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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 9

Lesson 3: AP Calculus AB 2011 free response- 2011 Calculus AB free response #1a
- 2011 Calculus AB Free Response #1 (b, c, & d)
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c

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# 2011 Calculus AB free response #6b

Derivative of a piecewise-defined function. Created by Sal Khan.

## Want to join the conversation?

- Why did you not use the average value equation? (1/(b-a))times the definite integral from a to b. Since it is a piecewise function, it would have to be split into two integrals. Would this work?(3 votes)
- Hi Morgan. You are one step ahead. Check out the next video and see the average value equation in action. 8-)(5 votes)

- i still didn't get why we didn't include x = 0?(2 votes)
- Notice that the two possible derivatives of this function are not equal at x=0. Try differentiating 1 -2sinx. You get -2cosx, which has derivative of -2 at x=0. If you differentiate e^(-4x), you get -4e^-4x, with derivative of -4 at x=0. This means the derivative is not continuous at x=0. Thus, the AP test writers added the condition of not having the derivative at x=0.(1 vote)

- I didn't understand a word of that video. I think I am in the wrong playlist. What's Calculus? Does it include Addition? I don't even get place value, much less "derivatives" (whatever those are).(0 votes)
- Hi RonaqSingh. I looked through your discussion page, and I'm pretty sure that you know about addition and place values. Someday in the not to distant future you will be doing these Calculus problems. Best of luck to you in your educational journey. ☺(2 votes)

## Video transcript

Part B. For x is not
equal to 0, express f prime of x as a
piecewise-defined function. Find the value of x
for which f prime of x is equal to negative 3. So the first thing you
might be wondering is, why did we even have to
take out x is equal to 0, or why is the derivative not
going to be defined there? And that's just
because you're going to see that the derivative is
going to be something different when we approach x is equal to
0 from the left, versus when we approach x is equal
to 0 from the right. And that's why they just
took it out of there for us. But let's just figure out
what that derivative is for all of the
other values of x. So f prime of x is equal
to, so for x is less than 0, we're going to take the
derivative of this first case. So the derivative
of 1 is just 0. The derivative of
negative 2 sine of x. Well, the derivative of sine
of x is just cosine of x. So it's going to be
negative 2 cosine of x for x is less than 0. And then for x is
greater than 0-- I'll do this in
another color, I'll do it in orange-- we have
this case right over here. And we'll just do the
chain rule, derivative of negative 4x with
respect to x is negative 4. And derivative of e to the
negative 4x with respect to negative 4x is just
e to the negative 4x. Sometimes you could say
this is a derivative of the inside times the
derivative of the outside, with respect to the inside. So either way, it's negative
4 e to the negative 4x for x is greater than 0. So we did the first
part, we expressed f prime of x as a
piecewise-defined function. We didn't define the derivative. I actually forgot
a parenthesis here. We didn't define the derivative
when x is equal to 0, because it's actually not
going to be defined there. Now let's do the second part. Find the value of x
for which f prime of x is equal to negative 3. And so if this wasn't
piecewise-defined, you'd very simply
just say, look, f prime of x is
equal to negative 3. You would take whatever
f prime of x is equal to, and you'd do some
algebra to solve for it. But here you're like,
which case do I use? I don't know if the x
that gets us to negative 3 is going to be less
than 0, or I don't know if it's going
to be greater than 0. So I don't know
which case to use. And one thing that
we realize is to look at these functions a
little bit and realize that cosine of x is
a bounded function. Cosine of x can only go between
positive 1 and negative 1. So negative 2
cosine of x can only go between positive
2 and negative 2. So it can never
get to negative 3. So if anything's ever
going to get to negative 3, it's going to have to be
this part of the derivative, or this part of the
derivative definition. So it's going to have to be
this thing right over here. And hopefully
there's some values of x greater than 0 where
this thing right over here is equal to negative 3. So let's try it out. Negative 4 e to the
negative 4x needs to be equal to negative 3. We can divide both
sides by negative 4. We get e to the negative
4x is equal to negative 3/4 divided by negative 4 is 3/4. We could take the natural
log of both sides, and we will get negative 4x
is equal to the natural log of 3/4. And just to be clear, what
I did here, you literally could put the natural log
here, natural log there, and you could put
the natural log there as well to see that step. This is saying, what power do
I have to raise e to, to get e to the negative 4x? Well, obviously, I
just need to raise e to the negative
4x power there. So this power is negative 4x. And then we just
took the natural log of the right-hand side as well. And then to solve for x,
we can divide both sides by negative 4. So you get x is equal to-- or
we could multiply both sides by negative 1/4, either way--
negative 1/4 natural log of 3/4. And what we need to do
is verify that this x. So we used this case right over
here, but we have to make sure that we can use this case,
that this x is greater than 0. And we might be
tempted right when we look at this to
say, wait, this looks like a negative number. But we have to remind ourselves
that the natural log of 3/4, since 3/4 is less than
e, the natural log of 3/4 is going to be a
negative number. It's going to be e to
some negative exponent. So since this is negative
and this part right over here is negative, you have a
negative times a negative, so this right over here
is going to be positive. So this is a positive
value right over here. So you would use this
case right over here. So that's our answer-- x
is equal to negative 1/4 natural log of 3/4. Or the derivative, we could
write f prime of negative 1/4 times the natural log of
3/4 is equal to negative 3. And we're done.