Day 1
X Discuss Course Organization
Today: Section 2.1 Limits: Graphical Approach
The Definition of limit
 symbol: lim f  x   L .
x c




spoken: “The limit, as x approaches c, of f(x) is L.”
less-abbreviated symbol: f(x)  L as x  c
spoken: “f(x) approaches L as x approaches c.”
usage: x is a variable, f is a function, c is a real number constant, and L is a real number
constant.
 meaning: as x gets closer & closer to c, but not equal to c, the value of f(x) gets closer &
closer to L (may actually equal L).
 Leave blank line to be filled in later
today we will do graphical approach
Examples of graph  description of limit behavior
Class Drill 1 Limits
project the class drill on one screen, then project the book on the other screen and show that the
class drill is similar to suggested problems 3-1#7,8,13,15
do row x = 1.
Fill in the blank line in the definition of limit:
Graphical Interpretation: The graph of 𝑓 appears to be heading for location (𝑥, 𝑦) = (𝑐, 𝐿).
Do row x = 4
Point out the difference between the idea of the existence of a y-value at x=a and the existence of
the limit as x  a.
do row x = -1.
Define one-sided limits
Re-cast the definition of limit using 3-part test involving one-sided limits
finish row x = -1
do row x=-3
Now: Examples of description of limit behavior  graph
***Important*** Exercise 3-1#40 Sketch a graph that satisfies all these conditions:
𝑓(1) = 3
lim− 𝑓(𝑥) = 2
𝑥→1
lim 𝑓(𝑥) = −2
𝑥→1+
.
.
Day 2
X Continuing Section 2.1 Limits Analytical Approach
Today: Analytical Approach to Limits
So far, we have done graphical approach to limits. Now do Analytical Approach.
Project Reference 2: Facts about Limits from Section 3-1
EX1: Let 𝑓(𝑥) = −7𝑥 2 + 13𝑥 − 25. Find lim 𝑓(𝑥). Discuss common mistakes
𝑥→−2
EX2: Find lim √24 + 𝑥 2
𝑥→5
Examples involving cancelling factors inside the limit (“algebraic simplification”)
Ex3: IMPORTANT Examples using rational function 𝑓(𝑥) =
𝑥 2 −2𝑥−3
𝑥−3
=
(𝑥−3)(𝑥+1)
(𝑥−3)
.
(A) find f(c) for c=0,1,2
(B) Find 𝑓(𝑐) for 𝑐 = 3. Note that we cannot cancel 0/0. So f(2) DNE
(C) Find lim 𝑓(𝑥) for 𝑐 = 0,1,2,3. Note most important step: we can cancel (x-3)/(x-3)
𝑥→𝑐
Discuss why this example makes sense. Let 𝑔(𝑥) = 𝑥 + 1. Make side-by-side graphs of 𝑓(𝑥)
and 𝑔(𝑥) on separate axes. Using the graphs, explain why the answers in question (A),(B) make
sense. (draw lines like those in the drawing on page 130, Example 2, figure 2) The functions 𝑓(𝑥)
and 𝑔(𝑥) are not the same function, but they do have the same limit.
Discuss hole in graph at 𝑥 = 3.
Redo the computations: 𝑓(3) 𝐷𝑁𝐸, but lim 𝑓(𝑥) = 4.
common mistake: lim 𝑓(𝑥) = lim
𝑥→3
another: lim 𝑓(𝑥) = lim
𝑥→3
𝑥→3
𝑥→3
(𝑥−3)(𝑥+1)
(𝑥−3)
𝑥→3
(𝑥−3)(𝑥+1)
=
(𝑥−3)
(3−3)(3+1)
=
(3−3)
(3−3)(3+1)
(3−3)
0
= 0 𝐷𝑁𝐸 (one mistake on this line.)
= (3 + 1) = 4 (two mistakes on this line!)
𝑥+13
Ex4: IMPORTANT Examples using rational function 𝑓(𝑥) = 2
.
𝑥 +13𝑥
(A) Find 𝑓(𝑐) for 𝑐 = 13, −13,0
(B) Find lim 𝑓(𝑥) for 𝑐 = 13, −13,0.
𝑥→𝑐
Discuss correct notation.
*** Important *** Notice that Theorem 4 on page 103 tells us that limit lim 𝑓(𝑥) 𝐷𝑁𝐸
𝑥→0
Ex5: Difference Quotient examples from Section 2.1.
Similar to #67: For 𝑓(𝑥) = 𝑥 2 − 6𝑥 + 5 find lim
ℎ→0
𝑓(4+ℎ)−𝑓(4)
ℎ
= 2. “difference quotient”.
Examples involving piecewise-defined functions.
−2𝑥 + 10, 𝑥 ≤ 3
Ex6: similar to #45 Let 𝑓(𝑥) = {
.
𝑥2, 𝑥 > 3
(A) Find lim 𝑓(𝑥). (B) Find 𝑓(3). (C) Explain with a graph.
𝑥→3
𝑥−11
Ex7: Example similar to #49 Let 𝑓(𝑥) = |𝑥−11|. (A) Find 𝑓(11).
(B) Find lim− 𝑓(𝑥). (C) Find lim+ 𝑓(𝑥). (D) Find lim 𝑓(𝑥). (E) Explain with a graph.
𝑥→11
.
𝑥→11
𝑥→11
Day 3
X Section 2.2 Limits Involving Infinity
Today: Infinite Limits and vertical asymptotes
1
EX1: Consider the function 𝑓(𝑥) = (𝑥−3)2 .
the y-value f(3) DNE.
According to Section 2.1 Thm 4, the lim 𝑓(𝑥) 𝐷𝑁𝐸 as well.
𝑥→3
But consider what happens for x-values near x=3. Make table of values. Draw graph. Observe as x
gets closer & closer to 3 but not equal to 3, y-values get more and more positive without bound.
Define lim 𝑓(𝑥) = ∞ to mean this. Introduce terminology of vertical asymptote.
𝑥→3
Important point: we have revised the definition of limit! In Section 2.2, the book says the limit
does not exist, but we write the symbol lim 𝑓(𝑥) = ∞. That’s nonsense. We have changed the
𝑥→3
definition of limit. With the old definition of limit (from section 2.1), the limit does not exist.
With the new definition of limit (from section 2.2), the limit does exist and it is infinity.
𝑥 2 −6𝑥+5
(𝑥−1)(𝑥−5)
Class Drill 3: Guessing Limits by Substituting in Numbers for 𝑓(𝑥) = 𝑥 2 −8𝑥+15 = (𝑥−3)(𝑥−5)
Remarks about Class Drill 3
Book often analyzes this kind of function with a calculator, but we can do it by hand.
Consider behavior at 𝑥 = 1.
The y-value 𝑓(1) = 0, point on graph graph at (𝑥, 𝑦) = (1,0). This is an x-intercept.
Consider behavior at 𝑥 = 5.
The y-value 𝑓(5) DNE but lim 𝑓(𝑥) = 2. So hole in graph at (𝑥, 𝑦) = (5,2).
𝑥→7
Consider behavior at 𝑥 = 3.
The y-value 𝑓(3) DNE. According to Section 2.1 Thm 4, the lim 𝑓(𝑥) 𝐷𝑁𝐸 as well.
𝑥→3
Left limit lim− 𝑓(𝑥) = −∞. (using our new definition of infinite limit from Section 2.2)
𝑥→3
Right limit lim+ 𝑓(𝑥) = ∞. (using our new definition of infinite limit from Section 2.2)
𝑥→3
The two-sided lim 𝑓(𝑥) 𝐷𝑁𝐸. (using either old or new definition of limit)
𝑥→3
So graph has vertical asymptote at 𝑥 = 3. Graph goes down on left side of asymptote, and up on
right side of asymptote.
Observations about Factors
Factored form of function gives us the x-values where important features of the graph occur.
Factored form is also the best form for determining the precise behavior at those x-values.
