# Question Video: Identifying and Using Aspects of a One-to-One Function Mathematics

Which of the following is not a one-to-one function on the intervel [0, ∞)? [A] 𝑓(𝑥) = |𝑥| [B] 𝑓(𝑥) = 𝑥² [C] 𝑓(𝑥) = 10 [D] 𝑓(𝑥) = 2𝑥 + 4 [E] 𝑓(𝑥) = 1/(𝑥 + 1)

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### Video Transcript

Which of the following is not a one-to-one function on the left-closed, right-open interval from zero to ∞?

We are given five choices to consider. Option (A) is 𝑓 of 𝑥 equal to the absolute value of 𝑥. Option (B) is 𝑓 of 𝑥 equal to 𝑥 squared. Option (C) is 𝑓 of 𝑥 equal to 10. Option (D) is 𝑓 of 𝑥 equal to two 𝑥 plus four. And option (E) is 𝑓 of 𝑥 equal to one over 𝑥 plus one. We should note that four of these functions are one to one on the interval from zero to ∞. And only one choice is not a one-to-one function on this interval. That’s the function we are looking for.

We recall that the meaning of this interval is all positive real numbers and including zero because of the square bracket we have around zero. We may sometimes express this interval using other notations. Instead of the round bracket around ∞, we may see a square bracket that is opening outwards or we may see this interval expressed as a compound inequality. We recall that a one-to-one function, also called an injective function, is defined as a function where each element of the range corresponds to exactly one element of the domain. In other words, exactly one input value is paired with each output value.

In this example, we are using the variable 𝑥 to represent our input values, which make up the domain of each function. And we are using 𝑓 of 𝑥 to represent the output values, which make up the range of our functions. If we were given a graph for each of these five functions or if we wanted to take the time to sketch the graphs ourselves, we could use the horizontal line test. We recall that a function is one to one if each horizontal line intersects the graph of a function at most once.

Therefore, we can use the horizontal line test to show that a function is not one to one if we see any horizontal line that crosses the graph at more than one point. For example, if we looked at the graph of 𝑓 of 𝑥 equal to the absolute value of 𝑥, which is a V-shaped graph with the vertex on zero, zero, we would see that this is not a one-to-one function. Because any horizontal line above the 𝑥-axis would intersect the graph at two points. However, we are only asked to consider this function on the left-closed, right-open interval from zero to ∞, which leaves us with only the blue highlighted part of our graph. This restricted function passes the horizontal line test. And we conclude that the absolute value function is one to one on the interval from zero to ∞.

Let’s say that we were not interested in graphing these five functions. We can instead use an algebraic approach to check if these functions are one to one. First, we recall that if two different input values, let’s call them 𝑥 sub one and 𝑥 sub two, satisfy the equation 𝑓 of 𝑥 sub one equals 𝑓 of 𝑥 sub two, then 𝑓 is not a one-to-one function. And we can say that 𝑓 is a one-to-one function if 𝑓 of 𝑥 sub one equals 𝑓 of 𝑥 sub two implies that 𝑥 sub one equals 𝑥 sub two. In other words, if we evaluated a function two times and got the same output value, this means that we must have used the same input value each time. We will use this criteria to check each of the five functions that we were given to see if they’re one to one.

We will now clear some space to verify algebraically that option (A) is a one-to-one function. We begin with two arbitrary elements of the domain, 𝑥 sub one and 𝑥 sub two. Each are greater than or equal to zero. And 𝑓 of 𝑥 sub one equals 𝑓 of 𝑥 sub two. If we can clearly demonstrate that 𝑥 sub one equals 𝑥 sub two in this case, then our function is one to one. However, if we demonstrate that 𝑥 sub one does not equal 𝑥 sub two, under these same conditions, the function being considered is not one to one.

