### 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.

Letβs 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 functionβs domain. We can therefore think of the range
of a function as a set of values the function produces, or its output values.

Letβs 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, letβs 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 weβre 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 wonβt always be
the case. We also observe at this point that,
on the graph of a function, the functionβs 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 weβre given its graph.

Letβs now consider a more general
square root function: π of π₯ is equal to the square root of π of π₯. This time, weβre not finding the
square root simply of π₯, but of a function of π₯, which weβre calling π of π₯. We therefore have a composite
function or a function of a function. It follows from what weβve 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 weβll see in
our examples.

Letβs begin then with an example in
which weβll 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.

Letβs now consider an example in
which weβll 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₯?

Weβre 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. Weβve 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 weβve
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.

Letβs 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 weβve considered so
far, weβve found the domain and range of composite square root functions when the
expression inside the square root is linear. In the next example, weβll 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, letβs 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 weβve found the smallest
possible value of π of π₯. And now letβs 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 itβs 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 weβve 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.

Letβs 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 weβve seen that, unlike the
square root function, there is in fact no restriction on the domain and range of the
cube root function.

Letβs 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.

Letβs 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.

Letβs 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 weβve 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.

Letβs 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.