Factors of form (𝑥 − 𝑐) in numerator alone cause an x-intercept in the graph at (𝑥, 𝑦) = (𝑐, 0).
(𝑥−𝑐)
Factors of form (𝑥−𝑐) in numerator and denominator with equal powers cause a hole in the graph
at 𝑥 = 𝑐.The y-coordinate of the hole is the real number lim 𝑓(𝑥)
1
𝑥→𝑐
Factors of form (𝑥−𝑐) in denominator alone cause a vertical asymptote in the graph at 𝑥 = 𝑐. The
behavior of the graph near the asymptote (which side goes up, which side goes down) can be
determined by finding the sign of the function just to the left & just to the right of 𝑥 = 𝑐.
.
Day 4
X Continuing Section 2.2 Limits Involving Infinity
Today: Limits at Infinity and Horizontal asymptotes
We will study three functions:
7𝑥 2 − 42𝑥 + 35
7(𝑥 2 − 6𝑥 + 5)
7(𝑥 − 1)(𝑥 − 5)
𝑓(𝑥) = 2
=
=
2𝑥 − 16𝑥 + 30 2(𝑥 2 − 8𝑥 + 15) 2(𝑥 − 3)(𝑥 − 5)
7𝑥 2 − 42𝑥 + 35
7(𝑥 2 − 6𝑥 + 5)
7(𝑥 − 1)(𝑥 − 5)
𝑔(𝑥) = 3
=
=
2𝑥 − 16𝑥 2 + 30𝑥 2𝑥(𝑥 2 − 8𝑥 + 15) 2𝑥(𝑥 − 3)(𝑥 − 5)
7𝑥 3 − 42𝑥 2 + 35𝑥 7𝑥(𝑥 2 − 6𝑥 + 5) 7𝑥(𝑥 − 1)(𝑥 − 5)
ℎ(𝑥) =
=
=
2𝑥 2 − 16𝑥 + 30
2(𝑥 2 − 8𝑥 + 15)
2(𝑥 − 3)(𝑥 − 5)
Consider the “end behavior” of each function. That is, what happens as 𝑥 → ∞?
EX1: Let’s take lim 𝑓(𝑥) and see what happens.
𝑥→∞
7
Using technique of identifying dominant terms, we find lim 𝑓(𝑥) = 2.
𝑥→∞
7
That is, as x values get more and more positive without bound, the y-values get closer to 𝑦 = 2.
7
Get computer graph. Note that graph of 𝑓 has a horizontal asymptote on the right at 𝑦 = 2.
EX2; Now find lim 𝑔(𝑥).
𝑥→∞
Using technique of identifying dominant terms, we find lim 𝑔(𝑥) = 0.
𝑥→∞
That is, as x values get more and more positive without bound, the y-values get closer and closer
to 𝑦 = 0.
Get computer graph. Note that graph of 𝑔 has a horizontal asymptote on the right at 𝑦 = 0.
EX3: Now find lim ℎ(𝑥).
𝑥→∞
Using technique of identifying dominant terms, we find that as x values get more and more
positive without bound, the y-values get larger and larger positive, without bound.
That is, lim ℎ(𝑥) = ∞.
𝑥→∞
Get computer graph. Note that graph of ℎ goes up on right.
Conclusion we have observed that
 The standard form of rational function is most useful for determining behavior as 𝑥 → ∞.
 If lim 𝑓(𝑥) = 𝑏, then the graph of 𝑓 has a horizontal asymptote on the right at 𝑦 = 𝑏.
𝑥→∞

That is, the line equation of the asymptote is 𝑦 = 𝑏.
If lim 𝑓(𝑥) = ∞, then the graph of 𝑓 goes up on the right. There is no horizontal
𝑥→∞
asymptote on the right.
Obvious question: are these the only possibilities?
Could have lim 𝑓(𝑥) = −∞, then the graph of 𝑓 goes down on the right.
𝑥→∞
Could have lim 𝑓(𝑥) = 𝐷𝑁𝐸 then the graph of 𝑓 does not have simple behavior on the right.
𝑥→∞
Also notice: In our graphs of 𝑓 and 𝑔, there was also a horizontal asymptote on the left with the
same line equation. This is a fact about Rational Functions. If 𝑓 is a rational function and if 𝑓 has
a horizontal asymptote on the right, then the graph will also have a horizontal asymptote on the
left at the same height.
This is not true of general functions that are not rational functions. Example Exponential.
Also if a graph of a rational function goes up on one end, it can go up or down on the other end.
Example y=x^2 or y=x^3. But the graph cannot have a horizontal asymptote on other end.
Quiz 1
.
Day 5
X Continuing Section 2.2 Limits Involving Infinity, continued
Today: More examples of limits involving infinity for Rational Functions
EX1 (horiz asymp, vert asymptote and hole) Sim to 2.2#63 Find all horiz and vert asympt:
(3𝑥 2 − 3𝑥 − 36) 3(𝑥 2 − 𝑥 − 12) 3(𝑥 + 3)(𝑥 − 4)
𝑓(𝑥) =
=
=
(2𝑥 2 − 6𝑥 − 8)
2(𝑥 2 − 3𝑥 − 4) 2(𝑥 + 1)(𝑥 − 4)
Do more than just that.
Explain effect of each factor on the graph behavior. Use lim terminology where appropriate.
Explain end behavior using limit terminology..
EX2 (horiz asympt, no vert asympt or hole) Sim to 2.2#63 Find all horiz and vert asympt:
(3𝑥 2 − 3𝑥 − 36) 3(𝑥 2 − 𝑥 − 12) 3(𝑥 + 3)(𝑥 − 4)
𝑓(𝑥) =
=
=
(2𝑥 2 − 6𝑥 + 8)
2(𝑥 2 − 3𝑥 + 4) 2(𝑥 2 − 3𝑥 + 4)
Do more than just that.
Explain effect of each factor on the graph behavior. Use lim terminology where appropriate.
Explain end behavior using limit terminology..
Summary of analytic approach to limits involving infinity
The standard form of a rational function f(x) is useful for finding lim 𝑓(𝑥) to determine
𝑥→∞
horizontal asymptotes and end behavior.
The factored form of f(x) is useful for finding x-intercepts, vertical asymptotes, holes.
Conceptual Questions
2.2#77 A rational function has at least one vertical asymptote
2.2#78 A rational function has at most one vertical asymptote
2.2#79 A rational function has at least one horizontal asymptote
2.2#80 A rational function has at most one horiz asymptote
2.2#81 A polynomial function of degree ≥ 1 has neither horiz nor vert asympt
2.2#82 The graph of a rational function cannot cross a horiz asymptote.
Similar: the graph of a rational function cannot cross a vertical asymptote
Applications problems Drug problems 2.2#89,90
2.2#89 Drug administered to a patient through an injection. The drug concentration in the
bloodstream is described by the function
5𝑡 2 (𝑡 + 50)
𝐶(𝑡) 3
𝑡 + 100
where 𝑡 is the time in hours after the injection and 𝐶(𝑡) is the drug concentration in the
bloodstream (in milligrams/milliliter) at time 𝑡. Find and interpret lim 𝐶(𝑡).
𝑡→∞
2.2#89 Drug administered to a patient through an IV drip. The drug concentration in the
bloodstream is described by the function
5𝑡(𝑡 + 50)
𝐶(𝑡) 3
𝑡 + 100
where 𝑡 is the time in hours after the drip was started and 𝐶(𝑡) is the drug concentration in the
bloodstream (in milligrams/milliliter) at time 𝑡. Find and interpret lim 𝐶(𝑡).
𝑡→∞
Question: do these two problems make sense?
Day 6
X Section 3.2: Continuity
Return to class drill about limits. Observe that to draw graph at holes & jumps, we have to lift our
pen. The concept of continuity describes this distinction in abstract mathematical terms.
