MARK KIRALY: Hi, Virge.
VIRGE CORNELIUS: Hi, Mark.
Last session.
So glad everybody could join us for this last session.
What are we doing today?
Oh--
MARK KIRALY: Calculus AB.
VIRGE CORNELIUS: Oh, right.
That-- and a free response question again.
I'm going to do this session this time.
I'm Virge Cornelius.
I teach at Lafayette High School in Oxford, Mississippi.
MARK KIRALY: And I'm Mark Kiraly.
I teach at Ryan High School in Denton, Texas.
VIRGE CORNELIUS: I think I've said it before,
but I'll say it again.
This is my favorite.
I just love these questions.
And this is what I would call a medley or a mixed
bag or a buffet.
I don't know.
MARK KIRALY: If I can just interrupt.
Virge has said this before in other videos.
I've said it.
VIRGE CORNELIUS: Oh.
MARK KIRALY: There's a link somewhere.
VIRGE CORNELIUS: Oh, yeah--
MARK KIRALY: Especially this one because it
is the multiple representation, there's so much to see.
You're going to want to download or pause the screen right now,
and take in all this information.
VIRGE CORNELIUS: Good.
Yes, this is all the given information.
This is the stem of the problem.
All the given information.
And in part A, what are we being asked to find?
Oh, we're being given even more information.
Let P be the function defined.
Yeah, let P be the function defined
by P of x equals a of x times b of x.
And we are supposed to be finding P prime of negative 3.
This brings me back to the very first session
actually because I think the very first thing
I did on the very first multiple choice involved a product rule.
And that's exactly what's going on here
is that I have to find the derivative of this product.
So without even worrying too much about what's in here,
I'm just going to jump in and get my pen
to move a little bit.
So I've got a--
MARK KIRALY: To be honest--
VIRGE CORNELIUS: Go ahead.
MARK KIRALY: When I was doing this one earlier today,
I did actually remember you doing--
in the first video and I--
VIRGE CORNELIUS: See?
Yeah.
So first times the derivative of the second.
So a times b prime plus second times
the derivative of the first and product rule is full
of multiplication and addition, both of which are commutative.
So you can do this and see this in many different ways.
Substituting in negative 3.
I'm going to go ahead and take a look at these different values.
Just a tip, I know that I have a lot of very--
I have a lot of bright students, and I
know there's a lot of bright students watching who can
do a lot of this in their head.
But I would encourage, you on the exam,
to write everything down because things
are worth certain points M I would
hate for you to not get all the points
that you definitely qualify.
MARK KIRALY: I so almost jumped in and said that,
but I felt like I was jumping into much already
so I'm so glad you said it.
VIRGE CORNELIUS: Yeah, of course,
I said that because we see students who lose points,
and we want everybody to get as many points as possible.
So a of negative 3.
This is my a function right here.
Function 0 and derivative values.
So a of negative 3 would be 9.
B prime of negative 3-- well here's
my graph of b of x and at negative 3,
I need to think about the slope of this line segment.
And let's see.
This seems like it's got a slope of 2 perhaps.
Up 2 over 1.
Yeah, so that's 9 times positive 2 plus B of negative 3.
B of negative 3 is going to be the y value when
x is equal to negative 3.
So that's negative 1.
I'm going to use a little parentheses for clarity.
And then a prime of negative 3 is going to be also negative 1.
I think my final answer here ends up being 19.
This is no calculator.
And I could leave it just like this.
That would be totally fine.
But I went ahead and confirmed that it's 19.
Any other comments there, Mark?
MARK KIRALY: No, pretty good.
VIRGE CORNELIUS: Besides the fact
that I'm saying exactly what's on your mind?
MARK KIRALY: Other than get out of my head.
[LAUGHTER]
VIRGE CORNELIUS: All right.
Part B says, let's see C be the function defined by C
of x is equal to 5b a of x.
This is tricky and I think we've seen this at least one
other time in our 8 sessions.
Is this product rule, or is this something else?
MARK KIRALY: I see a function in a function.
VIRGE CORNELIUS: Yes, it would be nice
if we maybe had different kinds of parentheses there
or something like that.
But I definitely need chain rule here.
So when I find the derivative--
C prime is what I'm looking for-- evaluated at negative 1,
I need to think about my chain rule.
So my outer function is b of something.
This 5 is just along for the ride.
That's what Mark always says.
I heard him say that in one of the other sessions.
So just bring it along for the ride.
So 5b prime of a of x.
So I take the derivative of my b function
with the a of x function still in there and then times
the derivative of that inside function.
So a prime of x.
So C prime of negative 1 is going
to be all of this with a negative 1
substituted in for x.
And let's see where I need to go.
A prime of negative 1.
Here's my a function.
A prime of negative 1 is negative 3.
MARK KIRALY: Yeah.
VIRGE CORNELIUS: I'm going to use parentheses for clarity
there.
A of negative 1 is what I need to figure out next
because it's inside that b prime so I need to figure out that.
