### Video Transcript

In this video, we will learn how to
find the inverse of a function by changing the subject of the formula.

In order to consider the inverse of
a function, we must first be familiar with functions themselves. We should recall that a function,
which we might call π, takes an input value π₯ and maps it to an output value π of
π₯. Itβs defined by an equation showing
how the output is calculated from the input. For example, the function π of π₯
equals two π₯ minus three takes a value of π₯. It then doubles it and subtracts
three to give an output value. If we were to input the value four
to this function, we would find that π of four is equal to two multiplied by four
minus three, thatβs eight minus three, which is equal to five. So the input four leads to the
output five.

We can also visualize this using a
diagram to show the steps involved in mapping an input value to an output value. We multiply four by two to give
eight and then subtract three to give five. If we were to plot the graph of π¦
equals π of π₯, we would, in this instance, get a straight line representing the
mapping all possible input values of π₯ to the output values π of π₯. We could then use this graph to
work out the output value for a given input value. For example, for an input value of
two, we would go up from two on the π₯-axis until we meet the line. We then go across from the line to
the π¦-axis, and we find that the output value for an input value of two is one.

We can also use this graph to
illustrate what we mean by the domain and range of a function. The domain of a function is the set
of all possible input values, which corresponds to all values on the π₯-axis for
which the graph can be plotted. In this case, the domain of the
function is the complete set of real numbers. The range of a function is the set
of all outputs, which are all the π¦-values on the graph. Because this line continues
infinitely in both directions, the range of this function is also the complete set
of real numbers.

Now that weβve recapped the
functions themselves, letβs think about their inverses. Suppose we had an output value π¦,
in this case π¦ equals three, and we wanted to find the input value corresponding to
it. Well, we could do this from the
graph. We could go across from the
π¦-value of three to the line and then down to the π₯-axis to find the corresponding
input value, which it happens is also three. But it would be nice to be able to
do this without needing to draw the graph. We could substitute this value into
our function as the output value and then solve for π₯. Doing so, we find that an output
value of three is indeed associated with an input value also of three.

But if we wanted to find the input
value for many output values, it would be more efficient to find a general equation
we can use rather than substituting each value and then solving the equation each
time. To do this, we could rearrange the
general equation π¦ equals π of π₯ to make π₯ the subject. Starting with the equation π¦
equals two π₯ minus three then, we can add three to both sides to give π¦ plus three
is equal to two π₯ and then divide both sides of the equation by two to give π¦ plus
three over two equals π₯. And weβve rearranged the formula to
make π₯ the subject. As before, if we substitute the
output or π¦-value of three, we get π₯ equals three plus three over two, which is
six over two, which is equal to three.

But we can use the same equation to
find the input value for any output value. For example, when π¦ is equal to
zero, π₯ is zero plus three over two, which is 1.5. The process weβve just gone through
demonstrates how to find the inverse of a function. Informally, the inverse of a
function is a function that undoes the original function. If a function π applied to an
input π₯ gives the output π¦, then the inverse of π, which we represent as π and
then a superscript negative one, takes the input value π¦ and gives the output value
π₯. So we get back to where we
started. Weβve seen that in this example, if
weβre given a value of π₯, we can find the value of π¦, which is equal to π of π₯,
using the equation π¦ equals two π₯ minus three. And if weβre given the value of π¦,
or π of π₯, we can find the value of π₯ using the equation π₯ equals π¦ plus three
over two.

We want to write the inverse
function as a function of π₯ though, not a function of π¦. We therefore swab the position of
π₯ and π¦ around, and we define π inverse of π₯ to be equal to π¦. We have that the inverse function
is π inverse of π₯ is equal to π₯ plus three over two. We can also consider the link
between the graphs of a function and its inverse. By swapping the roles of π₯ and π¦
in the function around, this means that the graph is flipped so that the π¦-axis for
the original function is in the position of the π₯-axis for the inverse and vice
versa.

The geometric transformation taking
place here is a reflection in the line π¦ equals π₯. We can see this if we add the line
π¦ equals π inverse of π₯ to our sketch. Because we swap the π₯- and
π¦-axes, it also follows that the domain of the function π of π₯ becomes the range
of the function π inverse of π₯. And the range of the function π of
π₯ is the domain of its inverse function.

