AP Calculus Textbooks
Optimization and the Mean
Value Theorem
Part of the “Calc for Idiots” Series
Neil Patel and Tom Kappil
June 2012
Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
Calculus for Idiots
Optimization and the Mean Value
Theorem
Table of Contents
1. Rationale
2. Optimization
3. Mean Value Theorem
4. Optimization Exercises
5. Mean Value Theorem Exercises
3
4
9
13
16
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
Rationale
a. We chose the topics based on their interesting nature, their complex and varied
applications to real life topics, and for its ease as an introduction to more difficult
calculus, making it prime material to teach.
b. To ensure that our textbook section can teach students who are at an AP level, we
will provide AP level problems, but accompanied by a level of instruction that would
aid 5th graders. This marriage of difficult and simplistic will provide ample
instruction to encourage further studies into calculus, as well as provide a strong
foundation on the conceptual to students.
c. The resources used to provide this level of education will include the world wide
web, Larson’s Calculus textbook, and our brains, with small additions for sources
such as the Khan Academy. Computer software such as Microsoft Office will also be
used.
d. Our appreciation for the chosen topics derives from the real world applicability of
said topics, and the fact that some other branches of calculus tend to be more
arcane, and less applicable to a student’s life. In addition to that, the subjects are
interesting, and provide challenge without being too difficult to understand.
e. Our enthusiasm for the chosen topic will demonstrated through excellent
discussion, examples of a spectacular quality and glowing language used to laud the
wonderful aspects of optimization and the mean value theorem, and through our
hard work, students will leave this section of the textbook with a deep appreciation
of the material.
f. Our textbook will different in the method of explanations for the topics we choose.
We will remove the convoluted explanations and provide easy to understand,
straightforward directions regarding how to solve the problems and provide a
conceptual understanding of the material without confusing the student. In addition
to this, we will encourage the student by a combination of severe sarcasm and
praise meant to simultaneously provide support and rejection
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
Optimization
This is a really easy application and extension of basic differentiation, and the first derivative test.
To do optimization problems, you should follow a few basic steps. First, identify what quantities
you are looking for, and what is already provided. This will allow you to know what to optimize for.
Second, look for the primary equation that contains a variable for the quantity you are maximizing.
Next, solve the equation so it includes just one variable (what you are optimizing for). You may
need secondary equations to solve for this. Finally, solve using basic differentiation.
Example 1:
You have 50 feet of fencing, and need it to cover the maximum amount of area. What are the
dimensions of the plot?
Step 1: Lay out variables and identify:
Step 2: Identify Primary Equation: this
allows you to know what you’re primary end
goal is
Step 3: Secondary Equations: these are any
equations that will help you solve the
primary equation for only 1 variable.
Usually, this will involve equations that
relate some of the variables to another
Step 4: Solve the primary equation for a
single unknown, usually the object to be
maximized, by plugging in the secondary
equations. This will give you a new working
primary equation
Step 5: Differentiate. In this case, the
primary equation solved out for A can be
differentiated. One differentiates in order to
use the First Derivative test. Because the
new working primary equation shows area
(the quantity to be maximized, as a result of
changes in width, the first derivative test
will give the width for which A is maximized
Step 6: First Derivative Test: This test allows
you to take critical points (zeros) of the
derivative to find relative maxima and
minima on a curve. Doing the test here
