Video Transcript
Given that 𝑓 of 𝑥 equals 𝑥
squared plus nine 𝑥 plus 14 over 𝑥 squared minus four divided by 𝑥 squared minus
49 over 𝑥 squared minus two 𝑥 and 𝑓 of 𝑎 equals four, find the value of 𝑎.
Let’s begin by inspecting our
function 𝑓 of 𝑥. It’s the quotient, that means
divide, of a pair of rational functions. Now, we remember that a rational
function itself is the quotient of a pair of polynomials. And when we’re finding the product
or quotient of two or more rational functions, we can look to see if we can simplify
by factoring one or more expressions in each of our rational functions.
Let’s begin by rewriting 𝑓 of 𝑥
by thinking about what we know about dividing by a fraction. To divide by a fraction, we can
multiply by the reciprocal of that fraction. In other words, we’re multiplying
our first rational function by 𝑥 squared minus two 𝑥 over 𝑥 squared minus 49. Once we’ve done this, we’ll be able
to simplify by cross canceling. Let’s begin by looking at the
numerator of our first fraction.
The coefficient of 𝑥 squared is
one, and there are no common factors between 𝑥 squared, nine 𝑥, and 14. So we have a pair of parentheses in
which 𝑥 is the first term within. We need to find a pair of numbers
such that they multiply, their product is 14, and they sum to make nine. Those two numbers are two and
seven. So, the expression 𝑥 squared plus
nine 𝑥 plus 14 is equal to 𝑥 plus two times 𝑥 plus seven.
Let’s now consider the denominator
of the fraction. If we look carefully, we see that
both 𝑥 squared and four are square numbers. And this is the difference between
them, so it’s the difference of two squares. We know that when we’re working
with an expression that’s the difference of two squares, such as 𝑥 squared minus 𝑎
squared, it factors to be 𝑥 plus 𝑎 times 𝑥 minus 𝑎. Since the square root of four is
two, this means 𝑥 squared minus four is 𝑥 plus two times 𝑥 minus two.
Let’s keep going. But this time, we’re going to look
at the numerator of our second rational function, 𝑥 squared minus two 𝑥. This time, we have a quadratic, but
there’s a common factor of 𝑥 between both our terms. This means 𝑥 sits outside the
parentheses, and we get 𝑥 times 𝑥 minus two. We’re going to do this one more
time for the expression 𝑥 squared minus 49, which once again is the difference of
two squares. And so, it factors to 𝑥 plus seven
times 𝑥 minus seven.
Now, let’s rewrite 𝑓 of 𝑥 using
all of our factored expressions. 𝑓 of 𝑥 is 𝑥 plus two times 𝑥
plus seven over 𝑥 plus two times 𝑥 minus two multiplied by 𝑥 times 𝑥 minus two
over 𝑥 plus seven times 𝑥 minus seven.
We’re now going to simplify by
dividing through by any common factors. We note that 𝑓 of 𝑎 is equal to
four; it’s defined. Since 𝑎 is defined, we know it’s
within the domain of our function. And so, we can continue by
simplifying by dividing.
First, we divide through by 𝑥 plus
two. Next, we can cross cancel a factor
of 𝑥 minus two and a factor of 𝑥 plus seven. This actually means that our first
fraction is simply one. And when we simplify 𝑓 of 𝑥, we
get 𝑥 over 𝑥 minus seven. So, to find the value of 𝑎, we use
the fact that 𝑓 of 𝑎 equals four. We’re going to substitute 𝑥 equals
𝑎 and 𝑓 of 𝑥 equals four into our new equation for 𝑓 of 𝑥. And so, we get four equals 𝑎 over
𝑎 minus seven.
To solve for 𝑎, let’s begin by
multiplying through by 𝑎 minus seven to get four times 𝑎 minus seven equals
𝑎. Next, let’s distribute our
parentheses by multiplying 𝑎 minus seven by four. That gives us four 𝑎 minus 28, and
that’s equal to 𝑎. Then, we subtract 𝑎 from both
sides. Three 𝑎 minus 28 equals zero. And then, we can add 28 to get
three 𝑎 equals 28. Finally, let’s divide through by
three. So, 𝑎 is 28 over three.
And now, we notice that by
simplifying before substituting, we ended up with a fairly straightforward equation
in 𝑎 to solve. If we had tried to substitute 𝑥
equals 𝑎 and 𝑓 of 𝑥 equals four into the original equation for 𝑓 of 𝑥, we would
have ended up with a very nasty equation in 𝑎. So, the value of 𝑎 is 28 over
three.
