Video Transcript
Which of the following functions
are equal?
We are given five pairs of
functions to consider. First, we notice that all five
options include the function 𝑛 sub one of 𝑥 equals 𝑥 cubed minus 729 over 𝑥
cubed plus nine 𝑥 squared plus 81𝑥, while each 𝑛 sub two function varies in at
least some small way from the other 𝑛 sub two functions. We note that all 𝑛 sub one and 𝑛
sub two functions given are rational functions, where both the numerator and the
denominator contain polynomials. In the case of our 𝑛 sub one
function, the numerator is a cubic binomial, that is, degree three with two
terms. And the denominator is a cubic
trinomial, which means degree three with three terms.
We will now recall what it means
for two functions to be equal. Rational functions 𝑛 sub one and
𝑛 sub two are equal if, number one, they have equal domains. We recall that the domain of a
rational function equals all real numbers minus the set containing the zeros of the
denominator. This means that for rational
functions 𝑛 sub one and 𝑛 sub two to fulfill the first requirement, their
denominators must contain the same real number zeros.
The second requirement for the
equality of rational functions 𝑛 sub one and 𝑛 sub two is that 𝑛 sub one equals
𝑛 sub two on their shared domain. This requirement is what we refer
to as the equivalence of 𝑛 sub one and 𝑛 sub two. Equivalent rational functions
simplify to give the same expression. In this context, equivalence and
equality do not mean the same thing, because equal rational functions require equal
domains, whereas equivalent rational functions are said to be equivalent on their
shared domain. In this question, we are asked to
identify which of the following functions are equal, which means that 𝑛 sub one and
𝑛 sub two have denominators that contain the same real number zeros and they
simplify to give the same expression. We will now clear some space
leaving a brief summary of these two requirements.
We recall that to find the real
number zeros of each denominator, we will need to factor or factorize those
denominators. And to simplify rational
expressions, we will need to factor or factorize their numerators and
denominators. This then allows us to cancel out
any shared factors as shown in the example below. Therefore, we will begin our work
by factorizing the numerator and the denominator of our 𝑛 sub one function. Once that is done, we will be able
to find the real number zeros of the denominator and simplify the expression. These findings will then be
compared to the properties of our various 𝑛 sub two functions. Let’s clear some space to show our
work.
In order to focus on just the 𝑛
sub one function for now, we will move all the 𝑛 sub two functions off to the
side. We will begin by factoring the
numerator. We pause to note that 729 is a
perfect cube number. We know this is true because it is
the product of the same integer three times: nine times nine times nine, that is,
nine cubed. Therefore, we can write 𝑥 cubed
minus 729 as 𝑥 cubed minus nine cubed. We call this the difference of
cubes. We call any polynomial of the form
𝑥 cubed minus 𝑎 cubed a difference of cubes. And its factorization is as
follows: 𝑥 minus 𝑎 times 𝑥 squared plus 𝑎𝑥 plus 𝑎 squared. In this case, we will substitute
the integer nine for the variable 𝑎. It follows that the first factor of
𝑥 cubed minus 729 is 𝑥 minus nine.
And the second factor is 𝑥 squared
plus nine 𝑥 plus 81, where the 81 comes from squaring our 𝑎-value, which was
nine. At this point, we might be
wondering if the second factor is prime. After exhausting all the factors of
81 and seeing that none of them add to the coefficient of the 𝑥-term, we are
confident in saying that this is a prime factor.
Now we will move on to find the
full factorization of the denominator. The first thing we notice is that
all three terms in the denominator contain a common factor of 𝑥. In fact, 𝑥 is the highest common
factor. Once we factored 𝑥 out of each
term, we are left with the factorization 𝑥 times 𝑥 squared plus nine 𝑥 plus
81.
Now that the denominator is
factored, we are ready to find the real number zeros of this denominator. We search with the zeros by setting
each factor equal to zero. The first equation is already
solved: 𝑥 equals zero. But the second equation is a little
bit more complicated. As determined earlier, this
quadratic polynomial is prime, which means that it cannot be factorized.
