Main content

## Calculus, all content (2017 edition)

### Unit 8: Lesson 1

AP Calculus AB questions- Sal interviews the AP Calculus Lead at College Board
- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c
- 2011 Calculus AB free response #1a
- 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

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 2011 Calculus AB free response #4b

Absolute maximum over an interval. Critical points and differentiability. Created by Sal Khan.

## Want to join the conversation?

- 06:26why is the derivative f(x) and not f(t)(6 votes)
- The Fundamental Theorem of Calculus states that if you are trying to find the derivative of g(x)= definite integral from a to x of f(t)dt, then you simply replace the x for every t in the inner function. Therefore, g'(x) = f(x).(1 vote)

- Is it possible to have a function that produce a circle and straight line?!(1 vote)
- Not a circle, because it would fail the verticle line test, but anything less than or equal to a semicircle is allowed, so are straight lines. I can't think of one single way to define this function, but it could be defined piecewise.(9 votes)

- was this a calculator allowed questions(2 votes)
- No, this questions does not allow calculators (someone confirm that I am right).(4 votes)

- Discuss the role of constant of integration in indefinite integrals, why we don't involve it in definite integrals? (Be precise)(2 votes)
- Maybe I can try it in a more
*general*example. Let**f(x)=d/dx F(x**

The indefinite integral ∫ f(x) dx = F(x) + c

Now, if I want to find the definite integral (area under the curve) ∫ f(x) dx from a to b, the fundamental theorem tells us that this is the same thing as the indefinite integral evaluated at the endpoints. So we have ( F(b) + c ) - ( F(a) + c ). The constant of integration is there in both minuend and the subtrahend. Now, what happens if we distribute the subtraction?

F(b) + c - F(a) - c. We can simplify this! The two constants are the same, so they subtract to 0. Thus, our definite integral simplifies to F(b) - F(a). Notice:*no constant*. Because of the canceling out, we tend to ignore the constant of integration when taking definite integrals because it will*always*simply subtract out.(3 votes)

- the problem asks for the absolute maximum and the |-8 - 2pi| is clearly the largest so should the value be x = -4 and not x = 5/2(2 votes)
- No. Absolute maximum means the highest point on the graph of g(x), not the point where the absolute value of g(x) is the highest.(3 votes)

- I think Sal's answer is right, but I'm not sure about the evaluation at x = -4... I think 2 pi should have been added, not subtracted.

The formula for g(x) is 2x + area under f(x) from 0 to -4. There is clearly more positive area than negative area, so it should be 2x + some positive number (2 pi).

I'm thinking that since Sal didn't directly calculate the area, he forgot to check signs? Am I wrong?(1 vote)- It is subtracted because they reversed the normal way of expressing the interval which reversed the sign. (Integral from 0 to -4 instead of Integral from - 4 to 0) He explains this in the previous video.(3 votes)

- Why is the answer x=5/2, when the max on the graph is x=0?(1 vote)
- at x=0, f(x) has a maximum, but we are looking for the maximum of g(x)(3 votes)

- what should i do to solve a problem for this video that i dont know with out watching the video agian or getting the problem wrong does any one have any tips for me(1 vote)
- Post the problem and your solution and we can tell you whether you're correct.(2 votes)

- Hey Sal, I'm curious to know why you didn't g(3/2), where the straight line on the right side of this graph has a zero, would that be the absolute maximum? It seems to me that g(x) would be greater at that point than at g(5/2) because g(5/2) has negative area from integration lowering the number.(1 vote)
- At g(3/2) = 2* (3/2) + integral(f(t),t,0,3/2) = 6/2 + 9/4 = 21/4 or 5.25

at12:44Sal shows that g(5/2) = 6, which is larger.

It makes since if you remember that a function will increase as long as its derivative is positive and the function decreases when the derivative is negative. To find the points where a function changes from increasing to decreasing or decreasing to increasing, can generally be found by setting the first derivative to 0. For this function that only happened at x = 5/2(1 vote)

- How do you know that it is a maximum? Couldn't it be a minimum?(1 vote)
- As you know when solving f'(x) = 0 you obtain the horizontal slopes of the functions (which could be a maximun OR a minimun), another property that is not very known is that if you solve the SECOND derivative f''(x) you can find the concavity of the function, when positive the function is concave upward and thus for f'(x) = 0 for f''(x)>0 we have a minimun, same for the opposite => f'(x) = 0 AND f''(x) < 0 thus we have a maximun.

