Question Video: Identifying One-to-One Functions Mathematics

Which of the following is a one-to-one function? [A] 𝑓(π‘₯) = |π‘₯| [B] 𝑓(π‘₯) = π‘₯Β² [C] 𝑓(π‘₯) = 5 [D] 𝑓(π‘₯) = π‘₯ + 2

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Video Transcript

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

We begin by recalling that 𝑓 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 𝑓 of π‘₯ sub one is equal to 𝑓 of π‘₯ sub two. Let’s 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 𝑓 of π‘₯ sub one equals the absolute value of negative two which is equal to two and 𝑓 of π‘₯ sub two, the absolute value of two, is also equal to two. Since 𝑓 of negative two is equal to 𝑓 of two, then the function the absolute value of π‘₯ is not one-to-one.

We can repeat this process for option (B), where 𝑓 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 𝑓 of π‘₯ sub one and 𝑓 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 𝑓 of π‘₯ sub one is equal to 𝑓 of π‘₯ sub two, the function is not one-to-one.

Let’s now consider the third option, 𝑓 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 doesn’t matter what values we choose for π‘₯ sub one and π‘₯ sub two. 𝑓 of π‘₯ sub one and 𝑓 of π‘₯ sub two will always be equal to five. This means that 𝑓 of π‘₯ equals five is not a one-to-one function.

Let’s now consider option (D) 𝑓 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 𝑓 of π‘₯ sub one is equal to 𝑓 of π‘₯ sub two and that π‘₯ sub one and π‘₯ sub two belong to the domain of 𝑓 imply that π‘₯ sub one is equal to π‘₯ sub two. If 𝑓 of π‘₯ sub one is equal to 𝑓 of π‘₯ sub two for the function 𝑓 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 𝑓 of π‘₯ sub one equals 𝑓 of π‘₯ sub two is only possible when π‘₯ sub one is equal to π‘₯ sub two.

We can therefore conclude that 𝑓 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).

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