### Video Transcript

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

We are given five choices to
consider. Option (A) is π of π₯ equal to the
absolute value of π₯. Option (B) is π of π₯ equal to π₯
squared. Option (C) is π of π₯ equal to
10. Option (D) is π of π₯ equal to two
π₯ plus four. And option (E) is π of π₯ equal to
one over π₯ plus one. We should note that four of these
functions are one to one on the interval from zero to β. And only one choice is not a
one-to-one function on this interval. Thatβs 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
β, 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 π₯ to represent our input values, which make up the domain of each
function. And we are using π of π₯ 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 π of π₯ equal to the absolute value of π₯, 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 π₯-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 β, 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 β.

Letβs 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, letβs call them π₯ sub one and π₯ sub two, satisfy the
equation π of π₯ sub one equals π of π₯ sub two, then π is not a one-to-one
function. And we can say that π is a
one-to-one function if π of π₯ sub one equals π of π₯ sub two implies that π₯ sub
one equals π₯ 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 theyβre 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, π₯ sub one and π₯ sub two. Each are greater than or equal to
zero. And π of π₯ sub one equals π of
π₯ sub two. If we can clearly demonstrate that
π₯ sub one equals π₯ sub two in this case, then our function is one to one. However, if we demonstrate that π₯
sub one does not equal π₯ sub two, under these same conditions, the function being
considered is not one to one.

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

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

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

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

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

To check the last option, we again
suppose that π of π₯ sub one equals π of π₯ sub two for the function π of π₯
equals one over π₯ plus one. It follows that one over π₯ sub one
plus one equals one over π₯ sub two plus one. Since π₯ sub one and π₯ 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 π₯ sub two equals π₯ 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 π of π₯ equals 10,
across the left-closed, right-open interval from zero to β. We found that two different input
values corresponded to the same output value, disqualifying this function from being
one to one.