Video Transcript
Discuss the differentiability of
the function 𝑓 at 𝑥 is equal to one given that 𝑓 of 𝑥 is equal to two 𝑥 plus
eight if 𝑥 is less than one and 𝑓 of 𝑥 is equal to 𝑥 squared plus nine if 𝑥 is
greater than or equal to one.
We’re given a piecewise-defined
function 𝑓 of 𝑥. We need to determine whether the
function 𝑓 is differentiable when 𝑥 is equal to one. There’s a few different ways of
doing this. For example, we could first check
whether our function 𝑓 of 𝑥 is continuous when 𝑥 is equal to one. Then if 𝑓 of 𝑥 is continuous when
𝑥 is equal to one, we can check the slope from the left and from the right when 𝑥
is equal to one by differentiating both pieces of our function. If these two slopes match, then we
can say 𝑓 of 𝑥 is continuous when 𝑥 is equal to one.
However, in this video, we’re going
to determine this directly from the definition of a derivative. Let’s start by recalling what we
mean by the derivative of 𝑓 of 𝑥 with respect to 𝑥 at a point 𝑥 naught. This is equal to the limit as ℎ
approaches zero of 𝑓 of 𝑥 naught plus ℎ minus 𝑓 of 𝑥 naught all divided by
ℎ. And this assumes this limit
exists. If this limit does not exist, then
we say that our function 𝑓 of 𝑥 is not differentiable at the point 𝑥 naught. In this case, we’re given the
piecewise function 𝑓 of 𝑥. And we want to know if it’s
differentiable at the point where 𝑥 is equal to one. So we’ll set 𝑥 naught equal to
one.
We can then substitute 𝑥 naught is
equal to one into our definition of the derivative. So determining whether 𝑓 of 𝑥 is
differentiable when 𝑥 is equal to one is the same is determining whether this limit
exists. However, we can’t directly evaluate
this limit because our function 𝑓 of 𝑥 is given as a piecewise function and 𝑥 one
is the endpoints of one of these intervals.
So we need to remember a fact about
limits. Instead of finding the limit as ℎ
approaches zero of this expression, we’ll find the limit as ℎ approaches zero from
the right of this expression and the limit as ℎ approaches zero from the left of
this expression. If both of these limits exist and
are equal, then we know our function 𝑓 of 𝑥 is differentiable at this point. However, if either limit does not
exist or these limits are not equal, then we know that our function is not
differentiable when 𝑥 is equal to one.
Let’s start by evaluating the limit
as ℎ approaches zero from the left of 𝑓 of one plus ℎ minus 𝑓 of one all divided
by ℎ. To evaluate this limit, we first
need to notice that ℎ is approaching zero from the left. This means that all of our values
of ℎ will be less than zero. And if ℎ is less than zero, then
one plus ℎ will be less than one. And we know from our piecewise
definition of the function 𝑓 of 𝑥 when 𝑥 is less than one, 𝑓 of 𝑥 is equal to
the linear function two 𝑥 plus eight. So in this case, to evaluate 𝑓 at
one plus ℎ, we just substitute 𝑥 is equal to one plus ℎ into the linear function
two 𝑥 plus eight. This gives us two times one plus ℎ
add eight.
We might be tempted to substitute
one into this linear function. But remember, we need to use the
piecewise definition of 𝑓 of 𝑥. When 𝑥 is equal to one, 𝑓 of 𝑥
is exactly equal to the quadratic 𝑥 squared plus nine. Substituting 𝑥 is equal to one
into our quadratic gives us one squared plus nine. So we now have to evaluate the
limit as ℎ approaches zero from the left of two times one plus ℎ plus eight minus
one squared plus nine all divided by ℎ.
To do this, we need to simplify the
expression inside of our limit. We’ll start with the first term in
our numerator. We’ll distribute two over the
parentheses. This gives us two plus two ℎ plus
eight. And of course, we can simplify this
since two plus eight is equal to 10. Next, we’ll evaluate the second
term in our numerator. We have one squared plus nine is equal to one plus nine. And of course, we can simplify one
plus nine to give us 10.
