### Video Transcript

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

Weβre given a piecewise-defined
function π of π₯. And weβre told that this function
π is differentiable when π₯ is equal to one. We need to use this to determine
the values of π and π. The first thing we notice about
this is when π₯ is equal to one, we can see weβre at the endpoints of our interval
of the piecewise-defined function. In other words, when π₯ is equal to
one, our function π of π₯ changes from being equal to negative π₯ squared plus four
to being equal to negative two ππ₯ minus π.

And at this point, thereβs a few
different methods we could use to try and answer this question. For example, we might be tempted to
directly use the definition of π being differentiable at π₯ is equal to one. And this would work. However, because our function π of
π₯ is defined piecewise and π₯ is equal to one is one of the endpoints of this
interval, we can actually do this in a simpler way.

First, we recall if a function is
differentiable at some point, then it must also be continuous at this point. In other words, since we know π is
differentiable at π₯ is equal to one, we know that π must also be continuous when
π₯ is equal to one. And we can see something
interesting about our function π of π₯. We can see both pieces of this
function are polynomials. And we know polynomials are
continuous for all real values of π₯. So our function π of π₯ is
piecewise continuous.

And for a piecewise continuous
function to be continuous at its endpoints, its endpoints must match. In other words, we know the limit
as approaches one from the left of π of π₯ must be equal to the limit as π₯
approaches one from the right of π of π₯.

Now, we could evaluate this limit
directly. However, we need to remember π of
π₯ is a piecewise continuous function. And because each piece is
continuous, we can evaluate each of these limits by using direct substitution. So we just substitute π₯ is equal
to one into negative π₯ squared plus four to evaluate the limit as π₯ approaches one
from the left of π of π₯. We get negative one squared plus
four.

And we can do the same to evaluate
the limit as π₯ approaches one from the right. We have negative two ππ₯ minus π
is continuous. So we can evaluate this limit by
using direct substitution. We just substitute π₯ is equal to
one into negative two ππ₯ minus π. This gives us negative two π times
one minus π. And because we know π is
continuous, we know these two limits have to be equal.

Letβs now simplify both sides of
this equation. First, negative one squared plus
four is equal to three. And we can simplify the right-hand
side of this equation to give us negative two π minus π. But this is only one equation with
two variables. So we need more information to find
the values of π and π. To do this, weβll want to
specifically use the fact that π is differentiable at π₯ is equal to one. And we know several different ways
of explaining that π is differentiable at π₯ is equal to one. But one of these is a lot easier to
work with for our function π of π₯.

We can see that both of the parts
of π of π₯ are defined as polynomials. And we already know how to
differentiate polynomials term by term by using the power rule for
differentiation. So instead of directly applying the
definition of a derivative to our function π of π₯, we can instead look at the
slope as π₯ approaches one from the left of π of π₯ and look at the slope as π₯
approaches one from the right of π of π₯. In other words, we know if π is
differentiable at π₯ is equal to one, then the slope as π₯ approaches one from the
left of π of π₯ must be equal to the slope as π approaches one from the right of
π of π₯. And we use this because we can
easily find an expression for π prime of π₯ when π₯ is less than one and when π₯ is
greater than one.

We just need to differentiate each
piece of π of π₯ separately. We get π prime of π₯ is equal to
the derivative of negative π₯ squared plus four with respect to π₯ if π₯ is less
than one and π prime of π₯ is equal to the derivative of negative two ππ₯ minus π
with respect to π₯ if π₯ is greater than one. And itβs worth reiterating at this
point weβre not stating what π prime of π₯ is equal to when π₯ is equal to one. Weβre just finding an expression
for the slope for all of the other values of π₯.

Now, we can evaluate both of these
derivatives by using the power rule for differentiation. We want to multiply by our
exponents of π₯ and reduce this exponent by one. First, the derivative of negative
π₯ squared plus four with respect to π₯ is equal to negative two π₯. Next, to differentiate our second
function, we could again use the power rule for differentiation term by term. However, this is also a linear
function. So we could just differentiate this
by taking the coefficient of π₯, which is negative two π. This gives us π prime of π₯ is
equal to negative two π₯ if π₯ is less than one and π prime of π₯ is equal to
negative two π if π₯ is greater than one.

Now, we can evaluate the slope as
π₯ approaches one from the left of π of π₯ and the slope as π₯ approaches one from
the right of π of π₯. First, when π₯ approaches one from
the left, we can see π prime of π₯ is equal to negative two π₯. And of course, negative two π₯ is a
continuous function. So we can evaluate this by using
direct substitution. We just substitute π₯ is equal to
one. We get negative two times one,
which is equal to negative two.

And we can do exactly the same as
π₯ approaches one from the right. This time, π prime of π₯ will be
equal to negative two π. But this time, negative two π is a
constant. So this is just equal to negative
two π. And remember, weβre told that π of
π₯ is differentiable when π₯ is equal to one. So the slope as π₯ approaches one
from the left of π of π₯ must be equal to the slope as π₯ approaches one from the
right of π of π₯. In other words, we can equate these
two values. We get negative two must be equal
to negative two π.

And if we divide both sides of this
equation through by negative two, we see that π must be equal to one. Now, to find our value π, we
substitute π is equal to one into our equation three is equal to negative two π
minus π. Substituting π is equal to one
gives us that three is equal to negative two times one minus π. And by simplifying and rearranging
this equation, we can see that π must be equal to negative five.

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