allows us to find width at which area is
maximized
Normally for a first derivative test, at this
point, you would check the W values around
𝐿 = length
𝑊 = 𝑤𝑖𝑑𝑡ℎ
𝐴 = 𝑎𝑟𝑒𝑎
Basic Area: L x W = Area
Perimeter: P = 2L + 2W = 50 ft
LxW=A
P = 75 = 2L + 2W
50-2W=2L
25-W = L
A = (25-W) x W = 25W – W2
A = 25W – W2
𝑑𝐴
𝑑𝐴
(𝐴) = (25W – W2 )
𝑑𝐿
𝑑𝐿
𝐴′ = 25 − 2𝑊
𝐴′ = 0 = 25 − 2𝑊
25 = 2W
W = 12.5
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Calculus BC Textbook
Optimization and the Mean Value Theorem
the critical value to determine if the critical
value is a maximum or minimum, but in this
case, because there is only one, we can
assume that it is the area-maximizing
number
Step 6: Finish the Problem: the question asks
for length and width of the finished plot, so
plugging in the only value for width received
into the secondary equation gives you the
length
An AP level Multiple Choice Example
Neil Patel, Tom Kappil
P = 2L + 2W = 50 ft
50 = 2L + 2(12.5)
25=2L
L=12.5
Which point on the graph of y = x3 + 2x is closest to (0, 1)
a. (0, 0)
b. (.25, .5156)
c. (.5, 1.125)
d. (1, 3)
e. (1.5, 6.375)
Solution: First understand that in this case, one must look for a relative minimum, not
maximum. Second, the primary equation is the distance formula (𝐷 =
√(𝑥1 − 𝑥2 )2 + (𝑦1 − 𝑦2 )2 ). Third, the realization must be that the distance formula is centered
about (0, 1), so plug that into the equation for the second set of (X, Y) coordinates. Finally,
realize that Y1 can be solved by plugging the original equation into the coordinate. From here,
differentiate, perform the first derivative test, and solve.
𝑑𝐷
𝑑𝐷
(𝐷) =
(√(𝑋 − 0)2 + ((𝑥 3 + 2𝑋) − 1)2 )
𝑑𝑋
𝑑𝑋
1
2
𝐷 ′ = ( ) ((𝑋 − 0)2 + ((𝑥 3 + 2𝑋) − 1) )−1/2 × (2𝑥 + (2)((𝑥 3 + 2𝑋) − 1) × (3𝑥 2 + 2))
2
Solve for D’ = 0 and you find that X = .5
Again, because it is the only point, one must assume it is the minimum. You then plug back into
the original equation, and find the y value of x = .5 to be 1.125
So the answer is (C)
AP Level Conceptual
Why does Optimization need to use the first derivative test?
Optimization works because you take multiple equations, and substitute them together until
you have one dependent and one independent variable in a primary equation. The primary
equation will have a maximum value, but we cannot mathematically determine it without
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
calculus. By taking the derivative, you will find the rate at which the dependent variable
changes. The maximum or minimum of any equation will be when the rate of change changes
sign, or, in easier terms, a zero on the graph of the derivative. Using zeros allows us to find the
critical numbers of the primary equation. The critical numbers are possible locations of relative
extreme, but only by testing it, with the first derivative test, can one be certain that a critical
value is indeed a maximum or minima, because one can test the sign of the derivative about the
point. The first derivative test is used to ensure that one is finding the relative maxima or
minima of an equation by testing the signs of the derivative
AP Level Free Response Question:
A box is constructed of a piece of cardboard of sides each measuring 16 inches. Squares of width
X are cut from the corners of the cardboard such that it will be folded up to form a box, giving
each side a new length of (16-2x). What should the width of the cut-out squares be as to
maximize the volume of the box?
This is another fairly straightforward problem; the only complex part is understanding how it
all works together. Volume of any rectangular solid will always be length times width times
height. In this case, volume = (16-2x) (16-2x) (x). From here, the problem becomes simple, it
was just understanding how to set it up.
𝑉 =𝐿 ×𝑊×𝐻
𝑉 = (16 − 2𝑥)(16 − 2𝑥)(𝑋) = (256 − 64𝑥 + 4𝑥 2 )(𝑥) = 4𝑥 3 − 64𝑥 2 + 256𝑥
𝑑𝑉
𝑑𝑉
(𝑉) =
(4𝑥 3 − 64𝑥 2 + 256𝑥) = 12𝑥 2 − 128𝑥 + 256 = 𝑉 ′
𝑑𝑥
𝑑𝑥
𝑉 ′ = 0 = 12𝑥 2 − 128𝑥 + 256
Solve using a graphing calculator, and one finds that x = 2.666 and x = 8
Because you have 2 possible x values, you must think to use the first derivative test. Because
you are maximizing volume, you want where the derivative changes from positive to negative.