So now, we have the limit as ℎ
approached zero from the left of two ℎ plus 10 minus 10 all divided by ℎ. Of course, we can keep simplifying
this. In our numerator, 10 minus 10 is
equal to zero. So in actual fact, this entire
limit just simplified to give us the limit as ℎ approaches zero from the left of two
ℎ divided by ℎ. And of course, we know ℎ is
approaching zero from the left. In particular, this means that ℎ is
not equal to zero. It’s just getting closer and closer
to zero. So we can cancel the shared factor
of ℎ in our numerator and our denominator, meaning we’re just left with the limit as
ℎ approaches zero from the left of the constant two. But two is a constant, so this
limit evaluates to give us two.
Therefore, as 𝑥 approaches one
from the left, the slope of our function 𝑓 of 𝑥 is equal to two. We now need to do exactly the same
thing from the right. So let’s clear some space and
evaluate the limit as ℎ approaches zero from the right of 𝑓 of one plus ℎ minus 𝑓
of one all divided by ℎ. This time, we have ℎ is approaching
zero from the right. This means we know all of our
values of ℎ will be bigger than zero. And if ℎ is just bigger than zero,
this means one plus ℎ will be bigger than one.
So once again, we need to use the
piecewise definition of our function 𝑓 of 𝑥 to evaluate 𝑓 of one plus ℎ. We know when 𝑥 is greater than or
equal to one, 𝑓 of 𝑥 is exactly equal to the quadratic 𝑥 squared plus nine. So because one plus ℎ is greater
than one, to evaluate 𝑓 at one plus ℎ, we substitute 𝑥 is equal to one plus ℎ into
this quadratic. This gives us one plus ℎ all
squared plus nine. We already found 𝑓 evaluated at
one in our previous limit. To do this, we use the piecewise
definition of 𝑓 of 𝑥.
We just substitute 𝑥 is equal to
one into the quadratic 𝑥 squared plus nine. This gave us one squared plus nine,
which we simplified to give us 10. So we now need to evaluate the
limit as ℎ approaches zero from the right of one plus ℎ all squared plus nine minus
10 all divided by ℎ. To do this, we can start
simplifying. First, we know that nine minus 10
is equal to negative one. Next, we want to distribute the
square over our parentheses.
Doing this by using the FOIL method
or by using binomial expansion, we get one plus two ℎ plus ℎ squared. Then we need to subtract one from
this and divide through by ℎ. And in fact, we can see we can keep
simplifying. In our numerator, we have one minus
one, which simplifies to give us zero. So this gives us the limit as ℎ
approaches zero from the right of two ℎ plus ℎ squared all divided by ℎ. And once again, this limit is as ℎ
is approaching zero from the right. ℎ is getting closer and closer to
zero, but ℎ is never equal to zero.
This means we can cancel the shared
factor of ℎ in our numerator and our denominator. This leaves us with the limit as ℎ
approaches zero from the right of two plus ℎ. And of course, as ℎ is approaching
zero from the right, our value of ℎ is approaching zero and our constant two remains
constant. So this limit just evaluates to
give us two. And now, we can see directly from
the definition of the slope that these two slopes are equal. And because these two values are
equal, we can conclude that 𝑓 of 𝑥 is differentiable at 𝑥 is equal to one.
It’s also worth reiterating at this
point, we could only use this method because we directly worked with the definition
of differentiability at a point. If we had instead wanted to work
with other rules of differentiation, such as the power rule for differentiation on
our function 𝑓 of 𝑥, we would also need to show that 𝑓 of 𝑥 is continuous when
𝑥 is equal to one. However, as we’ve seen, it’s not
necessary to use this method. We can just do this directly from
the definition of a derivative. Therefore, we were able to show the
function 𝑓 of 𝑥 is equal to two 𝑥 plus eight if 𝑥 is less than one and 𝑓 of 𝑥
is equal to 𝑥 squared plus nine if 𝑥 is greater than or equal to one is
differentiable at 𝑥 is equal to one.