Definition of Continuity (formal and informal)
do Class Drill 4 Limits and Continuity
Sign behavior of functions and using sign behavior to solve inequalities
Define continuous on an interval
Observations
Point out that polynomial functions are always continuous. So for polynomial functions, the
equation lim 𝑓(𝑥) = 𝑓(𝑎) is always true. This is where Limit Theorem 3 came from!
𝑥→𝑎
Point out that rational functions are continuous everywhere except at x-values that cause the
denominator to be zero. (These x-values can be the locations of holes or of vertical asymptotes.
Both are types of discontinuities.) So for rational functions, the equation lim 𝑓(𝑥) = 𝑓(𝑎) is
𝑥→𝑎
always true as long as x=a is in the domain. This is where the other part of Limit Theorem 3 came
from!
Point out that a function can only change sign by touching the x-axis and crossing it (an xintercept) or by jumping across the x-axis (an x-value where the function is discontinuous.)
Define partition numbers to be x-values where the y-value is zero or where the function is
discontinuous.
This gives us an insight into sign behavior: the sign behavior is unchanged on each interval
between partition numbers
Examples
Example: for 𝑓(𝑥) = 9𝑥 2 − 90𝑥 + 189 = 9(𝑥 − 3)(𝑥 − 7)
(a) Determine sign behavior of f. Point out difficulty of using sample numbers.
(b) solve 𝑓(𝑥) ≥ 0. present answer 3 ways.
Example: Solve 9𝑥 2 − 90𝑥 ≥ −189. Present answer three ways.
Example: Solve 𝑥 3 ≤ 49𝑥. Present answer three ways. Incorrect and correct solution.
Day 7
X continuing Section 2.3: Studying the sign behavior of functions
𝑥 2 −5𝑥
Example: 𝑓(𝑥) = 𝑥−4 . (a) Determine sign behavior of f. Point out difficulty of using sample
numbers. (b) solve 𝑓(𝑥) ≤ 0. present answer 3 ways.
Example:
(3𝑥 2 − 3𝑥 − 36) 3(𝑥 2 − 𝑥 − 12) 3(𝑥 + 3)(𝑥 − 4)
𝑓(𝑥) =
=
=
(2𝑥 2 − 6𝑥 − 8)
2(𝑥 2 − 3𝑥 − 4) 2(𝑥 + 1)(𝑥 − 4)
Solve inequality 𝑓(𝑥) ≥ 0. Present answer 3 ways.
EX1 Return to Mon Aug 31 Example 1 Sim to 2.2#63 Find all horiz and vert asympt:
(3𝑥 2 − 3𝑥 − 36) 3(𝑥 2 − 𝑥 − 12) 3(𝑥 + 3)(𝑥 − 4)
𝑓(𝑥) =
=
=
(2𝑥 2 − 6𝑥 − 8)
2(𝑥 2 − 3𝑥 − 4) 2(𝑥 + 1)(𝑥 − 4)
Do more than just that.
Explain effect of each factor on the graph behavior. Use lim terminology where appropriate.
Explain end behavior using limit terminology.
Now that we have a methodical way to anayze sign behavior, we have an easier way to
determine limit behavior at vertical asymptotes.
Quiz 2
.
Day 8
X Section 2.4 The Derivative
Today: Rates of Change
We will do a series of examples involving 𝑓(𝑥) = −𝑥 2 + 8𝑥 − 7 = −(𝑥 − 1)(𝑥 − 7).
(A) Draw the graph on paper.
(B) Draw the secant line that passes through the points (2,f(2)) and (4,(f(4)).
Δ𝑦
(C) Find slope of that secant line. Result: using 𝑚 = Δ𝑥 , we get 𝑚 = 2.
Confirm using Secant Tangent grapher at the following website:
http://www.personal.psu.edu/dpl14/java/calculus/secantlines.html.
Introduce Average Rate of Change
Project Reference 3 Rates of Change (p.3 of course packet) on the movie screen.
Discuss the definition of Average Rate of Change, and the fact that the secant line slope
calculation that we just did was an example of a calculation of an average rate of change. That is,
the average rate of change of 𝑓(𝑥) = −𝑥 2 + 8𝑥 − 7 from 𝑥 = 2 to 𝑥 = 4 is number 𝑚 = 2.
(D) Find avg rate of change of 𝑓(𝑥) = −𝑥 2 + 8𝑥 − 7 from 𝑥 = 2 to 𝑥 = 2 + ℎ where ℎ ≠ 0.
𝑓(2+ℎ)−𝑓(2)
Result: 𝑚 =
= 4 − ℎ. (Very important can cancel h/h because ℎ ≠ 0)
ℎ
(E) Make a new graph of f and Illustrate this quantity on that new graph.
Introduce the Tangent Line
(F) Draw a new graph and Draw the line tangent to the graph of 𝑓 at 𝑥 = 2.
Goal: Find the slope of that tangent line. That is, find the slope 𝑚 of the line tangent to the graph
of 𝑓(𝑥) = −𝑥 2 + 8𝑥 − 7 at 𝑥 = 2.
Discuss the fact that we don’t have two known points on the tangent line, so we can’t use our
Δ𝑦
slope formula 𝑚 = Δ𝑥 to find the slope of the tangent line.
View computer graph on Secant Tangent Grapher.
Turn on Tangent line.
Observe that by bringing right intersection point of Secant Line closer to left intersection point,
the secant line changes to look more & more like tangent line.
Observe that the slope of secant line seems to be getting closer and closer to the number m=4, and
it seems that this number m=4 is probably the slope of the tangent line.
Analytically, pulling the second point closer to the first corresponds to finding the limit
𝑓(2 + ℎ) − 𝑓(2)
𝑚 = lim
ℎ→0
ℎ
𝑓(2+ℎ)−𝑓(2)
Do that limit for our computation. We find 𝑚 = lim
= lim 4 − 0 = 4
ℎ
ℎ→0
ℎ→0
.
Official introduction of the tangent line
The line tangent to the graph of f at x=a is defined to be the line that has these two properties:
(1) The line touches the graph of f at x=a. This means that the point (x,y)=(a,f(a)) is on the line.
This point is called the point of tangency. (Note that this requires that f(a) must exist.)
𝑓(𝑥+ℎ)−𝑓(𝑥)
(2) The line has slope 𝑚 = lim
if this limit exists and is a real number.
ℎ
ℎ→0
Graphical interpretation: The tangent line (if it exists) is the line that passes through the point of
tangency (x,y)=(a,f(a)) and looks like it is going the same direction as the graph of f at that point.
Class Drill 5 on page 18 of Course Packet: Representations of Slopes
.
Day 9
X continuing Section 2-4
Recall Rates of Change from yesterday (Ref 3)
Introduce the idea of the derivative function, 𝑓′. Imagine a machine:
input: number “a”
Instructions inside machine: (find slope of the line tangent to graph of f at x=a)
output obtained graphically: number m that is slope of line tangent to the graph of f at x=a
𝑓(𝑎+ℎ)−𝑓(𝑎)
output obtained analytically: the number 𝑚 = 𝑓 ′ (𝑎) = lim
.
ℎ
ℎ→0
This machine is really called the derivative machine. Discuss using x instead of a.
Class Drill 6 Finding derivatives graphically using a ruler
Examples of computing 𝑓′(𝑥). analytically
𝑓(𝑥+ℎ)−𝑓(𝑥)
Analytic definition of the derivative: 𝑓 ′ (𝑥) = lim
2
′ (𝑥)
ℎ→0
ℎ
Example: 𝑓(𝑥) = 𝑥 − 2𝑥 − 3. Find 𝑓
analytically, using the definition of the derivative.
Write explanation for most important step: Cancelling h/h. Result: 𝑓 ′ (𝑥) = 2𝑥 − 2.
Observe that 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 3 and 𝑓 ′ (𝑥) = 2𝑥 − 2 match the graphs in class drill 6!
Related questions (A) Find the slope of the line that is tangent to the graph of f at x=2.
(B) Find the slope of the line that is tangent to the graph of f at x=0.
(C) Find the x-coordinates of all points on the graph of f that have horizontal tangent lines.