From: http://www.math.hmc.edu/calculus/tutorials/secondderiv/(1 vote)

## Video transcript

Part b. "Determine the x-coordinate
of the point at which g has an absolute maximum
on the interval negative 4 is less than or equal to x,
is less than or equal to 3. And justify your answer." So let's just think about
it in general terms. If we just think about
a general function over an interval, where it
could have an absolute maximum. So let me draw some
axes over here. And I'm speaking in
general terms first, and then we can go back
to our function g, which is derived from this
function f right over here. So let's say that these
are my coordinate axes, and let's say we care
about some interval here. So let's say this is the
interval that I care about. A function could look
something like this. And in this case,
its absolute maximum is going to occur at the
beginning of the interval. Or a function could look
something like this. And then the absolute
maximum could occur at the end
point of the interval. Or the other possibility
is that the function looks something like this. At which point,
the maximum would be at this critical point. And I say critical point, as
opposed to just a point where the slope is zero, because
it's possible to the functions not differentiable there. You could imagine a function
that looks like this, and maybe wouldn't be
differentiable there. But that point there still
would be the absolute maximum. So what we really
just have to do is evaluate g at the different
endpoints of this interval, to see how high it gets,
or how large of a value we get for the g
at the end points. And then we have to see if g has
any critical points in between. And then evaluate
it there to see if that's a candidate
for the global maximum. So let's just evaluate g
of the different points. So let's start off, let's
evaluate g at negative 4-- at kind of the lowest
end, or the starting point of our interval. So g of negative 4 is
equal to 2 times negative 4 plus the integral from 0
to negative 4 f of t dt. The first part is very easy, 2
times negative 4 is negative 8. Let me do it over here so
I have some real estate. So this is equal to negative 8. And instead of leaving this
as 0 to negative 4 f of t dt, let's change the bounds
of integration here. Especially so that we can get
the lower number as the lower bound. And that way, it
becomes a little bit more natural to think
of it in terms of areas. So this expression
right here can be rewritten as the
negative of the integral between negative
4 and 0 f of t dt. And now this expression
right over here is the area under f of
t, or in this case, f of x-- or the area under f
between negative 4 and 0. So it's this area
right over here. And we have to be careful,
because this part over here is below the x-axis. So this we would
consider negative area when we think of it
in integration terms. And this would be positive area. So the total area here is
going to be this positive area minus this area right over here. So let's think
about what this is. So this area-- This section
over here we did this in part a, actually, This section--
this is a quarter circle, so it's 1/4-- so these
are both quarter circles. So we could multiply 1/4 times
the area of this entire circle, if we were to draw the entire
thing all the way around. It has a radius of 3. So the area of the entire circle
would be pi times 3 squared or it would be 9 pi. And of course we're
going to divide it by 4-- multiply it by 1/4 to
just get this quarter circle right over there. And then this area
right over here, the area of the entire
circle, we have a radius of 1. So it's going to be
pi times 1 squared. So it's going to be
pi, and then we're going to divide it
by four, because it's only one fourth of that circle. And we're going
to subtract that. So we have negative pi
and we were multiplying it times 1/4 out here,
because it's just a quarter circle in either case. And we're subtracting
it, because the area is below the x-axis. And so this simplifies to--
this is equal to 1/4 times 8 pi, which is the
same thing as 2 pi. Did I do that right? 1/4 times 8 pi. So this all simplifies to 2 pi. So g of negative 4 is equal
to negative 8 minus 2 pi. So clearly it is a
negative number here. More negative than negative 8. So let's try the other bounds. So let's see what
g of positive 3 is. I'll do it over here so
I have some more space. So g of positive 3,
when x is equal to 3, that-- we go back to our
definition-- that is 2 times 3 plus the integral
from 0 to 3 f of t dt. And this is going to be
equal to 2 times 3 is 6. And the integral from 0
to 3 f of t dt, that's this entire area. So we have positive
area over here. And then we have an equal
negative area right over here because it's below the x-axis. So the integral from 0 to
3 is just going to be 0, you're going to have
this positive area, and then this negative
area right over here is going to completely
cancel it out. Because it's symmetric
right over here. So this thing is going
to evaluate to 0. So g of 3 is 6. So we already know that
our starting point, g of negative 4-- that when
x is equal to negative 4-- that is not where g
hits a global maximum. Because that's a
negative number. And we already
found the end point, where g hits a positive value. So negative 4 is
definitely not a candidate. x is equal to 3 is still in the
running for the x-coordinate where g has a global
absolute maximum. Now what we have
to do is figure out any critical points
that g has in between. So points in there where
it's either undifferentiable or its derivative is equal to 0. So let's look at
this derivative. So g prime of x-- we just take
