Video Transcript
The function 𝑓 of 𝑥 is equal to
𝑥 squared minus nine 𝑘 if 𝑥 is less than negative five and 𝑓 of 𝑥 is equal to
two 𝑥 plus 116 if 𝑥 is greater than negative five has a limit as 𝑥 approaches
negative five. What is the value of 𝑘?
The question gives us a
piecewise-defined function 𝑓 of 𝑥. And we’re told that this function
𝑓 of 𝑥 has a limit as 𝑥 approaches negative five. We need to find the value of 𝑘
that makes this possible. We recall that we say that the
limit as 𝑥 approaches 𝑎 of a function 𝑓 of 𝑥 exists if the following conditions
hold. We need the limit as 𝑥 approaches
𝑎 from the left of 𝑓 of 𝑥 and the limit as 𝑥 approaches 𝑎 from the right of 𝑓
of 𝑥 to both exist. And these two limits must be
equal. And we’re told that this is true
for our function 𝑓 of 𝑥 when our value of 𝑎 is equal to negative five. So let’s set 𝑎 equal to negative
five in our definition.
Since we know the limit as 𝑥
approaches negative five of 𝑓 of 𝑥 exists, all of our conditions must be true. The limit as 𝑥 approaches negative
five from the left of 𝑓 of 𝑥 and the limit as 𝑥 approaches negative five from the
right of 𝑓 of 𝑥 must both exist, and they must both be equal. So let’s try evaluating these
limits. We’ll start with the limit as 𝑥
approaches negative five from the left of 𝑓 of 𝑥. Since 𝑥 is approaching negative
five from the left, our values of 𝑥 will be less than negative five. And we can see from our piecewise
definition of the function 𝑓 of 𝑥, if our values of 𝑥 are less than negative
five, our function 𝑓 of 𝑥 is exactly equal to the function 𝑥 squared minus nine
𝑘.
So because our function 𝑓 of 𝑥 is
exactly equal to the function 𝑥 squared minus nine 𝑘 for all values of 𝑥 less
than negative five, their limits as 𝑥 approaches negative five from the left will
be the same. But now we can see this is just the
limit of a polynomial. We can evaluate this by direct
substitution. Substituting 𝑥 is equal to
negative five, we get negative five squared minus nine 𝑘. And since negative five squared is
equal to 25, we can simplify this expression to 25 minus nine 𝑘. Let’s do the same to evaluate the
limit as 𝑥 approaches negative five from the right of 𝑓 of 𝑥.
Since 𝑥 is approaching negative
five from the right, our values of 𝑥 will be greater than negative five. And again, from our piecewise
definition of the function 𝑓 of 𝑥 for values of 𝑥 greater than negative five, our
function 𝑓 of 𝑥 is equal to two 𝑥 plus 116. So because we’re taking the limit
as 𝑥 approaches negative five from the right, we can rewrite 𝑓 of 𝑥 as two 𝑥
plus 116. And we can now see we’re just
evaluating the limit of a linear function. We can do this by direct
substitution.
Substituting 𝑥 is equal to
negative five, we get two times negative five plus 116. And we can calculate this to give
us 106. Now, remember, since this limit
exists, we know the limit as 𝑥 approaches negative five from the left of 𝑓 of 𝑥
and the limit as 𝑥 approaches negative five from the right of 𝑓 of 𝑥 must both be
equal. This tells us that 25 minus nine 𝑘
is equal to 106. So we now just have to solve the
equation 25 minus nine 𝑘 is equal to 106. We’ll subtract 25 from both sides
of the equation. This gives us negative nine 𝑘 is
equal to 81. Then, we’ll just divide both sides
of the equation through by negative nine. This gives us that 𝑘 is equal to
negative nine.
So we’ve shown if the function 𝑓
of 𝑥 is equal to 𝑥 squared minus nine 𝑘 if 𝑥 is less than negative five and 𝑓
of 𝑥 is equal to two 𝑥 plus 116 if 𝑥 is greater than negative five has a limit as
𝑥 approaches negative five. Then the value of 𝑘 must be equal
to negative nine.