Since we are considering the function 𝑓 of 𝑥 equals absolute value of 𝑥, we can express the left side of our equation as absolute value of 𝑥 sub one and the right side of our equation as absolute value of 𝑥 sub two. Considering 𝑥 sub one and 𝑥 sub two are nonnegative values, the absolute value of 𝑥 sub one is 𝑥 sub one and the absolute value of 𝑥 sub two equals 𝑥 sub two. Therefore, 𝑥 sub one equals 𝑥 sub two. This verifies that our first option is a one-to-one function.

Now, we move on to option (B), where 𝑓 of 𝑥 equals 𝑥 squared instead of absolute value of 𝑥. This means that the left side of our equation is 𝑥 sub one squared and the right side of our equation is 𝑥 sub two squared. In order to examine how 𝑥 sub one relates to 𝑥 sub two, we will rearrange this equation so that it is factorable. By subtracting 𝑥 sub two squared from each side, we now have 𝑥 sub one squared minus 𝑥 sub two squared. And we recognize this as the difference of squares.

We recall that 𝑎 squared minus 𝑏 squared has the following factors: 𝑎 minus 𝑏 times 𝑎 plus 𝑏. We will use this factoring pattern to factor the left side of our equation. Therefore, 𝑥 sub one minus 𝑥 sub two times 𝑥 sub one plus 𝑥 sub two equals zero. For this equation to hold, we must have either 𝑥 sub one minus 𝑥 sub two equal zero or 𝑥 sub one plus 𝑥 sub two equal zero. By adding 𝑥 sub two to both sides of the first equation, we have 𝑥 sub one equals 𝑥 sub two. Since both 𝑥 sub one and 𝑥 sub two are nonnegative, the second equation is only possible if both 𝑥 sub one and 𝑥 sub two are equal to zero. So, in either case, we must have 𝑥 sub one equals 𝑥 sub two. Hence, 𝑓 of 𝑥 equaling 𝑥 squared is a one-to-one function on the interval from zero to ∞.

The next option is a constant function: 𝑓 of 𝑥 equals 10. A constant function maps all different inputs to the same output. Thus, 𝑓 of 𝑥 sub one will always equal 𝑓 of 𝑥 sub two. But we want to know if this equality implies that 𝑥 sub one equals 𝑥 sub two. As an example, let’s consider the situation where 𝑥 sub one equals zero and 𝑥 sub two equals one. Then, we see that 𝑓 of zero equals 10 and 𝑓 of one equals 10. This example illustrates two different inputs corresponding to the same output. It’s true that 𝑓 of 𝑥 sub one always equals 𝑓 of 𝑥 sub two. But 𝑥 sub one did not equal 𝑥 sub two. Therefore, this constant function is not one to one.

It looks like we have found our answer. But just to be sure, let’s quickly check the last two options. Given that 𝑓 of 𝑥 sub one equals 𝑓 of 𝑥 sub two and the function 𝑓 of 𝑥 equals two 𝑥 plus four leads us to the equation two 𝑥 sub one plus four equals two 𝑥 sub two plus four, then by subtracting four from each side of the equation and dividing by two, we have 𝑥 sub one equals 𝑥 sub two. Therefore, option (D) is a one-to-one function.

To check the last option, we again suppose that 𝑓 of 𝑥 sub one equals 𝑓 of 𝑥 sub two for the function 𝑓 of 𝑥 equals one over 𝑥 plus one. It follows that one over 𝑥 sub one plus one equals one over 𝑥 sub two plus one. Since 𝑥 sub one and 𝑥 sub two are nonnegative, we can be sure that both denominators are positive. This allows us to cross multiply. And finally, by subtracting one from each side of the equation, we have 𝑥 sub two equals 𝑥 sub one. Therefore, this last option is a one-to-one function.

Of the five options given, we have demonstrated that they are all one-to-one functions except 𝑓 of 𝑥 equals 10, across the left-closed, right-open interval from zero to ∞. We found that two different input values corresponded to the same output value, disqualifying this function from being one to one.