Harder example: find derivative of function 𝑓(𝑥) = −3𝑥 2 + 5𝑥 − 7. (maybe skip this one)
.
Day 10
X continuing Section 2.4 The Derivative. Today: More examples
Example 𝑓(𝑥) = √𝑥. Find the Derivative of 𝑓. That is, find 𝑓′(𝑥). using the def of deriv.
Now do a harder example involving a √𝑥-type function: Let 𝑓(𝑥) = 5 − 17√𝑥. Find 𝑓′(𝑥).
Example 𝑓(𝑥) = 1/𝑥. Find the Derivative of 𝑓. That is, find 𝑓′(𝑥). using the def of deriv
17
Now do a harder example involving a 1/x-type function: Let 𝑓(𝑥) = 5 − 𝑥 . Find 𝑓′(𝑥).
.
Day 11
X Section 2.5 Basic Differentiation Properties.
Today: Constant Function Rule; Power Rule
Review Def of deriv. In Section 2.5, we will learn rules for taking derivatives. We won’t use def.
Notation for Derivatives
The Constant Function Rule
Present the Rule
𝑑
Examples: (A) if 𝑓(𝑥) = 5 then 𝑓 ′ (𝑥) = 0. (B) 𝑑𝑥 5 = 0. Why does this make sense graphically?
The power rule
Present the power rule. Two equation form: If 𝑓(𝑥) = 𝑥 𝑛 , then 𝑓 ′ (𝑥) = 𝑛𝑥 𝑛−1 .
𝑑
Single Equation form: 𝑑𝑥 𝑥 𝑛 = 𝑛𝑥 𝑛−1 .
Example (1) 𝑓(𝑥) = 𝑥 3 .
Example (2) 𝑓(𝑥) = √𝑥. Compare to result of Friday using definition of derivative.
1
Example (3) 𝑓(𝑥) = 𝑥. Compare to result of Friday using definition of derivative.
Discuss importance of step 1: rewriting function then step 2: take derivative.
Discuss common incorrect notation.
Example (4) 𝑓(𝑥) = 𝑥. Use Power Rule. Then check to see if it makes sense graphically.
Example (5) 𝑓(𝑥) = 1. Compare to result obtained using Constant Function Rule.
Quiz 3
.
Day 12
X continuing Section 2.5 Basic Differentiation Properties, continued
Today: The Sum and Constant Multiple Rule
Present the rule
Example Find deriv of 𝑓(𝑥) = 5𝑥 2 − 3𝑥 + 7.
Present method that I want them to be able to do
Step 1: rewrite f as sum of terms of form constant*power function.
Step 2: identify multiplicative constants and use sum & const multiple rule. Simplify result to
eliminate negative exponents.
Compare to result of Wed, Sep 9 using definition of derivative.
17
Example Find derivative of 𝑓(𝑥) = 5 − 𝑥 .
Compare to result of Monday using definition of derivative.
Example Find derivative of 𝑓 (𝑥) = 5 − 17√𝑥.
Compare to result of Wed, Sep 9 using definition of derivative.
3
Class Drill 7: Find deriv of 𝑓(𝑥) =
Trick Problems
7 √𝑥
5
3
+ 11𝑥 2/5 .
2𝑥 5 −4𝑥 3 +2𝑥
Trick problem Similar to Suggested Problem: Let 𝑓(𝑥) =
. Find 𝑓′(𝑥).
𝑥3
Special Topics
The Equation for the Tangent Line
Review point-slope form (y-b) = m(x-a).
Remember what we know about the line tangent to graph of f at x = a.
Analytical description: tangent line is the line that has these two properties:
1) Touches graph at x=a. That point is called “point of tangency”. The y-coord is f(a). So point of
tangency has coordinates (x,y) = (a,f(a)).
2) Tangent line has slope 𝑚 = 𝑓′(𝑎). This is the KNOWN SLOPE of the tangent line.
Point Slope form of equation for tangent line: .(𝑦 − 𝑓(𝑎)) = 𝑓′(𝑎)(𝑥 − 𝑎).
tangent line problem: Let 𝑓(𝑥) = 𝑥 3 − 9𝑥 2 + 15𝑥 + 25 = (𝑥 + 1)(𝑥 − 5)2 .
(A) Find equation of the line tangent to graph at x=2.
(B) Find x-coordinates of all points on graph where tangent line is horizontal.
Class Drill 8: Questions about Tangent Lines
Day 13
X Section 2.7 Marginal Analysis
Discuss Definition of Demand, Price, Revenue, Cost, Profit from Reference 5
Marginal Quantities
Definition of Marginal Quantities from Reference 5
Examples involving Marginal Quantities
For the coming 7 examples,
use 𝑪(𝒙) = 𝟏𝟒𝟓 + 𝟏. 𝟏𝒙 and 𝑹(𝒙) = 𝟓𝒙 − 𝟎. 𝟎𝟐𝒙𝟐 .
Example A: (Similar to problem 2.7#9) Find the marginal cost function.
Example B: (Similar to problem 2.7#13) Find the marginal revenue function.
Example C: (Similar to problem 2.7#17) Find the marginal profit function.
Average Quantities
Discuss Definition of Average Quantities from Reference 5
Examples involving Average Quantities
Example 4: (Similar to problem 2.7#21) Find the average cost function.
Example 5: (Similar to problem 2.7#23) Find the marginal average cost function.
Estimation Problems
2.7#34 The total cost of producing x electric guitars is
𝐶(𝑥) = 1000 + 100𝑥 − 0.25𝑥 2 dollars.
(A) What is the cost of producing a batch of 50 guitars?
(B) What is the cost of producing a batch of 51 guitars?
(C) If batch size changes from x=50 guitars to x=51 guitars, what will be change in the cost of
producing a batch of guitars? That is, if 𝑥 = 51 and Δ𝑥 = 1, what is Δ𝐶? (exact value)
(Book wording: “Find exact cost of producing the 51st guitar.”)
Use Wolfram, result is Δ𝐶 = $74.75.
Idea of approximation: Exact Change is Δ𝑦 = 𝑓(𝑥2 ) − 𝑓(𝑥1 ).
Approximate change is 𝑓 ′ (𝑥1 ) ⋅ Δ𝑥.
important to display the relationship correctly, and to be clear about what one is presenting.
exact change = Δ𝑄 ≈ approximate change = 𝑄 ′ (𝑥1 ) × Δ𝑥.
(D) If the batch size changes from x=50 guitars to x=51 guitars, use the marginal cost function to
find an approximate value for the change in the cost of producing a batch of guitars. That is, Use
the marginal cost function to find an approximation for Δ𝐶.
(Book wording: “Use marginal cost to approximate the cost of producing the 51st guitar.”)
Result 𝐶 ′ (𝑥1 ) ⋅ Δ𝑥 = 75 ⋅ 1 = 75.
Compare approximate and Exact Results
.
Day 14
X Continuing Section 2.7 Marginal Analysis
today: more examples
2.7#44 Price p (in dollars) and demand x for a particular item are related by 𝑥 = 1000 − 20𝑝.
𝑥
(A) Find price p in terms of x, and domain. result: 𝑝 = − 20 + 50 Domain 0 ≤ 𝑥 ≤ 1000.
(B) Find the revenue 𝑅(𝑥) from the sale of x item. What is the domain of R?
𝑥
𝑥2
Result: 𝑅(𝑥) = 𝑥𝑝 = 𝑥 (− 20 + 50) = − 20 + 50𝑥. Domain is 0 ≤ 𝑥 ≤ 1000.
(C) Find the marginal revenue at a production level of 400 items and interpret the results.
𝑥
400
Result: 𝑅′(𝑥) = − 10 + 50. So 𝑅 ′ (400) = − 10 + 50 = −40 + 50 = 10. Therefore if company
changes from batch size of 400 to batch size of 401, the revenue will increase by approximately
$10. The book would say that this is approx the increase in revenue from sale of 401st item.
