### Video Transcript

Find the values of 𝑎 and 𝑏 given

the function 𝑓 is differentiable at 𝑥 is equal to one where 𝑓 of 𝑥 is equal to

negative 𝑥 squared plus four if 𝑥 is less than or equal to one and 𝑓 of 𝑥 is

equal to negative two 𝑎𝑥 minus 𝑏 if 𝑥 is greater than one.

We’re given a piecewise-defined

function 𝑓 of 𝑥. And we’re told that this function

𝑓 is differentiable when 𝑥 is equal to one. We need to use this to determine

the values of 𝑎 and 𝑏. The first thing we notice about

this is when 𝑥 is equal to one, we can see we’re at the endpoints of our interval

of the piecewise-defined function. In other words, when 𝑥 is equal to

one, our function 𝑓 of 𝑥 changes from being equal to negative 𝑥 squared plus four

to being equal to negative two 𝑎𝑥 minus 𝑏.

And at this point, there’s 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 𝑓 being differentiable at 𝑥 is equal to one. And this would work. However, because our function 𝑓 of

𝑥 is defined piecewise and 𝑥 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 𝑓 is

differentiable at 𝑥 is equal to one, we know that 𝑓 must also be continuous when

𝑥 is equal to one. And we can see something

interesting about our function 𝑓 of 𝑥. We can see both pieces of this

function are polynomials. And we know polynomials are

continuous for all real values of 𝑥. So our function 𝑓 of 𝑥 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 𝑓 of 𝑥 must be equal to the limit as 𝑥

approaches one from the right of 𝑓 of 𝑥.

Now, we could evaluate this limit

directly. However, we need to remember 𝑓 of

𝑥 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 𝑥 is equal

to one into negative 𝑥 squared plus four to evaluate the limit as 𝑥 approaches one

from the left of 𝑓 of 𝑥. We get negative one squared plus

four.

And we can do the same to evaluate

the limit as 𝑥 approaches one from the right. We have negative two 𝑎𝑥 minus 𝑏

is continuous. So we can evaluate this limit by

using direct substitution. We just substitute 𝑥 is equal to

one into negative two 𝑎𝑥 minus 𝑏. This gives us negative two 𝑎 times

one minus 𝑏. And because we know 𝑓 is

continuous, we know these two limits have to be equal.

Let’s 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 𝑎 minus 𝑏. But this is only one equation with

two variables. So we need more information to find

the values of 𝑎 and 𝑏. To do this, we’ll want to

specifically use the fact that 𝑓 is differentiable at 𝑥 is equal to one. And we know several different ways

of explaining that 𝑓 is differentiable at 𝑥 is equal to one. But one of these is a lot easier to

work with for our function 𝑓 of 𝑥.

We can see that both of the parts

of 𝑓 of 𝑥 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 𝑓 of 𝑥, we can instead look at the

slope as 𝑥 approaches one from the left of 𝑓 of 𝑥 and look at the slope as 𝑥

approaches one from the right of 𝑓 of 𝑥. In other words, we know if 𝑓 is

differentiable at 𝑥 is equal to one, then the slope as 𝑥 approaches one from the

left of 𝑓 of 𝑥 must be equal to the slope as 𝑓 approaches one from the right of

𝑓 of 𝑥. And we use this because we can

easily find an expression for 𝑓 prime of 𝑥 when 𝑥 is less than one and when 𝑥 is

greater than one.

We just need to differentiate each

piece of 𝑓 of 𝑥 separately. We get 𝑓 prime of 𝑥 is equal to

the derivative of negative 𝑥 squared plus four with respect to 𝑥 if 𝑥 is less

than one and 𝑓 prime of 𝑥 is equal to the derivative of negative two 𝑎𝑥 minus 𝑏

with respect to 𝑥 if 𝑥 is greater than one. And it’s worth reiterating at this

point we’re not stating what 𝑓 prime of 𝑥 is equal to when 𝑥 is equal to one. We’re just finding an expression

for the slope for all of the other values of 𝑥.

Now, we can evaluate both of these

derivatives by using the power rule for differentiation. We want to multiply by our

exponents of 𝑥 and reduce this exponent by one. First, the derivative of negative

𝑥 squared plus four with respect to 𝑥 is equal to negative two 𝑥. 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 𝑥, which is negative two 𝑎. This gives us 𝑓 prime of 𝑥 is

equal to negative two 𝑥 if 𝑥 is less than one and 𝑓 prime of 𝑥 is equal to

negative two 𝑎 if 𝑥 is greater than one.

Now, we can evaluate the slope as

𝑥 approaches one from the left of 𝑓 of 𝑥 and the slope as 𝑥 approaches one from

the right of 𝑓 of 𝑥. First, when 𝑥 approaches one from

the left, we can see 𝑓 prime of 𝑥 is equal to negative two 𝑥. And of course, negative two 𝑥 is a

continuous function. So we can evaluate this by using

direct substitution. We just substitute 𝑥 is equal to

one. We get negative two times one,

which is equal to negative two.

And we can do exactly the same as

𝑥 approaches one from the right. This time, 𝑓 prime of 𝑥 will be

equal to negative two 𝑎. But this time, negative two 𝑎 is a

constant. So this is just equal to negative

two 𝑎. And remember, we’re told that 𝑓 of

𝑥 is differentiable when 𝑥 is equal to one. So the slope as 𝑥 approaches one

from the left of 𝑓 of 𝑥 must be equal to the slope as 𝑥 approaches one from the

right of 𝑓 of 𝑥. In other words, we can equate these

two values. We get negative two must be equal

to negative two 𝑎.

And if we divide both sides of this

equation through by negative two, we see that 𝑎 must be equal to one. Now, to find our value 𝑏, we

substitute 𝑎 is equal to one into our equation three is equal to negative two 𝑎

minus 𝑏. Substituting 𝑎 is equal to one

gives us that three is equal to negative two times one minus 𝑏. And by simplifying and rearranging

this equation, we can see that 𝑏 must be equal to negative five.

Therefore, if the function 𝑓 of 𝑥

is equal to negative 𝑥 squared plus four if 𝑥 is less than or equal to one and 𝑓

of 𝑥 is equal to negative two 𝑎𝑥 minus 𝑏 if 𝑥 is greater than one is

differentiable when 𝑥 is equal to one, then we’ve shown that 𝑎 must be equal to

one and 𝑏 must be equal to negative five.