Instead of trying to immediately
solve this quadratic, let’s stop to find the discriminant. We recall that a quadratic with a
positive discriminant has two real zeros, whereas a discriminant of zero means it
has one real zero. And a negative discriminant means
that it has no real zeros. Using the 𝑎-value of one, the
𝑏-value of nine, and the 𝑐-value of 81 gives us a discriminant of negative
243. This means that the quadratic 𝑥
squared plus nine 𝑥 plus 81 gives us no real number zeros. Therefore, having found that the
denominator has only one real number zero, the domain of 𝑛 sub one is all real
numbers minus the set containing zero.
Now we will go back to look at the
factorization of the numerator and the denominator of 𝑛 sub one to see if its
expression can be simplified. This is where we notice the shared
factor of 𝑥 squared plus nine 𝑥 plus 81. After canceling the shared factor,
we have the simplified expression 𝑥 minus nine over 𝑥. We will now clear some space and
leave a brief summary of what we have learned about the function 𝑛 sub one.
Let’s consider the first 𝑛 sub two
function from option (A). The factorization of this
denominator is 𝑥 times 𝑥 squared minus 63. The 𝑛 sub two function that is
equal to the 𝑛 sub one function will also have a domain of all real numbers minus
the set containing zero. So let’s find the zeros of the
denominator. After setting each factor equal to
zero, we have found three zeros: 𝑥 equals zero, 𝑥 equals positive three times the
square root of seven, and negative three times the square root of seven. This leads us to conclude that this
𝑛 sub two function has a domain of all real numbers minus the set containing zero,
three square root of seven, and negative three square root of seven, which is not
equal to the domain of 𝑛 sub one. Therefore, we eliminate option
(A).
Next, we will consider the 𝑛 sub
two function from option (B). As before, we will first factorize
the denominator of our new 𝑛 sub two function. The factors of the denominator are
𝑥 times 𝑥 squared plus 63. And we will set each of these
factors equal to zero. The first zero of this denominator
is the same as the zero of the denominator of 𝑛 sub one. We also note that 𝑥 squared plus
63 gives us no other real number zeros because the square root of a negative is not
a real number. Therefore, this 𝑛 sub two function
has the same domain as the 𝑛 sub one function. Therefore, the first requirement of
equal rational functions is fulfilled.
Now we need to see if this 𝑛 sub
two function simplifies to give the same expression as 𝑛 sub one. This is easily confirmed by
canceling out the shared factor of 𝑥 squared plus 63. And therefore, the function
simplifies to 𝑥 minus nine over 𝑥, making 𝑛 sub two equivalent to 𝑛 sub one. And since we already showed that
their domains are equal, we have found the correct answer to be choice (B).
We can now take a quick look at the
remaining three options to determine why these 𝑛 sub two functions are not equal to
the 𝑛 sub one function. In option (C), we determine that 𝑛
sub two does not equal 𝑛 sub one because 𝑛 sub two cannot simplify to the
expression 𝑥 minus nine over 𝑥. In option (D), we determine that 𝑛
sub two does not equal and sub one because it does not have the same real number
zeros in the denominator as the 𝑛 sub one function. And finally in option (E), we
determine that 𝑛 sub two does not equal 𝑛 sub one because, once again, it does not
have the same real number zeros in the denominator as the 𝑛 sub one function.
In summary, we eliminated option
(A) because the domain of 𝑛 sub two did not equal the domain of 𝑛 sub one. We eliminated option (C) because
the 𝑛 sub two function could not simplify to the expression 𝑥 minus nine over
𝑥. We eliminated option (D) for the
same reason that we eliminated option (A). And similarly, we eliminated option
(E) because the domain of 𝑛 sub two did not equal the domain of 𝑛 sub one.
In conclusion, out of the five
options given, the only pair of functions that were equal were 𝑛 sub one of 𝑥
equals 𝑥 cubed minus 729 over 𝑥 cubed plus nine 𝑥 squared plus 81𝑥 and 𝑛 sub
two of 𝑥 equals 𝑥 minus nine times 𝑥 squared plus 63 all over 𝑥 cubed plus
63𝑥.