(D) Find the marginal revenue at a production level of 650 items and interpret the results.
650
Result: 𝑅 ′ (650) = − 10 + 50 = −65 + 50 = −15. Therefore if company changes from batch
size of 650 to batch size of 651, revenue will decrease by approx $15. The book would say that
this is approximately the decrease in revenue from the sale of the 651st item.
problem 2.7#48 A company makes toasters. Marketing dept estimates a weekly demand of 300
toasters at price of $25 per toaster, and a weekly demand of 400 toasters at price of $20 per.
Financial dept estimates fixed weekly costs are $5000 and variable costs are $5 per toaster.
(A) Assume linear relationship between x and p. Find equation that expresses p as a function of x,
and find domain. Result: We want equation of line through two known points (𝑥, 𝑝) = (300,25)
20−25
−5
1
and (𝑥, 𝑝) = (400,20). We find 𝑚 = 400−300 = 100 = − 20. Use point-slope form: (𝑝 − 25) =
1
1
− 20 (𝑥 − 300). Convert to slope intercept form: 𝑝 = − 20 𝑥 + 40. The domain is 0 ≤ 𝑥 ≤ 800.
(B) Find revenue function in terms of x and state its domain.
1
1
Solution: : 𝑅(𝑥) = 𝑥𝑝 = 𝑥 (− 20 𝑥 + 40) = − 20 𝑥 2 + 40𝑥. The domain is 0 ≤ 𝑥 ≤ 800.
(C) Assume cost function is linear. Find cost function. Solution: 𝐶(𝑥) = 5𝑥 + 5000.
(D) Graph the cost function and revenue function on the same coordinate system for 0 ≤ 𝑥 ≤
800. Find the break-even points and indicate regions of loss and profit.
Solution: Downward-facing parabola, skewered by line. Break-even points at x=200 and x=500.
Corresponding values of price are p=6000 and p=7500.
1
1
(E)Find Profit function. Res: 𝑃(𝑥) = (− 𝑥 2 + 40𝑥) − (5𝑥 + 5000) = − 𝑥 2 + 35𝑥 − 5000.
20
20
(F) Evaluate the marginal profit at x=325 and x=425 and interpret the results.
1
Solution: First we find the marginal profit function. 𝑃′(𝑥) = − 10 𝑥 + 35,
1
Using this, 𝑃′ (325) = − 10 (325) + 35 = −32.5 + 35 = 2.5. This tells us that if batch size
increases from 325 toasters per week to 326 per week, profit will increase by roughly $2.5.
1
Using this, 𝑃′ (425) = − 10 (425) + 35 = −42.5 + 35 = −7.5. This tells us that if batch size
increases from 425 toasters per week to 426 per week, profit will decrease by roughly $7.5.
Day 15 is Exam 1
Day 16
X Section 3.1: The constant e
Interest formulas. Start with “Simple Interest”
Simple interest formula: 𝐴 = 𝑃 + 𝑃𝑟𝑡 = 𝑃(1 + 𝑟𝑡). Introduce P,r,t,A Graph equation.
Example #1: Deposit $1000 into bank account with 2% simple interest. What will the balance be
after 5 years? Solution: 𝐴 = 1000(1 + .02 ⋅ 5) = 1000(1 + .1) = 1000(1.1) = $1100.
Periodically Compounded Interest
Discuss the idea of compound interest. Graph it.
𝑟 𝑚⋅𝑡
Present general formula for Periodically Compound Interest: 𝐴 = 𝑃 (1 + 𝑚) .
Example #2: Deposit $1000 into bank account with 2% interest compounded yearly. What will be
the balance after 5 years? Show how it turns out 𝐴 = 1000(1 + .02)5 ≈ $1104.08.
Example #3: Deposit $1000 into bank account with 2% interest compounded monthly. What will
be the balance after 5 years?
.02 12⋅5
Show how it turns out 𝐴 = 1000 (1 + )
≈ $1105.08.
12
Example #4: Deposit $1000 into bank account with 2% interest compounded daily. What will be
.02 365⋅5
the balance after 5 years? Result: 𝐴 = 1000 (1 + 365)
≈ $1105.17.
More Frequent Compounding
Observe: $1100 < $1104.08 < $1105.08 < $ 1105.17 for compounding (never, m=1,12,365)
𝑟 𝑚⋅𝑡
Question: What will happen to the value of a as 𝑚 → ∞? That is, what is lim 𝑃 (1 + 𝑚)
𝑚→∞
?
Day 17
X Continuing Section 3.1
Today: The constant e and continually compounded interest
Recall Question from yesterday:
𝑟 𝑚⋅𝑡
What will happen to value of A as 𝑚 → ∞? That is, what is lim 𝑃 (1 + 𝑚)
𝑚→∞
1 𝑛
?
Related question: What is lim (1 + 𝑛) ? Investigate with table.
𝑛→∞
1 𝑛
Explore values of (1 + 𝑛) as 𝑛 → ∞.
y-values seem to be getting closer & closer to a number in the vicinity of 2.7182
Based on this, we might guess that


1 𝑛
The limit lim (1 + 𝑛) exists.
𝑛→∞
The value of the limit is near 2.7182
.
BIG FACTS FROM HIGHER MATH




1 𝑛
The limit lim (1 + 𝑛) does exist.
𝑛→∞
The symbol “e” is used to denote the real number that is the value of the limit. That
1 𝑛
is lim (1 + 𝑛) = 𝑒.,
𝑛→∞
The number e is between 2 & 3. That is, 2 < e < 3
The value of e is near 2.7182 but e is irrational. (e cannot be written exactly as a fraction
or as a terminating decimal, or as a repeating decimal.
.
𝑥 𝑛
𝑥 𝑛
Related limit: lim (1 + 𝑛) ? Answer lim (1 + 𝑛) = 𝑒 𝑥 .
𝑛→∞
Discuss graph of 𝑦 = 2𝑥 , 3𝑥 , 𝑒 𝑥 .
𝑛→∞
𝑟 𝑛𝑡
𝑟 𝑛𝑡
Related limit: lim 𝑃 (1 + 𝑛) ? Answer lim 𝑃 (1 + 𝑛)
𝑛→∞
𝑛→∞
= 𝑃𝑒 𝑟𝑡 .
That answers our question from above.
𝑟 𝑚⋅𝑡
Fact: lim 𝑃 (1 + 𝑚)
𝑚→∞
= 𝑃𝑒 (𝑟𝑡) .
Discuss graph of 𝑦 = 𝑃𝑒 (𝑟𝑡) .
Inspired by this, invent a bank account that uses the formula 𝐴 = 𝑃𝑒 (𝑟𝑡) to compute its balance.
Call it “continuously-compounded interest”. Add this to our lists of types of interest.
Example #5: Deposit $1000 into bank account with 2% interest compounded Continuously. What
will be the balance after 5 years? Result: 𝐴 ≈ $1105.17.
Solve the equation 𝐴 = 𝑃𝑒 (𝑟𝑡) . for 𝑡 in terms of the others.
Example 6: Deposit $937 into account with 2.3% interest compounded continuously. What is the
balance after 7 years?Solution: 𝐴 = 𝑃𝑒 (𝑟𝑡) = 937𝑒 (7⋅0.023) ≈ $1100.68.
Example 7: Deposit $937 into account with 2.3% interest compounded continuously.How long
𝑙𝑛(
1200
)
937
after initial deposit until the Balance has grown to 1200?Solution: 𝑡 =
≈ 10.75 years.
0.023
Example 8: Deposit some money into account with 2.3% interest compounded continuously. How
long until the balance doubles?
Example 9: If you want an account with continuously compounded interest to double in 20 years,
what interest rate will you need?
.
Day 18
X Section 3.2 Derivatives of Exponential and Logarithmic Functions
Today: Exponential functions
New Rules
𝑑
New rule: Exponential Function Rule #1 𝑑𝑥 𝑒 (𝑥) = 𝑒 (𝑥)
This rule is proven using calculus above the level of this course.