the derivative of this business up here. Derivative of 2x is 2. Derivative of this definition
going from 0 to x of f of t dt-- we did that in part a, this
is just the fundamental theorem of calculus-- this
is just going to be plus f of x right over there. So it actually
turns out that g is differentiable over
the entire interval. You give any x value
over this interval, we have a value for f of x.
f of x isn't differentiable everywhere, but
definitely f of x is defined everywhere,
over the interval. So you'll get a number here,
and obviously two is just two, and you add two to it, and
you get the derivative of g at that point in the interval. So g is actually differentiable
throughout the interval. So the only critical
points would be where this derivative
is equal to 0. So let's set this
thing equal to 0. So we want to solve the
equation-- I'll just rewrite it actually-- So we want to solve
the equation g prime of x is equal to 0, or 2 plus
f of x is equal to 0. You can subtract
2 from both sides, and you get f of x is
equal to negative 2. So any x that satisfies f
of x is equal to negative 2 is a point where the
derivative of g is equal to 0. And let's see if f of x is equal
to negative 2 at any point. So let me draw a line
over here at negative 2. We have to look at it visually,
because there's only given us this visual
definition of f of x. Doesn't equal negative 2,
doesn't equal to negative 2, only equals negative
2 right over there. And it looks like we're at about
2 and 1/2, but let's get exact. Let's actually figure out
the slope of the line, and figure out what
x value actually gives f of x equal
to negative 2. And we could figure out the
slope of this line fairly visually-- or figure out the
equation of this line fairly visually, we can
figure out its slope. If we run 3-- if our
change in x is 3, then our change in y,
our rise, is negative 6. Change in y is
equal to negative 6. Slope is rise over run, or
change in y over change in x. So negative 6 divided
by 3 is negative 2. It has a slope of negative 2. And actually, I could
have done that easier. Where if we go forward
one, we go down by 2. So we could have seen that
the slope is negative 2. So this part of
f of x, we have y is equal to negative
2x plus-- and then the y-intercept is
pretty straightforward. This is at 3-- 1, 2, 3. Negative 2x plus 3. So part of f of x
where clearly it equals negative 2 at some point
of that-- this part of f of x is defined by this line. Obviously this part of f of x
is not defined by that line. But to figure out
the exact value, we just have to figure
out when this line is equal to negative 2. So we have negative 2x plus
3 is equal to negative 2. And remember, this isn't--
this is what f of x is equal to, over the interval
that we care about. If we were talking
about f of x over there, we wouldn't be able to
put a negative 2x plus 3, we would have to have some form
of equation for these circles. But right over here, this is
what f of x is, and now we can solve this pretty
straightforwardly. So we can subtract
3 from both sides, and we get negative 2x
is equal to negative 5. Divide both sides
by negative 2, you get x is equal to negative
5 over negative 2, which is equal to 5/2. Which is exactly
what we thought it was when we looked
at it visually. It looked like we were
at about 2 and 1/2, which is the same thing as 5/2. Now we don't know what this is. We don't know if this
is an inflection point. Is this a maximum? Is this a minimum? So really we just
want to evaluate g at this point to see if it gets
higher than when we evaluate g at 3. So let's evaluate g at 5/2. So g at 5/2 is going
to be equal to 2 times 15/2 plus the integral
from 0 to 5/2 of f of t dt. So this first part right over
here, the 2's cancel out. So this is going
to be equal to 5. And then plus the
integral from 0 to 5/2. Now, you might be able
to do it visually, but we know what the value is
of f of t over this interval, we already figured out
the equation for it over this interval. So it's the integral of
negative 2x plus 3 dt. And then let's
just evaluate this. Let me get some real estate. So this is-- let me
draw a line here., so we don't get confused. So this is going to
be equal to 5 plus and then I take
the antiderivative. The antiderivative of negative
2x is negative x squared. So we have negative x squared. And the antiderivative of
3 is just going to be 3x. So plus 3x. And we're going to
evaluate it from 0 to 5/2. So this is going to
be equal to 5 plus-- and I'll do all this
stuff right over here. I'll do this stuff in green. So when we evaluate
it at 5/2, this is going to be
negative 5/2 squared. So it's going to be
negative 25 over 4 plus 3 times 5/2,
which is 15 over 2. And then from that,
we're going to subtract this evaluated at 0. But negative 0 squared
plus 3 times 0 is just 0. So this is what
it simplifies to. And so what do we
have right over here? So let's get our ourselves
a common denominator. Looks like a common denominator
right over here could be 4 So this is equal to--
5 is the same thing as 20 over 4 minus 25 over
4 and then plus 30 over 4. So 20 plus 30 is
50 minus 25 is 25. So this is equal to 25 over 4. And 25 over 4 is the
same thing as 6 and 1/4. So when we evaluate our
function at this critical point, at this thing where the
slope, or the derivative, is equal to 0, we
got 6 and 1/4, which is higher than six, which is
what g was at this end point. And it's definitely higher
than what g was a negative 4. So the x-coordinate
of the point at which g has an absolute maximum,
on the interval negative 4 to three is x is equal to 5/2.