So test the intervals between 0 and 2.66, between 2.66 and 8, and then 8 to infinity. Only at
x=2.66 does the sign change from positive to negative, so it is the volume maximizing point
Finally, find the volume of the box, by plugging in x = 2.666 into the volume equation, to find
that the maximum volume is 303.407 square inches
Example Using a Graphing Calculator
Looking at the above problem, it required a fair bit of work by hand. Besides finding zeros using
the calculator (which one could do since Trigonometry), the calculator was not used at all.
However, on the AP multiple choice or free response calculator sections, it is far simpler to use a
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
graphing calculator to solve all the issues out. Once you have found the working primary
equation, all the work is far simpler on the calculator.
With the working primary equation already established, one should plug it directly into Y1.
From there, enter into Y2 the command to take the derivative of Y1. Tell the graphing calculator
to graph only Y2, and then find the zeros. The graphing calculator even takes care of the first
derivative test, because you can see visually where the derivative changes from positive to
negative, without having to do any sort of calculation.
Step 1: Enter Equations:
Step 2: Graph ONLY the derivitave:
Step 3: Find the zeros:
Step 4: 1st derivitve test. The derivite only changes from positive to negative at 2.666, so it must
be the maximum
Overall, using the calculator made the problem far simplier, and took less time than by hand. It
also removes the possiblity of human error while differentiating.
Real-World Applicability:
Optimization is one of the most applicable sub-branches of calculus to real life. While other
branches of calculus may skew to the unintelligible for people not using it daily, optimization is
fairly simple to understand and highly important to use. With it, one can determine how to
maximize efficiency or profit, or minimize loss or resources used if one can derive the need of
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
certain products into a simple equation. In fact, optimization is also an Algebra 2 topic, but
instead of taking the derivative, we were trusted to use the maximum and minimum features on
a calculator. Overall, optimization is applicable to real life because allows a person to find the
best method to do something. Optimization, unlike things like the integration of polar
equations, is accessible to everyday life because it takes simple equations and finds the
combination of inputs that will benefit production the most. In other words, it simply works, for
almost everything. If you need to discover how to maximize the area enclosed by a fence with
limited fencing, or to minimize length of a connecting cord, or the amount of paper wasted,
optimization is useful. It requires just a simple formula, and supporting information to find
maximum and minima. Anything where a resource is in limited supply optimization can be
used.
Contributing Mathematicians
Pierre de Fermat was highly influential in solving optimization problems, as he created the
Fermat’s Theorem, which became the First derivative test. Without this realization that any
local maxima/minima occurs when the derivative changes sign and equals 0, optimization
would not work as a subject of calculus. Fermat was a brilliant mathematician who contributed
greatly to the field and development of calculus. But without the efforts of Fermat to and his
theorem, optimization would not work, simply because the connection between the derivatives,
changing signs on the derivative, and maxima and minima would not be realized.
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
Mean Value Theorem for Idiots
The Mean Value Theorem is an extension of Rolle’s Theorem; it is important to understand
Rolle’s before we jump to Mean Value.
o
Rolle’s Theorem
 Requires a function that is:
 continuous on the closed interval [a,b]
 and differentiable on the open interval (a,b)
 Rolle’s Theorem states that if f(a) and f(b) are the same, then there is a
value of c within the interval (a, b) that has a derivative of 0
If you didn’t understand what you just read above, then clearly calculus isn’t the subject for you.
Being one of the simplest concepts in the course, Rolle’s Theorem is actually something that can be
understood without any real prior calculus knowledge…its pure logic.
A ROLLER COASTER THAT GOES UP AND COMES BACK DOWN MUST HAVE A HIGHEST POINT
(maximum where derivative equals 0)
The Mean Value Theorem: We now take the Rolle’s Theorem and use it more generally to make
assumptions about different functions. Mean Value still has the same requirements:
1) The function is continuous on the closed interval [a,b]
2) The function is differentiable on the open interval (a,b)

If these two conditions are met, then there exists a number c in the interval whose
derivative is equal to the slope of the line created by the points a and b (secant line).