So a of negative 1 is positive 3.
And now I've got to figure out b prime of positive 3
and I'll be almost done.
So here's positive 3, and b prime would be the derivative.
So I need to think about the slope of that line segment.
And so that line segment has a slope
of-- looks like it's negative, and it
looks like I'm going down 3 to the right 1.
Would you agree, Mark.
MARK KIRALY: Yeah, it ends up-- if you do the other one,
it ends up being 6 down total and 2 over so--
VIRGE CORNELIUS: True.
True, true.
So I'm seeing 5 times negative 3 times negative 3.
And then this is where we could leave it, OK?
You can't leave it like this because you still
got to evaluate b prime of 3.
But, again, I would probably just
go ahead and say 45 here because that
is the final, final answer.
And if this showed up on multiple choice,
you would have to get it down to that 45
anyway, so it's good to practice.
Even though you might figure it out
and then cross it out like Mark just did in a previous video.
Part C. Same stem.
Lots of information to digest, and it's
important to read all this.
In the essence of time, I'm expecting
that you've paused your video and you've really
read all of this carefully.
It says, let V of x be the function defined by V of x
is equal to a inverse of x.
Find V prime of negative 2.
These questions are just wonderful because you can just
hop around all these ideas of the inverse function
and the relationships with the derivatives
and things like that.
I know that V and a are inverses,
and I have a lot of information about a.
But I need to find the derivative at--
a V at negative 2.
So this is what I do anyway.
I think to myself, OK, I have a of x,
and I have V of x and I know that the relationship is
that these are inverses.
I'm trying to figure out V prime of negative 2.
So negative 2 is an x value on the function
V. So that means it had to have been what, Mark?
MARK KIRALY: A y value.
The x and y are would get inverted, right?
VIRGE CORNELIUS: Yep, that's right.
So I guess I want to see what's going on here at--
ooh, there's a lot of negative 2's in this table.
Huh, I bet we did that on purpose.
MARK KIRALY: --but it's a, a of x equals negative 2.
VIRGE CORNELIUS: Yeah, well--
so we want that to be the y value on the a function.
So we want this negative 2.
So we know that the x value is 1.
So we know that the companion point or the mapped point
is negative 2, 1 on the V of x function.
Now, the slope at 1 on the a function is negative 4.
And the cool thing about inverses
is that their slopes are reciprocal.
And I don't know if I'm going to get in trouble from Mark
by saying this but if you think about switching
the x's and the y's--
MARK KIRALY: That's exactly what I do.
Yeah.
VIRGE CORNELIUS: Yeah--
MARK KIRALY: Dy, dx on one is dx, dy on the other.
VIRGE CORNELIUS: So it's nothing to do
with perpendicular lines and opposite reciprocal
or whatever.
It's just that you're inverting that slope.
You're not changing the sign to be positive or negative
or whatever.
So this slope-- V prime of negative 2--
would be the reciprocal of the negative 4.
So I would get negative one fourth.
Good?
MARK KIRALY: Very good job.
VIRGE CORNELIUS: Thank you.
Thank you.
OK.
Last question.
On the last question.
MARK KIRALY: [INAUDIBLE] V be for very good.
VIRGE CORNELIUS: Oh, this is so small.
I can barely read it.
OK.
Getting so old here.
All right.
Part D says, find the value of the limit
as x approaches 1 of this extremely complicated quotient.
Show the work that leads to your answer.
Well, if I have a limit, I should probably
do direct substitution and just kind of see
what happens, right?
What happens if I substitute in a 1?
Let's see.
The limit as--
MARK KIRALY: I just want to say, I always plug it
into the denominator first--
VIRGE CORNELIUS: OK.
MARK KIRALY: --because if you get
a number other than 0 in the denominator,
direct sub's going to work.
VIRGE CORNELIUS: OK.
So what you're saying is you're doing 1 squared.
Putting in a 1 there.
MARK KIRALY: Yep.
VIRGE CORNELIUS: Plsu cosine of pi times 1.
Put in a 1 there.
Cosine of pi is negative 1.
1 squared is 1.
So I have 1 plus negative 1 in the denominator or 0.
So maybe I do need to check the numerator.
MARK KIRALY: So I know-- undefined is a possibility,
right?
VIRGE CORNELIUS: Undefined as a possibility.
MARK KIRALY: It does not-- the limit doesn't exist.
The evaluation might be undefined.
VIRGE CORNELIUS: Mm-hmm.
So I better check and see what's going on in the numerator.
So in the numerator--
I'm just going to kind of scratch it around down here.
I've got negative 5 plus the integral
from 0 to substituting in 1 for x of b of tdt.
So the integral from 0 to 1 of my b of x function--
I've pretty much done a lot of derivatives on here
or found y values.
Now, I have to figure out what is that area from x equals
0 to x equals 1 between the b function and the x-axis.
It's looking like 1 times 5 is 5.
MARK KIRALY: Sure does.
