out graph (B). In graph (D), however, the range is indeed the set of all nonnegative real numbers, as the graph is above and on the π₯-axis. We also note that this graph does indeed have the correct shape for a square root function. And as it also has the correct domain and range for this square root function, we can be satisfied that graph (D) is the graph of π of π₯ equals the square root of one minus two π₯.

In the examples weβve considered so far, weβve found the domain and range of composite square root functions when the expression inside the square root is linear. In the next example, weβll find the domain and range of a composite square root function where instead the expression inside the square root involves the absolute value function.

Consider the function π of π₯ equals the square root of four minus the absolute value of π₯ minus five. Part one, find the domain of π of π₯, and part two, find the range of π of π₯.

In this problem, we have a composite square root function of the form π of π₯ equals the square root of π of π₯. We recall that the domain of a composite square root function is the set of all values of π₯ for which π of π₯ is nonnegative. For this function, this gives the inequality four minus the absolute value of π₯ minus five is greater than or equal to zero.

To solve this inequality, we first need to isolate the absolute value function. We can subtract four from each side of the inequality and then multiply or divide each side of the inequality by negative one, remembering that when we do we need to reverse the direction of the inequality sign. So we have the absolute value of π₯ minus five is less than or equal to four. If the absolute value of π₯ minus five is less than or equal to four, this means that the distance of the expression π₯ minus five from zero must be no greater than four. Or in other words, π₯ minus five is greater than or equal to negative four and less than or equal to four.

To solve this double-sided inequality, we add five to each part, giving π₯ is greater than or equal to one and less than or equal to nine. So we found that the domain of the function π of π₯ is the closed interval from one to nine.

Now, letβs consider the range of π of π₯. As the range of a function is the set of all possible values of that function, we can obtain the range by considering what the largest and smallest values the function can take are. As our function π of π₯ is a square root function, we know that it cannot output negative values. Therefore, as a minimum, π of π₯ is greater than or equal to zero. Zero would be the smallest value of the function if it is possible to produce zero from a value in the domain of π of π₯.

To see whether zero is indeed a possible output value for this function, we would need the expression under the square root to be equal to zero because the square root of zero is zero. This would give the equation four minus the absolute value of π₯ minus five is equal to zero. Solving this absolute value equation gives the absolute value of π₯ minus five equals four. This means π₯ minus five is either equal to negative four or four, leading to π₯ equals one or π₯ equals nine. Both of these values are in the domain of the function π of π₯, which means it is possible to achieve zero using a value in the domain.

So weβve found the smallest possible value of π of π₯. And now letβs consider what the largest possible value is. The largest value of π of π₯ will correspond to the largest value of π of π₯, which will in turn correspond to the smallest possible value of the absolute value of π₯ minus five. The absolute value function is always nonnegative, so its smallest value occurs when it is equal to zero. This occurs when π₯ is equal to five, which is in the domain of our function.

At this point, π of π₯ will be four minus zero, which is four, and π of π₯, remember, is the square root of π of π₯. So itβs the square root of four, which is two. The largest value of π of π₯ then is two. As π of π₯ is a continuous function, its range will be everything from its smallest value to its largest value. So it is the closed interval from zero to two.

So weβve completed the problem. The domain of π of π₯ is the closed interval from one to nine. And the range is the closed interval from zero to two.

Letβs now consider the domain and the range of the cube root function π of π₯ equals the cubed root of π₯. The graph of the cube root function looks like this. Unlike the square root function, we notice that the function extends to both the left and the right side of the π¦-axis, indicating that the cube root can take any real numbers as its input. This suggests that the domain of the cube root function is the open interval from negative β to positive β.

We also note that there are sections of the graph both above and below the π₯-axis, indicating that the cube root function outputs both positive and negative values. The values of the function tend to positive and negative β as the π₯-values tend to positive and negative β, respectively. This indicates that the range of the cube root function is the set of all real numbers. So weβve seen that, unlike the square root function, there is in fact no restriction on the domain and range of the cube root function.

Letβs just consider why this is the case. If there is a real value π¦ such that π¦ is equal to the cube root of π₯, then it follows that π₯ is equal to π¦ cubed. But we know that if we cube positive values, we get a positive answer, whereas if we cube negative values, we get a negative answer. So this time, π₯ can take both positive and negative values as well as zero, and so can π¦, the value of the function.

Letβs now consider one final example in which we determine the domain and range of a function which involves both the square and cube roots.

Consider the function π of π₯ equals the cubed root of 125 minus the square root of two π₯ plus three. Part (a): Find the domain of π of π₯. Part (b): Find the range of π of π₯.

We have here a cube root function and then a square root function in the expression within the cube root. We recall first that the domain of a cube root function is the set of all real numbers. So as there is no restriction on the domain of the cube root function, we only need to consider the restriction for the square root function.

The domain of a square root function must be nonnegative. So we have the inequality two π₯ plus three is greater than or equal to zero. Solving this inequality leads to π₯ is greater than or equal to negative three over two. So we find that the domain of the function π of π₯ is the left-closed, right-open interval from negative 1.5 to β, because we can only evaluate the entire expression under the cube root when π₯ is in this interval.

Letβs now consider the range of π of π₯. We can write our function as π of π₯ equals the cube root of 125 minus π, where π is equal to the square root of two π₯ plus three. We know that the range of a square root function is the set of all nonnegative values. And so π is greater than or equal to zero. The largest value of π of π₯ will correspond to the smallest value of π, which is zero. So the largest value of π of π₯ is the cube root of 125 minus zero, which is the cube root of 125, which is five.

The smallest value of π of π₯ will correspond to the largest value of π. So as π tends to β, π of π₯ will tend to the cube root of negative β, which itself tends to negative β. As π of π₯ is a continuous function, its range will therefore include all the values between its smallest and largest value, which is the left-open, right-closed interval from negative β to five.

So weβve completed the problem and found the domain and range of this fairly complicated radical function, which involves both square and cube roots. The domain is the left-closed, right-open interval from negative 1.5 to β. And the range is the right-open, left-closed interval from negative β to five.

Letβs now summarize the key points from this video. Firstly, the domain and range of the square root function π of π₯ equals the square root of π₯ are both the left-closed, right-open interval from zero to β. We saw that the domain of a function corresponds to the part of the horizontal axis at which the graph exists, whereas the range corresponds to the part of the vertical axis at which the graph exists.

For a composite square root function π of π₯ equals the square root of π of π₯, the domain is the set of values of π₯ for which π of π₯ is greater than or equal to zero. We also saw that, for the cube root function π of π₯ equals the cube root of π₯, the domain and range are each the set of real numbers, which we can express as the open interval from negative β to β. For more complicated square or cube root functions, the range of the function will depend on the values in its domain.

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