### Video Transcript

Which of the following is a
one-to-one function? Is it (A) π of π₯ equals the
absolute value of π₯, (B) π of π₯ equals π₯ squared, (C) π of π₯ equals five, or
(D) π of π₯ equals π₯ plus two?

We begin by recalling that π of π₯
is not a one-to-one function if there are two different π₯-values from the domain,
π₯ sub one and π₯ sub two, satisfying π of π₯ sub one is equal to π of π₯ sub
two. Letβs consider each possible option
with respect to this condition. Option (A) is the absolute value
function. And we know that two numbers of the
same size but with opposite signs have the same absolute value. For example, if we let π₯ sub one
equal negative two and π₯ sub two equal two, then π of π₯ sub one equals the
absolute value of negative two which is equal to two and π of π₯ sub two, the
absolute value of two, is also equal to two. Since π of negative two is equal
to π of two, then the function the absolute value of π₯ is not one-to-one.

We can repeat this process for
option (B), where π of π₯ is the square function. As with the absolute value
function, two numbers of the same size but with opposite signs have the same squared
value. If we once again let π₯ sub one be
negative two and π₯ sub two be two, then π of π₯ sub one and π of π₯ sub two are
both equal to four. The square of negative two and the
square of two are both equal to four. Once again, as π of π₯ sub one is
equal to π of π₯ sub two, the function is not one-to-one.

Letβs now consider the third
option, π of π₯ equals five. This is a constant function. Whichever number we input to a
constant function, the output will always be the same. In this case, it doesnβt matter
what values we choose for π₯ sub one and π₯ sub two. π of π₯ sub one and π of π₯ sub
two will always be equal to five. This means that π of π₯ equals
five is not a one-to-one function.

Letβs now consider option (D) π of
π₯ is equal to π₯ plus two. In order to prove that this
function is one-to-one, we recall that a function is one-to-one if the conditions
that π of π₯ sub one is equal to π of π₯ sub two and that π₯ sub one and π₯ sub
two belong to the domain of π imply that π₯ sub one is equal to π₯ sub two. If π of π₯ sub one is equal to π
of π₯ sub two for the function π of π₯ equals π₯ plus two, then π₯ sub one plus two
must be equal to π₯ sub two plus two. Subtracting two from both sides of
this equation, we see that π₯ sub one is equal to π₯ sub two. This tells us that π of π₯ sub one
equals π of π₯ sub two is only possible when π₯ sub one is equal to π₯ sub two.

We can therefore conclude that π
of π₯ equals π₯ plus two is one-to-one. The only one of the functions from
the list that is one-to-one or injective is option (D).