VIRGE CORNELIUS: So I've got negative 5 plus 5.
This ends up being 0 as well.
MARK KIRALY: Yeah.
VIRGE CORNELIUS: So my denominator went to 0.
My numerator went to 0.
Sounds a little like little L'Hopital's rule
here, doesn't it?
So it says show the work that leads to your answer.
I think I'm going to do a little bit more than that.
I'm going to really justify why I'm allowed to be--
or can invoke L'Hopital's rule here.
And I'm going to do my favorite thing.
I had a sandwich for dinner tonight,
but I didn't have a 0 sandwich.
Right now I'm going to make a 0 sandwich.
So I'm going to say that the limit,
as x approaches 1 of my numerator--
copying everything down-- is equal to 0.
And the limit, as x approaches 1 of my denominator,
also equals 0.
MARK KIRALY: And if they both are headed to 0--
VIRGE CORNELIUS: Mm-hmm.
MARK KIRALY: --the indeterminate form, 0 over 0,
we can use L'Hopital's rule.
Yay.
VIRGE CORNELIUS: Mm-hmm.
And L'Hopital's rule says that I can now
take the derivative of the numerator
and the derivative of the denominator and try again.
Try direct substitution again.
So the derivative of my numerator-- derivative of 5--
is going to be 0.
I'll just write that down.
Plus-- I know I did this on an earlier problem in an earlier
session-- the derivative of this definite integral
is fundamental theorem so it's just going to be b of x.
Derivative in my denominator, derivative of x squared is 2x.
Plus derivative of cosine pi x is going to require chain rule.
It's also going to require that we
remember derivative of cosine.
So derivative of cosine is negative sine of pi x.
And then times pi.
Chaining out that derivative of the pi x.
When I substitute in 1, I get b of 1 divided
by 2 times 1 and then here--
well, I guess I'll just write it.
Minus pi sine of pi.
Sine of pi is what, Mark?
MARK KIRALY: 0 again.
VIRGE CORNELIUS: Mm-hmm.
So I have 2 minus 0 in the denominator, which is 2.
B of 1, I need to refer to my graph for that.
So b of 1 is going to be the y value and x is equal to 1.
So that is 5.
Mm-hmm.
Final answer, 5 halves.
MARK KIRALY: That is so neat.
VIRGE CORNELIUS: I'm sure you're thinking about this
but I want to make sure that we say it.
Always do direct substitution on limits
first because don't assume that L'Hopital's rule is
going to apply.
Sometimes it's just a direct substitution,
and you just get your answer.
MARK KIRALY: Or even if you get a 0 denominator
with a non-zero number in the numerator.
Now you can say you're only doing infinity
or negative infinity or--
VIRGE CORNELIUS: True.
Mm-hmm.
Yeah.
So definitely don't jump ahead on limits.
Just try that direct substitution,
analyze what it is you're dealing with first.
All right--
MARK KIRALY: --some takeaways.
VIRGE CORNELIUS: Ready for takeaways.
MARK KIRALY: We've had this takeaway before,
but it's worth repeating.
VIRGE CORNELIUS: Yeah, you have to read carefully.
Really hear what all these words are
saying so you form an image of what's going on.
And also, of course, write carefully.
Show your work.
Justify your answers.
You've learned so much since the beginning of this class.
MARK KIRALY: You are so much better at algebra
than you were last September or August
or if you started in January.
I know people start the course at different times of year.
But oh my goodness, you've learned so much.
VIRGE CORNELIUS: Mm-hmm.
And trig.
MARK KIRALY: Yeah.
So Virge and I did as much as we could to give you a good six
representations and these are the six representations we did.
If you haven't watched all the videos,
you're joining us in session 8, go back.
You can find a rate in, rate out, a graphical stem, a diff,
eq, particle machinery volume.
But as Virge wanted us to point out,
it's not like some of the other AP tests.
We can't guarantee it'll be these six topics.
VIRGE CORNELIUS: But these six topics do show up quite a bit.
MARK KIRALY: It will be on the exam.
Maybe just--
VIRGE CORNELIUS: Yes.
MARK KIRALY: --response.
VIRGE CORNELIUS: Yes.
And then we have made much longer videos in 2021 and 2022.
Full 45 minutes each.
So if you still feel like you need more review,
definitely start accessing those.
Those are available in AP Classroom
and your teacher doesn't even need to assign them.
MARK KIRALY: --more videos.
VIRGE CORNELIUS: Yeah.
Yeah.
Your teacher doesn't need to assign them.
You can just take the initiative and do it yourself
and we believe in you.
I believe in every single one of my students
that they can pass this exam.
I know they can but they have got
to practice because I can't be in there with them.
So they're going to have to practice.
MARK KIRALY: Good luck.
VIRGE CORNELIUS: Good luck.
MARK KIRALY: Thank you for joining us.
VIRGE CORNELIUS: Thank you.
Do a great job.
Bye, Mark.
MARK KIRALY: Bye, Virge.