Question Video: Determining the Injectivity of a Function Mathematics

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‘₯ 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, we’re asked to determine whether a given function 𝑓 of π‘₯ is a one-to-one function. And to answer this question, we’ll start by recalling what is meant by a one-to-one or injective function. It’s a function where each element in the domain of the function corresponds to exactly one element of the range of the function. And we’re given the domain of this function in the question: it’s the set of all real values excluding one. And there’s many different ways of answering this question. We’ll go through two of these.

First, let’s 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 𝑓, then the only way for 𝑓 evaluated at π‘₯ one to be equal to 𝑓 evaluated at π‘₯ two is for π‘₯ one and π‘₯ two to be equal because 𝑓 is injective. If we can show this is true for all values of π‘₯ one and π‘₯ two in the domain of 𝑓, we can prove it’s injective. So let’s try and solve 𝑓 evaluated at π‘₯ one is equal to 𝑓 evaluated at π‘₯ two by using the function 𝑓 of π‘₯ in the question.

Using the definition of the function 𝑓, 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 𝑓. 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 𝑓 is not an injective function. However, if we can prove they must be equal, then we can conclude 𝑓 is an injective function. To do this, let’s simplify the equation. We’ll 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, we’ve shown for any two values π‘₯ one and π‘₯ two in the domain of our function 𝑓, if 𝑓 evaluated at π‘₯ one is equal to 𝑓 evaluated at π‘₯ two, we must have π‘₯ one and π‘₯ two are equal. Every element in the domain of our function 𝑓 corresponds to exactly one element of the range. 𝑓 is an injective function. Therefore, we can conclude that the answer is yes.

However, it’s not always possible to directly prove that a function is one to one by using this method. So we’ll show a second method by using the sketch of the graph 𝑦 is equal to 𝑓 of π‘₯. Since 𝑓 of π‘₯ is one divided by π‘₯ minus one, we can see we’re subtracting one from the input values of the reciprocal function. Therefore, the graph of 𝑓 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.

We’re 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 𝑦 is equal to 𝑐 for some real value 𝑐 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 𝑦 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 β€œIs the function 𝑓 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?,” the answer is yes.

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