Does this make sense graphically?
𝑑
New rule: Exponential Function Rule #2 𝑑𝑥 𝑒 (𝑘𝑥) = 𝑐𝑒 (𝑘𝑥)
This rule is also proven using calculus above the level of this course.
The book introduces this fact only in exercises, in exercises 3.2 # 61, 62.
The book never presents this in a list of derivative rules, never gives it a theorem number.
That is a shame, because it is one of the most-used derivative rules.
𝑑
Use Rule #2 to prove Exponential Function Rule #3 𝑑𝑥 𝑏 (𝑥) = 𝑏 (𝑥) ⋅ ln(𝑏).
Review Derivative Rules so far in Reference 4 on page 4 of course packet
Example: Find the derivatives of the following functions: Have students work on these in pairs.
(A) 𝑓(𝑥) = 5𝑒 (𝑥) .
(B) 𝑓(𝑥) = 5𝑒 (7𝑥) .
(C) 𝑓(𝑥) = 5 ⋅ 7(𝑥) .
(D) 𝑓(𝑥) = 5𝑒 (7) . Notice, this is just a constant function
(E) 𝑓(𝑥) = 5𝑥 𝑒 . Notice: This is just a power function, not an exponential
More difficult example involving exponential function
Tangent Line Example
One more example involving exponential function: 𝑓(𝑥) = 11𝑒 (𝑥) + 23𝑥.
Find equation of line tangent to graph of f at x=0.
Find equation of line tangent to graph of f at x=1.
Rate of Change Examples
investment of $P earns interest at an interest rate of r compounded continuously.
(A) What is the value at t years
What is the instantaneous rate of change of the balance at time t years?
investment of $1000 earns interest at an annual rate of 2% compounded continuously.
(A) What is the value at 5 years
What is the instantaneous rate of change of the balance at time 5 years?
Value of a truck is given by the function 𝑆(𝑡) = 174,000(0.9)(𝑡)
(A) What is the purchase price of the truck?
(B) What is the value of the truck after 5 years?
(C) What is the rate of change of the value of the truck at time t=5 years?
(D) Write a sentence summarizing the results of (B) and (C). That is, “interpret them.”
Day 19
Fri Continuing Section 3.2 Derivatives of Exponential and Logarithmic Functions
Today: Derivatives of Logarithmic Functions
Review definition of logarithm function and shape of logarithm Graph.
New Derivative Rules
𝑑
1
Log Function Rule #1: 𝑑𝑥 ln(𝑥) = 𝑥. Does this make sense graphically?
Log Function Rule #2:
𝑑
1
log 𝑏 (𝑥) = 𝑥 ln(𝑏)
𝑑𝑥
Review Derivative Rules so far in Reference 4 on page 4 of course packet
Examples: Find the derivatives of the following functions:
(A) 𝑓(𝑥) = 12 ln(𝑥).
(B) 𝑓(𝑥) = 12 log13 (𝑥).
(C) 𝑓(𝑥) = 12 log(𝑥).
(D) 𝑓(𝑥) = 12 ln(13).
(E) 𝑓(𝑥) = 12 ln(13𝑥).
13
Class Drill 9 Derivs of: (A) 𝑓(𝑥) = 12 ln ( 𝑥 ). (B) 𝑓(𝑥) = 12 ln(𝑥13 ). (C) 𝑓(𝑥) = 12𝑥 ln(13).
tangent line problem: (D) Find equation of line tangent to 𝑓(𝑥) = 5 + ln(𝑥 3 ) at 𝑥 = 𝑒 2 .
Quiz 4
.
Day 20
X Section 3.3: Derivatives of Products and Quotients
The product rule
present the product rule
 Obvious thing: d(f*g)/dx = f’*g’ is WRONG!
 Present the correct rule
Review Derivative Rules so far in Reference 4 on page 4 of course packet
Example 𝑓(𝑥) = (−3𝑥 2 + 5𝑥 − 7)(3𝑥 − 2).
Find 𝑓′(𝑥) using the product rule. Don’t simplify.
Example 𝑓(𝑥) = (−3𝑥 2 + 5𝑥 − 7)𝑒 (𝑥) . (A) Find 𝑓′(𝑥) and simplify.(B) 𝑓′(0). (C) 𝑓′(1).
Example 𝑓(𝑥) = 5𝑥 7 ⋅ ln(𝑥). (A) find 𝑓′(𝑥) and simplify. (B) 𝑓′(1). (C) 𝑓′(𝑒).
.
Day 21
X Continuing Section 3.3 Derivatives of Products and Quotients
The Quotient rule
present the quotient rule
 The obvious thing: d(f/g)/dx = f’/g’ is WRONG!
 Present the correct rule
Review Derivative Rules so far in Reference 4 on page 4 of course packet
(3𝑥+5)
Example (similar to 3.3#25) 𝑓(𝑥) = (𝑥 2 −3). Find 𝑓′(𝑥).
𝑒 (𝑥)
Example (similar to 3.3#31) 𝑓(𝑥) = (𝑥 2 −3). Find 𝑓′(𝑥).
Class Drill 10: Don’t forget the easy derivative rules (Section 3.3) Four parts!
Trick Problems
𝑥 4 +4
Example (similar to 3.3#73): 𝑓(𝑥) = 4 .
𝑥
Find 𝑓′(𝑥) by (A) using quotient rule and by (B) first simplifying 𝑓.
Introduce Example (similar to 3.3#87): 𝑓(𝑥) =
.
3𝑥 5 −2𝑥 7
5
√𝑥 3
. Find 𝑓′(𝑥). Do not do this problem.
Day 22
X Continuing Section 3.3 Derivatives of Products and Quotients
today: More difficult Quotient Rule problems
Tangent Line Problems
2𝑥
Example (similar to 3.3#65): 𝑓(𝑥) = 3(𝑥). Find equation of line tangent to f at 𝑥 = 3.
𝑥
Example (similar to 3.3#69): 𝑓(𝑥) = 𝑥 2 +25. (A) find 𝑓′(𝑥). (B) find 𝑓′(0)
(C) find the x-coordinate of all the points on the graph of 𝑓 that have horizontal tangent lines.
Approximation Problem
(based on suggested exercise 3.3#93)
7000𝑡
Sales of a game are described by the function 𝑆(𝑡) = 𝑡+6
where 𝑡 is the time (in months) since the game was introduced and 𝑆(𝑡) is the total sales at time 𝑡.
(A) Find 𝑆(4).
(B) Find 𝑆′(𝑡). Show all details clearly, use correct notation, and simplify your answer.
(C) Find 𝑆′(4).
(D) Write a brief interpretation of the answers from (A) and (C). That is, explain what the answers
tell us. Include the correct units in your explanation.
(E) Use the results of (D) to estimate the total sales after 5 months.
(F) How many games will eventually sell?
.
Day 23
X Section 3.4 The Chain Rule
introduce the chain rule, used for taking derivatives of “nested functions” Functions of form
𝑑
𝑜𝑢𝑡𝑒𝑟(𝑖𝑛𝑛𝑒𝑟(𝑥)). The Chain Rule: 𝑑𝑥 𝑜𝑢𝑡𝑒𝑟(𝑖𝑛𝑛𝑒𝑟(𝑥)) = 𝑜𝑢𝑡𝑒𝑟 ′ (𝑖𝑛𝑛𝑒𝑟(𝑥)) ⋅ 𝑖𝑛𝑛𝑒𝑟 ′ (𝑥).
Review Derivative Rules so far in Reference 4 on page 4 of course packet
Today: examples where the outer function is a power function.
Book solves such problems using what it calls the “Generalized Power Function. I won’t call it
that. We’ll just use the chain rule.
example 1 (similar to 3.4#21): For 𝑓(𝑥) = 2(3𝑥 4 + 5𝑥 2 + 6)7 find 𝑓 ′ (𝑥).
