### Video Transcript

Is the function π of π₯ equal to
one over π₯ minus one a one-to-one function, where π₯ is an element of the set of
real numbers minus the set containing one?

In this question, we are asked to
determine whether the function π of π₯ is a one-to-one function. First, we recall that for π of π₯
to be a function, each element in the domain corresponds to exactly one element of
the range. If the function is one to one or
injective, then we also have that each element of the range corresponds to exactly
one element of the domain.

In other words, for two elements of
the domain π₯ one and π₯ two, if π of π₯ one is equal to π of π₯ two, then π₯ one
must equal π₯ two. If π₯ one does not equal π₯ two,
then the function cannot be one to one because two different elements of the domain
would correspond to the same element from the range. This means that a function such as
π of π₯ equal to π₯ squared is not a one-to-one function over the real numbers,
because four, an element of the range, corresponds to two elements from the domain,
negative two and positive two.

So, for the given function, π of
π₯ equal to one over π₯ minus one, we must determine when π of π₯ one equals π of
π₯ two whether π₯ one must equal π₯ two. So letβs set up the initial
equation, π evaluated at π₯ one is equal to π evaluated at π₯ two, by using the
definition of the given function. This gives us one over π₯ one minus
one equal to one over π₯ two minus one. And remember, we choose π₯ one and
π₯ two to be elements of the domain of π, which 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 must conclude that
π is not an injective function. However, if we are able to show
they must be equal, then we can say π is an injective function. We notice that the numerators are
equal. So, to simplify this equation, we
can set the denominators equal as well. So, we have π₯ one minus one is
equal to π₯ two minus one.

Then, we can simplify our equation
further by adding one to both sides. This gives us π₯ one is equal to π₯
two. Therefore, weβve shown that for any
two values in the domain, if π evaluated at π₯ one is equal to π evaluated at π₯
two, we must have that π₯ one and π₯ two are equal, meaning every element of the
range corresponds to a single unique element of the domain.

Thus, we conclude that the answer
is yes. π is a one-to-one or injective
function, where π₯ is an element of the real numbers excluding one.