### Video Transcript

Suppose that π of π₯ is equal to
negative one plus three over π₯ if π₯ is less than or equal to one and negative π₯
cubed plus three if π₯ is greater than one. What can be said about the
differentiability of π at π₯ equals one?

Here, we have been given a
piecewise defined function and asked to check the differentiability. If we call our value of one π₯
nought, our general process for this type of question is first to check for
continuity at π₯ nought. And then if this is satisfied, we
check that the limit as π₯ approaches π₯ nought of our derivative exists, which
means checking the left and right limits both exists and agree. We first begin with the continuity
condition stated here. Since this is a piecewise defined
function, weβll need to check that both the left and the right limits as π₯
approaches one of π of π₯ both exist and agree. When π₯ is less than one, our
function is negative one plus three over π₯. By direct substitution, our
left-sided limit then evaluates to two. When π₯ is greater than one, our
function is negative π₯ cubed plus three. And by the same approach, the
right-sided limit also evaluates to two.

Since these two limits both exist
and agree, we can also see that the normal limit exists and is equal to two. Moving onto π of one, the function
itself here is defined by our first subfunction since π₯ is equal to one. And in fact, weβve already
substituted one into our first subfunction here, which gave us an answer of two. This means that we can say that the
limit as π₯ approaches one of π of π₯ and π of one are both equal to two. We have, therefore, satisfied the
continuity condition and our function is continuous when π₯ is equal to one. Letβs now move on to considering
the derivative of our function.

Since our function is defind
peacewise, the derivative of our function will also be defined piecewise. And weβll need to differentiate
both of our subfunctions. And here is worth noting that were
not yet making any claims as the derivative when π₯ is equal to one, since thatβs,
in fact, what weβre trying to find right now. One of the tools that we can use to
help us differentiate our subfunctions is the power rules stated here. To help us apply this rule more
easily, we can reexpress three over π₯ as three π₯ to the power of minus one. Applying the rule, we find that π
dash of π₯ is equal to negative three over π₯ squared if π₯ is less than one and
negative three π₯ squared if π₯ is greater than one.

We now need to consider the limit
as π₯ approaches one of π dash of π₯. Weβll do so by considering the
left- and the right-sided limits in a process similar to the one that we just
reformed for the continuity condition. By direct substitution of one, we
find that both of these limits are equal to negative three. And since they exist and agree, the
limit as π₯ approaches one of π dash of π₯ is also equal to negative three. Since this limit exists, we know
that our derivative also exists at π₯ equals one. Weβve now completed all the steps
of our process. And from these, we can conclude
that the function π is differentiable at the point where π₯ is equal to one.