2
example 2 (similar to 3.4#59): For 𝑓(𝑥) = (3𝑥 4 2 7 find 𝑓 ′ (𝑥).
+5𝑥 +6)
example 3 (similar to part of 3.4#71): For 𝑓(𝑥) = 3√𝑥 2 − 3𝑥 + 21 Find 𝑓 ′ (𝑥).
Harder problems:
example 3 (similar to 3.4#71): For 𝑓(𝑥) = 3√𝑥 2 − 3𝑥 + 21
(A) Find the equation of the line tangent to the graph of 𝑓 at 𝑥 = 4.
(B) Find x-coordinate of all points on the graph of 𝑓 that have a horizontal tangent line.
Example 4: (similar to 3.4#67) 𝑓(𝑥) = 𝑥 3 (𝑥 − 7)4 find 𝑓 ′ (𝑥) and find all values of x where the
tangent line is horizontal. Discuss common mistake.
Quiz 5
Day 24
Tue Section 3.4 The Chain Rule, Continued
Class Drill 11 Don’t forget the easy derivative rules, part II (Section 3.4) Only two parts.
Today: Examples where the outer function is not a power function
Example where the outer function is a logarithmic function
example 5 (similar to 3.4#35): For 𝑓(𝑥) = 7 ln(5𝑥 2 − 30𝑥 + 65) Find 𝑓 ′ (𝑥).
Examples where the outer function is an exponential function
𝑑
Discuss the Exponential Function Rule #2 𝑑𝑥 𝑒 (𝑐𝑥) = 𝑐 ⋅ 𝑒 (𝑐𝑥) . Recall that we introduced it on
Wed, Sep 23. It could be proven at that time only using calculus techniques beyond the level of
this course. Example 6: Prove it again using Chain Rule
Example 7 (similar to 3.4#43) Chain Rule problem involving exponentials:
2
2
𝑓(𝑥) = 𝑒 (−𝑥 +4𝑥−4) = 𝑒 (−(𝑥−2) )
(A) Find the equation of the line tangent to the graph of 𝑓 at x=0.
(C) Find the x-values of all points with horiz tangent line
(D) Illustrate on computer
(E) Discuss “Bell Curve”. The general form of a function with a bell-shaped curve graph is:
𝑓(𝑥) = 𝑒 (𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙)
where the polynomial has degree=2 and a negative leading coefficient.
Day 25
X Rate of Change Class Drills
Class Drill 12a: Rate of Change Problem (Exponential Function)
Class Drill 12b: Rate of Change Problem (Rational Function with Peak)
Class Drill 12c: Rate of Change Problem (Rational Function with Horizontal Asymptote)
Class Drill 12d: Rate of Change Problem (Square Root Function)
Day 26
X Rate of Change Class Drills
Class Drill 12a: Rate of Change Problem (Exponential Function)
Class Drill 12b: Rate of Change Problem (Rational Function with Peak)
Class Drill 12c: Rate of Change Problem (Rational Function with Horizontal Asymptote)
Class Drill 12d: Rate of Change Problem (Square Root Function)
Day 27 is Exam 2
After that is a week of Spring Break
Day 28
X Section 4.1 1st Derivatives and Graphs
Today: horizontal tangents, increasing and decreasing functions
correspondence. Be sure to keep f’ on left and f on right!!
 If 𝑓′ is positive at 𝑥 = 𝑐 then line tangent to graph of 𝑓 at 𝑥 = 𝑐 tilts upward
 If 𝑓′ is negative at 𝑥 = 𝑐 then line tangent to graph of 𝑓 at 𝑥 = 𝑐 tilts upward
 If 𝑓′ is zero at 𝑥 = 𝑐 then line tangent to graph of 𝑓 at 𝑥 = 𝑐 is horizontal
Definition of “𝑓 is increasing on interval (𝑎, 𝑏)”: If 𝑎 < 𝑥1 < 𝑥2 < 𝑏 then 𝑓(𝑥1 ) < 𝑓(𝑥2 ).
correspondence. Be sure to keep f’ on left and f on right!!
 If 𝑓′ is positive on (𝑎, 𝑏) then f increasing on (𝑎, 𝑏).
 If 𝑓′ is negative on (𝑎, 𝑏) then f is decreasing on (𝑎, 𝑏).
 If 𝑓′ is zero on (𝑎, 𝑏) then f is constant on (𝑎, 𝑏).
Discuss Reference 6 Derivative Relationships from course packet. (Project on screen)
Graphs of f  graph of f ’
Class Drill 14 Determining shape of graph of 𝑓′ by studying shape of graph of 𝑓.
(Class drill is exercise 4.1 #84 graph of 𝑓  sign behavior of 𝑓′  graph of 𝑓′ )
Analytical example: Formula for 𝑓  information about increasing/decreasing behavior of 𝑓.
Example: 𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 − 12𝑥 + 5. Find x-coordinates where graph of f has horizontal
tangent line. Find intervals of increase & decrease.
information about 𝑓′  graph of 𝑓.
Do exercise 4.1#62 function values for 𝑓 and sign chart for 𝑓′  graph of 𝑓.
Given information about 𝑓 and 𝑓′.
 Function 𝑓 is continuous for (−∞, ∞).
𝑥
−2 −1 0 1 2
 Function values (
).
𝑓(𝑥) 1
3 2 1 −1
𝑥 (−∞, −1) −1 (−1,1) 1 (1, ∞)
 Sign chart for 𝑓′ (
).
𝑓′
+++
0 −−− 0 −−−
Sketch the graph of 𝑓.
exercise 4.1#66 function values for 𝑓 and sign information for 𝑓′  graph of 𝑓.
Given information about 𝑓 and 𝑓′.
 Function 𝑓 is continuous for (−∞, ∞).
 Function values 𝑓(−2) = −1, 𝑓(0) = 0, 𝑓(2) = 1.
 Derivative values 𝑓 ′ (−2) = 0, 𝑓′(2) = 0.
 Derivative sign 𝑓 ′ (𝑥) > 0 on (−∞, −2), (−2,2), (2, ∞).
Sketch the graph of 𝑓.
exercise 4.1#76 graph of 𝑓′  graph of 𝑓. Given the following graph of 𝑓′,
𝑓′(𝑥)
𝑥
(A) Find the 𝑥-coordinates of all points on the graph of 𝑓 that have horizontal tangent lines.
(B) Find the intervals on which 𝑓 is increasing.
(C) Find the intervals on which 𝑓 is increasing.
(D) Sketch a possible graph of 𝑓.
.
Day 29
Tue Continuing Section 4.1 1st Derivatives and Graphs
Today: Local Extrema and the 1st Derivative Test
Local Extrema
Define local max and local min notice that function f need not be continuous
Class Drill 15 Studying Graph Behavior
Observation: If 𝑓 has a local max or min at 𝑥 = 𝑐, then the following three things are true:
(1) 𝑓′(𝑐) = 0 or 𝑓′(𝑐) 𝐷𝑁𝐸.
(2) 𝑓(𝑐) exists. That is, the number 𝑥 = 𝑐 is in the domain of the function 𝑓.
(3) 𝑓′ changes sign at 𝑥 = 𝑐.
Define words: 𝑥 = 𝑐 is a partition number for 𝑓′ mean 𝑥 = 𝑐 has property (1).
Define words: 𝑥 = 𝑐 is a critical number for 𝑓 mean 𝑥 = 𝑐 has properties (1) and (2).
With terminology, we can say abbreviate our observation stated earlier:
If 𝑓 has local max or min at 𝑥 = 𝑐, then 𝑥 = 𝑐 is a crit num for 𝑓 and 𝑓′ changes sign at 𝑥 = 𝑐.
Is the converse true? That is, is it true that
If 𝑥 = 𝑐 is a crit num for 𝑓 and 𝑓′ changes sign at 𝑥 = 𝑐, then 𝑓 has local max or min at 𝑥 = 𝑐?