𝒇′ (𝒄) =
𝒇(𝒃) − 𝒇(𝒂)
𝒃−𝒂
The Mean Value Theorem is useful in real world problem solving in terms of finding the
specific value of c that has the same slope as the secant line. An example of this would be to find the
instantaneous rate when give proper information. However, the Mean Value Theorem serves a
much larger purpose in helping to prove other theorems in calculus. Because it is so broad in
nature, the Mean Value Theorem can help prove a variety of statements that allow us to make the
generalizations we make in calculus; for example, the theorem is very closely related to the
Fundamental Theorem of Calculus.
Rolle’s Theorem was first published by the French mathematician Michel Rolle. This is
ironic because before finding the theorem, he was one of the most vocal critics of calculus and
believed it gave inaccurate results and was based on unsound reasoning. Rolle’s Theorem was later
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
used as a basis for the Mean Value Theorem. Mean Value was first proven be the famous
mathematician Joseph-Louis Lagrange. This was one of the many great milestones in his
mathematical career, who at the time was considered the “lofty pyramid of the mathematical
sciences”.
Examples
A. ANALYTICAL EXAMPLE----The graph of the equation 1/3𝑥 2 − 3 is continuous and
differentiable on the interval (0, 4). Find the value of c that has a derivative of the secant
line.
To solve this question, we first find the derivative of the equation using our previous
2
knowledge of calculus. This turns out to be (3 𝑥). We then find the slope of the secant line. f(0)=
-3 while f(4)=2.3333. Using the Mean Value theorem, we know the value of c has the same slope
∆𝑦
as the secant line so to find this slope we use the equation ∆𝑥 = 𝑠𝑙𝑜𝑝𝑒
2.3333− −3
=
4−0
1.33325
2
Finally, we set this slope equal to the derivative equation (3 𝑥) to find our c value.
2
( 𝑥) = 1.33325
3
X= 1.999998= about 2
B. AP MULTIPLE CHOICE----The equation – 𝑥 2 − 𝑥 + 6 is continuous and differentiable
between x=-2 and x=2. Which of the following matches the correct derivative and c value
that represents to the secant line?
a. -2x-1, 6
b. -2x-1, 0
c. -1/3x-1, 6
d. -1/3x-1, 0
e. -2x-6, 2
To solve this AP level multiple question, we need to find the slope of the tangent line. To do
this, we take the derivative of the function. The derivative of this equation is -2x-1. This
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
automatically removes answer choices C and D. Next, we find the y values for our two
bounds. f(-2)=4 and f(2)=0. Using this information, we find the slope of the secant using the
∆𝑦
4−0
basic algebra equation: ∆𝑥 = 𝑠𝑙𝑜𝑝𝑒. −2−2 = −1 Finally, to find c, we set the derivative of the
equation equal to this slope and solve for x.
-1=-2x-1 -------> x=c= 0
B is the Correct Answer
C. AP LEVEL FREE RESPONSE
1
Given the equation f(x)= (𝑥 2 − 2𝑥 − 3) 𝑥+2 is continuous and differentiable between x=-1
and x=8
a. Find the slope of the secant line created by the two bounds
b. Find the derivative of the f(x)
c. Describe how you would find the equation (slope intercept form) for the tangent
line at point c.
a.
∆𝑦
= 𝑠𝑙𝑜𝑝𝑒
∆𝑥
0 − 4.5 1
=
−1 − 8 2
b.
1
1
𝑓 ′ (𝑥) = (2𝑥 − 2) (𝑥+2) + (− (𝑥+2)2 )(𝑥 2 − 2𝑥 − 3) To come to this
answer, we would use the multiplication rule combined with the power
and chain rule.
