### Video Transcript

Suppose π of π₯ is equal to
negative six π₯ minus four, if π₯ is less than or equal to negative one, and three
π₯ squared if π₯ is greater than negative one. What can be said about the
differentiability of the function at π₯ equals negative one?

Here, we have been given a
piecewise defined function composed of two subfunctions, a binomial and a
monomial. Both of these subfunctions on their
own or whatβs known as smooth and will be defined over all the real numbers. In fact, on their own, we can say
that all polynomials are smooth, which means that theyβre differentiable on the real
numbers. For a piecewise defined function,
however, we must check the point at which are two subfunctions join. That is, in this case, π₯ equals
negative one. To begin the process, let us
differentiate both of our subfunctions using the power rule.

The derivative of negative six π₯
minus four is negative six times π₯ to the power of zero, which is, of course, just
negative six. The derivative of three π₯ squared
is two times three π₯, which is, of course, six π₯. We can represent this more
succinctly by saying that π dash of π₯ is equal to a negative six, if π₯ is less
than negative one, and six π₯ if π₯ is greater than negative one. It should be noted that even though
we see a less than or equal to inequality symbol here, we are not yet making any
claims as to the value of our derivative when π₯ is equal to negative one, since
this is what weβre trying to find at the moment.

Considering the derivative either
side of the point where our two subfunctions join, we might try to move forward by
substituting the value of π₯ equals negative one into the two subfunctions that we
just found for π dash of π₯. Doing so, we would find that both
of our values are negative six. We may then conclude that since
these two values are equal, that our function is differentiable at the point where
π₯ is equal to negative one. Unfortunately, this is not the
case. And to see why. Let us recall the definition of the
derivative as a limit.

Here weβve shown the limit which
defines the derivative. To move forward, we recognize that
since our function is defined differently on either side of the point where π₯
equals negative one, the left- and the right-sided limits will, therefore, have
different expressions. Remembering that our π₯ nought is
equal to negative one, we can begin by expressing the left limit as follows. Now, clearly, we cannot simply
substitute β equals zero into this expression. Otherwise, weβll be left with the
indeterminate form of zero over zero. Instead, we know that when π₯ is
less than negative one, π of π₯ is equal to negative six π₯ minus four. Hence π of negative one plus β
will evaluate to two minus six β.

By similar reasoning, when π₯ is
equal to negative one π of π₯ takes the same subfunction. And therefore, π of negative one
evaluates to two substituting these in we find that the expression for our
left-sided limit becomes two minus six β minus two all over β, which simplifies to
negative six β over β. Now at this stage, since we know
that β is approaching zero and not equal to zero, weβre allowed to cancel the common
factor on the top and bottom half of our quotient. We are then left with the limit as
β approaches zero from the left of negative six which, of course, is just equal to
negative six.

Now for the right-sided limit, by
similar reasoning as previously, we know that when π₯ is greater than negative one,
π of π₯ is defined by the subfunction three π₯ squared. Hence, in this case, π of negative
one plus β evaluates to three minus six β plus three β squared. For π of negative one, we should
be careful not to use our three π₯ squared again. Since when π₯ is equal to negative
one, π is defined by our other subfunction. Weβve already found the value of
this to be two earlier. Again, we go through the same
process of substituting these in and simplifying the expression for our limit. This time we reach a different
result. Looking at the first term of our
limit, weβll have the limit as β approaches zero from the positive direction of one
over β, which is equal to infinity. By extension, this means our entire
right-sided limit is also equal to infinity.

We know that this is a particular
way of expressing that the limit does not exist. If the right limit does not exist,
then this also tells us that the normal limit does not exist. And hence, the derivative is not
defind. The reason this is the case is that
our graph actually has a discontinuity at π₯ equals negative one. And if we were to draw it, we would
see this. Since we have concluded that the
derivative is undefined at π₯ equals negative one, weβre also in a position to say
that the function is not differentiable at π₯ equals negative one. And in fact, this is the answer to
our question.