### Video Transcript

Is the function π of π₯ is equal
to one divided by π₯ minus one a one-to-one function, where π₯ is an element of the
reals minus the set including one?

In this question, weβre asked to
determine whether a given function π of π₯ is a one-to-one function. And to answer this question, weβll
start by recalling what is meant by a one-to-one or injective function. Itβs a function where each element
in the domain of the function corresponds to exactly one element of the range of the
function. And weβre given the domain of this
function in the question: itβs the set of all real values excluding one. And thereβs many different ways of
answering this question. Weβll go through two of these.

First, letβs determine if this is
an injective function just by looking at its definition. To do this, we note that in
one-to-one functions, every element in the domain corresponds to a unique element in
the range. Another way of saying this is if π₯
one and π₯ two are elements of the domain of an injective function π, then the only
way for π evaluated at π₯ one to be equal to π evaluated at π₯ two is for π₯ one
and π₯ two to be equal because π is injective. If we can show this is true for all
values of π₯ one and π₯ two in the domain of π, we can prove itβs injective. So letβs try and solve π evaluated
at π₯ one is equal to π evaluated at π₯ two by using the function π of π₯ in the
question.

Using the definition of the
function π, this gives us one divided by π₯ one minus one is equal to one divided
by π₯ two minus one. And remember, we choose π₯ sub one
and π₯ sub two to be elements of the domain of π. This means both of them are real
numbers and neither of them are equal to one. If we can find distinct values of
π₯ one and π₯ two which balance both sides of the equation, we can conclude π is
not an injective function. However, if we can prove they must
be equal, then we can conclude π is an injective function. To do this, letβs simplify the
equation. Weβll take the reciprocal of both
sides. This gives us π₯ sub one minus one
is equal to π₯ sub two minus one. We can simplify further by adding
one to both sides of the equation. This gives us that π₯ sub one is
equal to π₯ sub two.

Therefore, weβve shown for any two
values π₯ one and π₯ two in the domain of our function π, if π evaluated at π₯ one
is equal to π evaluated at π₯ two, we must have π₯ one and π₯ two are equal. Every element in the domain of our
function π corresponds to exactly one element of the range. π is an injective function. Therefore, we can conclude that the
answer is yes.

However, itβs not always possible
to directly prove that a function is one to one by using this method. So weβll show a second method by
using the sketch of the graph π¦ is equal to π of π₯. Since π of π₯ is one divided by π₯
minus one, we can see weβre subtracting one from the input values of the reciprocal
function. Therefore, the graph of π of π₯ is
the graph of the reciprocal function translated one unit to the right, so the shape
of this function is the same as the reciprocal curve. However, our vertical asymptote is
at π₯ is equal to one.

Weβre now ready to determine
whether this function is a one-to-one function by using the horizontal line
test. We recall this tells us if every
horizontal line of the form π¦ is equal to π for some real value π intersects the
curve at most once, then the function is a one-to-one function. Alternatively, if we can find one
horizontal line which intersects our curve multiple times, we can conclude it is not
a one-to-one function.

However, we can see from the
diagram every single horizontal line will intersect the curve at most once. For example, every single
horizontal line above or below the π₯-axis intersects once. And the π₯-axis itself, the line π¦
is equal to zero, is a horizontal asymptote. The curve does not intersect the
π₯-axis, and therefore this is a one-to-one function. Hence, to answer the question βIs
the function π of π₯ is equal to one divided by π₯ minus one, where π₯ is an
element of the real numbers excluding one, a one-to-one function?,β the answer is
yes.