We can formalize the general
process of finding an inverse function as follows. First, we let π¦ equal the function
π of π₯. Then, we rearrange to make π₯ the
subject. We then swap π₯ and π¦ around so
that instead of having π₯ is equal to some function of π¦, we have π¦ is equal to
some function of π₯. Finally, we define this new
function of π₯ to be π inverse of π₯. Letβs now consider some examples in
which weβll find the inverse of a function algebraically, beginning with a linear
function.

Find π inverse of π₯ for π of π₯
equals a half π₯ plus three.

Weβve been given a function π of
π₯ and asked to find an expression for its inverse. We begin by letting π¦ equal π of
π₯. We then want to rearrange this
equation to make π₯ the subject. The first step is to subtract three
from each side, giving π¦ minus three equals a half π₯. We can then multiply both sides of
the equation by two, giving a two multiplied by π¦ minus three is equal to π₯. And distributing the parentheses on
the left-hand side, we have two π¦ minus six is equal to π₯. Now, this is π₯ as a function of
π¦, but we want to write the inverse function as a function of π₯. We therefore swap the positions of
π₯ and π¦ around. So where we had π¦, we swap this
for π₯, and where we had π₯, we change this for π¦. So we now have π¦ is equal to two
π₯ minus six.

The final step is to define this
new function of π₯ to be π inverse of π₯. So we have that π inverse of π₯ is
equal to two π₯ minus six. This is the function that maps all
of the old output values back to their original inputs. We can see this if we take an input
value, for example, four. When we apply the function π of π₯
to the value four, we get the output value five. And when we apply this inverse
function that weβve just found to the output value five, we have that π inverse of
five is equal to two multiplied by five minus six, thatβs 10 minus six, which gives
the original input value of four.

Itβs also worth pointing out that
the two steps of rearranging to make π₯ the subject and then swapping π₯ and π¦
around can be done in either order. If you prefer, you can swap π₯ and
π¦ at the beginning and then rearrange to make π¦ the subject. Weβll get the same answer, which is
that π inverse of π₯ is equal to two π₯ minus six.

Letβs now consider a slightly more
complicated example in which the function weβre finding the inverse of is a square
root function. Weβll also find the domain of this
inverse function.

Find π inverse of π₯ for π of π₯
equals the square root of π₯ plus three and state the domain.

To find the inverse function, weβll
begin by letting π¦ equal π of π₯. So we have π¦ is equal to the
square root of π₯ plus three. We then rearrange to make π₯ the
subject. We begin by subtracting three from
each side to give π¦ minus three equals the square root of π₯. We then square both sides of the
equation or raise each side to the power of two to give π¦ minus three squared is
equal to π₯. We could distribute the
parentheses, but thereβs no real need to. We now have π₯ as a function of π¦,
but we want the inverse function to be a function of π₯. So we swap π₯ and π¦ around to give
π₯ minus three squared is equal to π¦. Finally, we define the inverse
function π inverse of π₯ to be equal to the expression for π¦. So we have that π inverse of π₯ is
equal to π₯ minus three all squared.

Weβre also asked to find the domain
of the function π inverse. We recall that the set of all input
values to the function π inverse is the same as the set of all output values of the
function π. Or, in other words, the domain of
the function π inverse is the same as the range of the original function π. So letβs consider the function π
of π₯, and we want to determine its range or set of all output values. We know that the square root
function produces nonnegative values. The minimum value it can produce is
zero when π₯ itself is equal to zero. And thereβs no limit to the maximum
value it can produce. If we take the square root of a
really big positive number, we get another really big positive number.

We then add three to this
value. This means that the minimum value
of π of π₯ will be zero plus three, which is three. And as before, there is no maximum
value. The range of the function π of π₯
is therefore all values greater than or equal to three. This set of values is then the same
as the domain for the function π inverse. But as these are input values for
π inverse, they are values of π₯. We can state then that the domain
of π inverse of π₯ is π₯ is greater than or equal to three. So weβve completed the problem. π inverse of π₯ is equal to π₯
minus three squared, and the domain is π₯ greater than or equal to three.

Weβve now seen two examples of how
we can find the inverse of a function algebraically. But it isnβt always possible to
find the inverse of a function over its entire domain. Letβs consider the function π of
π₯ equals two π₯ squared plus three. The graph of π¦ equals π of π₯
looks like this. It is a positive parabola with a
π¦-intercept of three. If we have an input value π₯, we
can find an output value π¦ by going up to the graph and then across to the
π¦-axis. But what if we want to go the other
way?