1
2
c. First we would set the derivative equation found above to . That would
allow us to find the c value. We then would plug the c value into the
original equation to get both the x and y coordinates. Finally we plug in
both the coordinates and the slope at c, which would be (1/2), into point
slope form. Last, we convert the point slope equation to slope intercept
form
D. CONCEPTUAL EXAMPLE----If the function 𝑓(𝑥) = 5𝑥 − 4 is continuous and differentiable
within the interval [-4, 8], what is the c value between these bounds?
This is a simple, but trick question. Because the function given to us is that of a line, the slope at any
point on the line is exactly the same as the secant line. Therefore, there are an infinite number of c
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
values within this interval because every x is equal to c. In order to answer this simple question on
an example without wasting valuable time, it’s important to understand the conceptual aspect of
the Mean Value Theorem.
E. GRAPHICAL EXAMPLE----- Graph the function 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 and use the zeros find the
relative minimum of the graph. Use Rolle’s Theorem (Mean Value Theorem) to explaining
the reasoning for your answer and how it helps prove that a relative minimum exists.
To answer this question, first we would graph the equation on the calculator. We
calculate the zeros to exist at x=0 and x=3. Since we know that the graph is
continuous and differentiable within this interval, we can use Rolle’s Theorem to
prove the fact that the graph has a point where the derivative that equals to zero
within that interval. Also, because we see the graph is dipping down, we know that
at that point we know that a relative minimum exists. Using this reasoning as proof
for existence, we set the derivative equal to zero to find this point. To make it easier,
we can use the zero function of the graphing calculator to help us find our answer
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
Exercise Section:
Optimization
Analytical
1. You need to make a prism with a square base and an open top, and are limited to 100 square
feet of material. What should the dimensions of the box such that the volume is maximized?
2. What point on the curve f(x) = x2 is closest to (0,1)
3. Given the curve f(x) = 4-x2, find the point on the curve closest to the origin.
4. Given 50 feet of fencing, would you create an equilateral triangle, a square or a circle to
maximize area?
5. A farmer needs to enclose 500 square feet of land in rectangle, one side of which is backed up
directly to the barn, and thus requiring no fencing. What dimensions for the rectangle will
minimize the amount of fence used?
6. A rectangle is formed from the area under the curve of the equation y= (-) 2.5 x + 17 in the first
quadrant. What are the dimensions of the rectangle such that area of the rectangle is
maximized?
7. A right circular cylinder is designed to hold 45 cubic inches of liquid. What radius for the can
would minimize the amount of surface area on the bottle?
8. A right cylindrical package can have a maximum combined length and perimeter of a cross
section of 185 inches. Assuming circular cross sections, what dimensions of the package will
maximize volume?
9. A solid is formed by adding half a sphere on the top and bottom of a right cylinder and total
volume of 50 cubic inches. What is the radius of the cylinder so that total surface area is
minimized?
10. Find the point on the curve 𝑦 = 3𝑥 2 + ∛𝑥 closest to (0, 7)
11. You are tasked with creating a rectangular prism with 450 cubic feet of volume and a square
while minimizing surface area. What are the dimensions of the box?
12. Two posts, one 9 ft tall, and one 20 ft tall, that are 15 feet apart. The two posts are secured by
ropes tied down to the same stake in the ground. How far away from the first pole should the
tie be inserted to minimize material used in the tie-downs?
13. A window is made by adding a semicircle on top of a rectangular window. If you only have 45 ft
of material to enclose the window, what should be the dimensions of the rectangle and the
semicircle?
14. A rectangle is inscribed inside a circle of radius 7. What are the dimensions of the rectangle such
that its area is maximized?
15. A 5 inch length of wire will be cut in order to form a square and an equilateral triangle. Where
should the wire be cut to maximize areas of the two shapes?
