Question Video: Determining the Injectivity of a Function | Nagwa Question Video: Determining the Injectivity of a Function | Nagwa

Question Video: Determining the Injectivity of a Function Mathematics • Second Year of Secondary School

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Is the function 𝑓(π‘₯) = 1/(π‘₯ βˆ’ 1) a one-to-one function, where π‘₯ ∈ ℝ βˆ’ {1}?

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Video Transcript

Is the function 𝑓 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 𝑓 of π‘₯ is a one-to-one function. First, we recall that for 𝑓 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 𝑓 of π‘₯ one is equal to 𝑓 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 𝑓 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, 𝑓 of π‘₯ equal to one over π‘₯ minus one, we must determine when 𝑓 of π‘₯ one equals 𝑓 of π‘₯ two whether π‘₯ one must equal π‘₯ two. So let’s set up the initial equation, 𝑓 evaluated at π‘₯ one is equal to 𝑓 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 𝑓, 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 𝑓 is not an injective function. However, if we are able to show they must be equal, then we can say 𝑓 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, we’ve shown that for any two values in the domain, if 𝑓 evaluated at π‘₯ one is equal to 𝑓 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. 𝑓 is a one-to-one or injective function, where π‘₯ is an element of the real numbers excluding one.

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