### Video Transcript

Find the values of π and π given
the function π is differentiable at π₯ is equal to negative one where π of π₯ is
equal to nine π₯ plus five if π₯ is less than negative one and π of π₯ is equal to
ππ₯ squared plus ππ₯ minus four if π₯ is greater than or equal to negative
one.

Weβre given a piecewise-defined
function π of π₯. Weβre told that this
piecewise-defined function is differentiable when π₯ is equal to negative one. We need to use this to determine
the values of π and π. And thereβs a few different ways of
doing this. We could do this directly from the
definition of the derivative of a function when π₯ is equal to negative one. But thereβs a simpler method by
using our definition of the function π of π₯. In this case, we can see our
function π of π₯ is defined piecewise. We can see the first piece is nine
π₯ plus five and the second piece is ππ₯ cubed plus ππ₯ minus four.

In other words, in both cases, our
function π of π₯ is defined by polynomials which we know are continuous for all
real values of π₯. When a piecewise-defined function
is defined by continuous function, we call it piecewise continuous. And in fact, thereβs something even
more useful. We know how to differentiate
polynomial functions by using the power rule for differentiation. We could use this to find an
expression for π prime of π₯, everywhere except where π₯ is equal to negative
one. Then, we can use this for an easier
method to check that our function π of π₯ differentiable when π₯ is equal to
negative one.

We just need to check that our
function is continuous when π₯ is equal to negative one and the slope as π₯
approaches negative one from the left of π of π₯ is equal to the slope as π₯
approaches negative one from the right of π of π₯. But in this case, we already know
our function π is differentiable when π₯ is equal to negative one. This gives us two pieces of
information about our function π. First, if π is differentiable when
π₯ is equal to negative one, then it also must be continuous when π₯ is equal to
negative one. Second, weβll also use the piece of
information that π is differentiable at π₯ is equal to negative one.

Letβs start with our first piece of
information, π is continuous when π₯ is equal to negative one. We know π is a piecewise
continuous function. And in fact, we can see π₯ is equal
to negative one is the endpoint of our interval. And a piecewise continuous function
can only be continuous at the endpoints of its interval if the endpoints match
up. So, for our function π, the
endpoints must match up. We could do this formally by using
our left-hand and right-hand limit. In other words, we would check the
limit as π₯ approaches negative one from the left of π of π₯ is equal to the limit
as π₯ approaches negative one from the right of π of π₯.

But remember, weβve already said
that π of π₯ is a piecewise continuous function. So, we could just evaluate these
limits directly by substituting π₯ is equal to negative one into both of our
functions. Because these are continuous, we
can just evaluate by direct substitution. This gives us nine times negative
one plus five is equal to π times negative one all squared plus π times negative
one minus four. And we can simplify this
expression.

Nine times negative one plus five
simplifies to give us negative four and then eight times negative one squared plus
π times negative one minus four simplifies to give us π minus π minus four. And then, we can just add four to
both sides of this equation, and this gives us zero is equal to π minus π. If we add π to both sides of this
equation, we can see that π must be equal to π. So because π must be continuous at
negative one, weβve shown that π must be equal to π.

We now want to use the fact that
our function π of π₯ must be differentiable when π₯ is equal to negative one. We could do this by directly using
the definition of a derivative at π₯ is equal to negative one. However, π of π₯ is defined in
terms of polynomials, and we know how to differentiate polynomials by using the
power rule for differentiation. So, instead of directly using the
definition of a derivative when π₯ is equal to negative one, instead, we can use the
fact that because π must be differential when π₯ is equal to negative one, we must
have the slope as π₯ approaches negative one from the left is equal to the slope as
π₯ approaches negative one from the right.

And we can in fact evaluate both of
these by just using the power rule for differentiation. So, weβll use this to find an
expression for π prime of π₯. We get π prime of π₯ is equal to
the derivative of nine π₯ plus five with respect to π₯ if π₯ is less than negative
one and π prime of π₯ is equal to the derivative of ππ₯ squared plus ππ₯ minus
four with respect to π₯ if π₯ is greater than negative one. And itβs important to know we donβt
know what happens when π₯ is equal to negative one.

And itβs also important to point
out since we know that π is equal to π, weβve just substituted this directly into
our expression. Next, we need to evaluate both of
these derivatives by using the power rule for differentiation. We need to multiply by the exponent
of π₯ and then reduce this exponent by one. In fact, in our first function, we
have a linear function. So, its slope will just be the
coefficient of π₯, which is nine. Differentiating our second piece
term by term by using the power rule for differentiation, we get two ππ₯ plus π if
π₯ is greater than negative one.

Now, we can use this to find the
slope of π of π₯ as π₯ approaches negative one from the left and the slope of π of
π₯ as π₯ approaches negative one from the right. Letβs start with the slope of π of
π₯ as π₯ approaches negative one from the left. Since our values of π₯ are
approaching negative one from the left. We know our values of π₯ are less
than negative one. So, our slope is just equal to the
constant nine. So, our first expression simplifies
to give us nine. We can do the same to find the
slope as π₯ approaches negative one from the right.

Since π₯ is approaching negative
one from the right, our values of π₯ will be greater than negative one. And we know the slope of our
function when π₯ is greater than negative one is equal to two ππ₯ plus π. And, of course, two ππ₯ plus π is
a linear function. So, we can evaluate this limit by
using direct substitution. We substitute π₯ is equal to
negative one into this expression. This gives us two π times negative
one plus π, which of course we can simplify to give us negative two π plus π
which is equal to negative π. But remember, weβre told our
function π is differentiable when π₯ is equal to negative one.

So, the slope as π₯ approaches
negative one from the left of π of π₯ must be equal to the slope as π₯ approaches
negative one from the right of π of π₯. So, we can just equate these
values. We get negative π must be equal to
nine. And if we multiply through by
negative one, we see that π is equal to negative nine. And for completion, remember, we
showed that π must be equal to π. So, we can also say that π must be
equal to negative nine.

Therefore, if the function π of π₯
is equal to nine π₯ plus five of π₯ is less than negative one and π of π₯ is equal
to ππ₯ squared plus ππ₯ minus four if π₯ is greater than or equal to negative one
is differentiable when π₯ is equal to negative one, then weβve shown that the value
of π must be equal to negative nine and the value of π must also be equal to
negative nine.