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
Multiple Choice
1. You only have 160 square inches of material, and need to create a right cylinder. What is the
radius of the cylinder to maximize volume?
a. 5.164
b. 2.913
c. 5.046
d. -2.913
e. 7.136
2. What point on the curve 𝑦 = 3𝑥 1/2 is closest to the point (1, 8)?
a. (3.204, 5.370)
b. (4.229, 6.169)
c. (5.345, 6.942)
d. (-1, 7)
e. (4.537, 6.390)
3. You have 79 feet of fencing to enclose a rectangular area. What are the length and width of the
box?
a. 39.5 x 0
b. 6.28 x 33.21
c. 8.89 x 30.611
d. 19.75 x 19.75
e. 38.1 x 1
4. A rectangular page is to contain 30 square inches of print, with margins of 1 inch on each side.
Find the dimensions of the page so that the least amount of paper is used.
a. 9.745 x 5.87
b. 3.29 x 25.24
c. 7.48 x 7.48
d. 11.65 x 5.1
e. 3 x 32
5. A piece of cardboard of dimensions 15 inches by 15 inches has squares of length x cut from each
side, so that the sides can be folded into a box. What measurement of x maximizes volume?
a. 7.5
b. 5
c. 15
d. 2
e. 2.5
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
Free Response Question (calculator enabled)
A figure is made in the first quadrant by the line 𝑦 = −(𝑥 3 ) + 2𝑥 2 − 2𝑥 + 5
a. A rectangle is drawn underneath the graph in the first quadrant between the y axis, the x axis,
and the graph. What point on the graph will maximize the area underneath the curve?
b. Which point on the curve is closest to the point (1, 1)
c. The area under the curve in the first quadrant bounded by the x and y axis’s is the base of a
rectangular solid with square cross sections taken perpendicular to the x axis. Find the volume
of the solid
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
Mean Value Theorem
Analytical
Find the value of c for the following functions given that the functions are continuous and differentiable
within the given interval.
1)
2)
1 2
𝑥
2
1 3
𝑥
4
+ 3, [-2,4]
− 𝑥 [−5, 0]
3) √5𝑥 − 3 [0,7]
4) ln(𝑥)2 [1,9]
5)
𝑥 2 −1
[2,6]
𝑥
Find the equation of the Tangent line at point c given that the functions are continuous and
differentiable within the given interval.
2
1) 𝑓(𝑥) = 𝑥 3 [0, 2]
2) 𝑓(𝑥) = 𝑥 2 [-4,6]
3) 𝑓(𝑥) = 𝑥(𝑥 2 +3x-4) [4,8]
4) √3 − 𝑥 [-8,-2]
5) 𝑓(𝑥) = sin 𝑥 [0,
3𝜋
]
2
Multiple Choice
For the Function 3𝑥 2 − 2𝑥 + 2, what is the value of c that meets the conditions of MVT within the
interval [-2, 3]?
a. 2
b. 1
c.
1
2
1
d. − 2
e. 4
Which of the functions would the Mean Value Theorem not apply between the interval [-2, 2]?
a.
b.
c.
d.
e.
𝑥 2 −4
𝑥−2
2
𝑥 −9
5𝑥 + 2
9𝑥
𝑥 2 − 2𝑥 + 3
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Calculus BC Textbook
Optimization and the Mean Value Theorem
Neil Patel, Tom Kappil
How many c values exist with the interval [0, 9] for the function 9x-3?
a.
b.
c.
d.
e.
1
2
4
Infinite
0
What is the y intercept of the tangent line at c, for the equation 4𝑥 2 + 3𝑥 − 2 for the interval [-3,1]?
a.
b.
c.
d.
e.
-6
6
-5
8
-1
If both bounds of an interval have the same y value, which of the follow is/are true about the function
with that interval, given that the function is continuous and differentiable?
a.
b.
c.
d.
e.
I.
There is a point within the interval where the tangent line has a slope of 0
II.
There must be a relative maximum between the interval
III.
Rolle’s theorem cannot be applied
I only
II only
III only
I and II only
I and III only
AP FREE RESPONSE
For the f(x) = – 𝑥 3 + 4𝑥 2 − 3𝑥 + 4 within the interval [0,4] answer the following questions
a. State the conditions of the mean value theorem and explain whether the theorem can be
applied to the given function and interval
b. Find the derivative of the function
c. Find the slope of the secant line
d. Find the equation of the tangent line at point c
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