Suppose we have the output value
five, and we want to find the input value that maps to this. If we go across from the π¦-value
of five to the graph, we can see that there are, in fact, two points that have a
π¦-coordinate of five. This means that thereβre two input
values, which in fact are positive and negative one, associated with this output
value. This means that the inverse
function isnβt well defined. If we have an output value of five,
how do we know whether this has come from an input value of one or negative one?

This leads to an important
condition for the inverse of a function to exist. The inverse of a function only
exists over its entire domain if the function is one-to-one. This means that every input to the
function produces a unique output so that when we go the other way using the inverse
function, every input to that function also produces a unique output. If a function isnβt one-to-one over
its entire domain, we can restrict the domain to a set of values over which it is
one-to-one and find the inverse only for that restricted domain. For example, for this quadratic
function, we could restrict the domain to be only π₯-values greater than or equal to
zero. Or, indeed, we could use π₯-values
less than or equal to zero. And then the function would be
one-to-one over this restricted domain, and so it would be possible to find its
inverse.

Letβs consider now an example of
identifying whether a function is invertible over its entire domain.

Which of the following functions
does not have an inverse over its whole domain? (a) π of π₯ equals two π₯. (b) π of π₯ equals two to the
power of π₯. (c) π of π₯ equals one over
π₯. Or (d) π of π₯ equals π₯
squared.

To answer this question, we need to
consider the conditions under which a function has an inverse. A function only has an inverse over
its entire domain if it is a one-to-one function. This means that every input to the
function produces a unique output so that when we go the other way using the inverse
function, every input to this function also produces a unique output. We can work out which of these
functions isnβt one-to-one by sketching graphs of each function. For option (a), the graph of π¦
equals π of π₯ or π¦ equals two π₯ is a straight line graph through the origin. This is clearly a one-to-one
function over its entire domain. Every input has a unique
output. And so going the other way, the
same will be true.

More generally though, we can
determine whether a function is one-to-one by drawing horizontal and vertical lines
on its graph. If each horizontal and each
vertical line intersects the graph a maximum of one time, then the function is
one-to-one. The graph of π¦ equals two to the
power of π₯ is an exponential graph passing through the point zero, one with a
horizontal asymptote along the π₯-axis. This is also a one-to-one function
over its entire domain because every vertical line intersects the graph once and
every horizontal line intersects the graph a maximum of one time.

For the graph of π¦ equals one over
π₯, thereβre two branches, one in the first quadrant, where π₯ and π¦ are both
positive, and one in the third quadrant, where π₯ and π¦ are both negative. We can see, though, that the
function is one-to-one over its entire domain. That just leaves option (d)
then. The graph of π¦ equals π₯ squared
is a positive parabola with its minimum point at the origin. We can see that if we draw
horizontal lines across the graph, some of these lines will intersect the graph in
more than one place. This means that for certain output
values, there are two possible input values, and so the function π of π₯ isnβt
one-to-one. This means we canβt find the
inverse of π of π₯ over its entire domain because we donβt know which of those
input values the output value should be mapped to by the inverse function. Our answer is that the function
which does not have an inverse over its whole domain is π of π₯ equals π₯
squared.

Letβs now summarize the key points
from this video. Informally, we can think of the
inverse of a function as the function which undoes the original function. More formally, for a value π₯ in
the domain of the function π, if π of π₯ is equal to π¦, then π inverse of π¦ is
equal to π₯. Values in the domain of π of π₯
form the range of π inverse of π₯ and vice versa. Values in the range of π of π₯
form the domain of π inverse of π₯. We saw that the inverse of a
function only exists if the function is one-to-one so that each input value is
mapped to a unique output value. And then going the other way, the
inverse function maps each output value back to a unique input value.

We also saw that to find the
inverse of a function π¦ equals π of π₯, we need to rearrange the formula to make
π₯ the subject. And then we swap π₯ and π¦
around. These two steps can be performed in
either order. We can also swap π₯ and π¦ first
and then rearrange to make the new π¦ the subject. Finally, we also saw that due to
this swapping of π₯ and π¦, the graphs of a function and its inverse are reflections
of one another in the line π¦ equals π₯.