Answer: No. Consider Class Drill 15 Examples # 9,10. Both functions have 𝑥 = 7 as critical
number and for both functions, 𝑓′ changes sign at 𝑥 = 7, but #9 has local max and #10 does not.
The problem with examples #9,#10 is that 𝑓 is discontinuous at 𝑥 = 7.
The First Derivative Test
Introduce first derivative test for local extremum at x=c
If
(1) 𝑓′(𝑐) = 0 or 𝑓′(𝑐) 𝐷𝑁𝐸.
(2) 𝑓(𝑐) exists. That is, the number 𝑥 = 𝑐 is in the domain of the function 𝑓.
(3) 𝑓 is continuous at 𝑥 = 𝑐.
(4) 𝑓′ changes sign at 𝑥 = 𝑐.
then 𝑓 has a local max or local min at 𝑥 = 𝑐.
Notice that this test will detect all local extrema that occur at locations where 𝑓 is continuous.
So for instance, in Class Drill 15, it would detect the local extrema in examples # 1,2,11.
But the test will not detect local extrema that occur at locations where 𝑓 is not continuous.
So for instance, in Class Drill 15, it would not detect the local max in example #9.
Using the First Derivative Test
Class Drill 16 Using the First Derivative test (sign chart for 𝑓′  graph of 𝑓)
Similar to 4.1#41:Find partition numbers of 𝑓′, critical number of 𝑓, intervals of incr/decr, and
local extema.
𝑓(𝑥) = −𝑥 4 + 50𝑥 2 + 7
𝑓 ′ (𝑥) = −4𝑥 3 + 100𝑥 = −4𝑥 (𝑥 2 − 25) = −4𝑥(𝑥 + 5)(𝑥 − 5)
. (local max at 𝑥 = −5 and 𝑥 = 5. Local min at 𝑥 = 0.)
.
Day 30
X Continuing Section 4.1 1st Derivatives and Graphs
Today:More sophisticated examples involving the 1st Derivative Test
[Example 1] Similar to Suggested Exercise 4.1#43
Find partition numbers of 𝑓′, critical number of 𝑓, intervals of incr/decr, and local extema.
Function 𝑓(𝑥) = 𝑥𝑒 (−𝑥) with derivative 𝑓 ′ (𝑥) = −(𝑥 − 1)𝑒 (−𝑥) .
(𝑓′ has partition number at 𝑥 = 1. 𝑓 has critical number at 𝑥 = 1. Local max at 𝑥 = 1.)
[Example 2] Similart to Suggested Exercise 4.1#29
Find partition numbers of 𝑓′, critical number of 𝑓, intervals of incr/decr, and local extema.
Function 𝑓(𝑥) = 1/(𝑥 − 7)2 with derivative 𝑓 ′ (𝑥) = −2/(𝑥 − 7)3 .
(𝑓′ has partition number at 𝑥 = 7. But 𝑓 has no critical numbers! No local extrema.)
[Example 3] Similar to Suggested Exercise 4.1#85
Find partition numbers of 𝑓′, critical number of 𝑓, intervals of incr/decr, and local extema.
Function 𝑓(𝑥) = 𝑥 + 4/𝑥 with derivative 𝑓 ′ (𝑥) = 1 − 4/𝑥 2 .
(𝑓′ has partition number at 𝑥 = −2,0,2. But 𝑓 has critical numbers only at 𝑥 = −2,2.
Local max at 𝑥 = −2, local min at 𝑥 = 2.)
[Example 4] (Function from Class Drill 12(b), similar to suggested exercise 4.1#97)
A drug is administered by pill. The drug concentration (in milligrams per milliliter) in the
bloodstream t hours after the pill is taken is given by the formula
0.14𝑡
𝐶(𝑡) = 2
𝑓𝑜𝑟 0 ≤ 𝑡 ≤ 21
𝑡 +1
(A) find intervals of increase & decrease.
(B) find t-coordinate of local extrema.
(C) Find local extrema. Round to three decimal places.
[Example 5] Similar to book Example, but not similar to any suggested exercises. Skip?
Find partition numbers of 𝑓′ and critical number of 𝑓.
10
Function 𝑓(𝑥) = 5(𝑥 − 7)2/3 with derivative 𝑓 ′ (𝑥) = 3(𝑥−7)1/3 .
(Critical number at 𝑥 = 7. Local min at 𝑥 = 7)
Day 31
X Section 4.2: Concavity
Today: Concavity and 1st Derivative
Definition of concave up at x=c
words: “f is concave up at x=c.”
meaning: The graph of 𝑓 has a tangent line at x=c, and for x-values near c, the graph of f stays
above the tangent line.
picture:
Definition of concave up on an interval
Definition of inflection point
Class Drill 17 Identifying Three Kinds of Graph Behavior
Relationship between 1st derivative and concavity
Study Examples that show concave up  f ’ increasing, etc
Class Drill 18: Using a graph of 𝑓′ to get information about 𝑓.
Quiz 6
Day 32
X Continuing Section 4.2: Concavity
Today: Concavity and 2nd Derivatives
Friday, we discussed that concave up on interval (𝑎, 𝑏)  𝑓′ increasing on (𝑎, 𝑏)
This leads us to want to know how to figure out when 𝑓′ is increasing or decreasing.
This leads us to want to know how to figure out when the derivative of 𝑓′ is positive or neg
This leads us to want to consider the derivative of 𝑓′.
Introduce 2nd derivative
Example: find the 2nd derivative of the function 𝑓(𝑥) = ln(𝑥 2 + 6𝑥 + 13). Point out to students
that book Example 3 considers intervals of concavity and inflection points of a function like this.
Example: find the 2nd derivative of the function 𝑔(𝑥) = 𝑥𝑒 (−𝑥) . (Useful for 4.2#89)
Example similar to sugg exercise 4.2#89 An ice cream company found that the price demand
equation for quarts of their ice cream is 𝑝 = 3𝑒 (−𝑥) for 0 ≤ 𝑥 ≤ 5, where 𝑝 is the price of a quart
(in dollars) and 𝑥 is the number of thousands of quarts that they sell each week.
(A) Find the Revenue Function
(B) Find the intervals of increase & decrease and the local extrema for the Revenue Function.
(C) Find the intervals of concavity and the inflection points for the Revenue Function
Day 33
X Continuing Section 4.2: Concavity
Today: Curve Sketching
curving graph  description of sign behavior of (𝑓, 𝑓′, 𝑓").
examples: (𝑓, 𝑓′, 𝑓") = (+, −, +) and (𝑓, 𝑓′, 𝑓") = (−, +, −).
Do a whole bunch of them.
Do exercise 4.2#42 function values for 𝑓 and sign charts for 𝑓 ′ , 𝑓′′  graph of 𝑓.
Given information about 𝑓, 𝑓 ′ , 𝑓′′.
 Function 𝑓 is continuous for (−∞, ∞).
𝑥
−4 −2 −1 0 2 4
 Function values (
).
𝑓(𝑥) 0 −2 −1 0 1 3
𝑥 (−∞, −2) −2 (−2,2) 2 (2, ∞)
 Sign chart for 𝑓′ (
).
𝑓′
−−−
0 +++ 0 +++
𝑥 (−∞, −1) −1 (−1,2) 2 (2, ∞)
 Sign chart for 𝑓′′ (
).
𝑓′
+++
0 −−− 0 +++
Sketch the graph of 𝑓.
Class Drill 19: Sketching a Graph of a Function from Information about its Derivatives
Given the following information:
𝑓(−2) = −2
𝑓(0) = 1
𝑓(2) = 4
𝑓 ′ (−2) = 0
𝑓 ′ (2) = 0
𝑓 ′ (𝑥) > 0 𝑜𝑛 (−2,2)
𝑓 ′ (𝑥) < 0 𝑜𝑛 (−∞, −2) 𝑎𝑛𝑑 (2, ∞)
𝑓"(0) = 0
𝑓"(𝑥) > 0 𝑜𝑛 (−∞, 0)
𝑓"(𝑥) < 0 𝑜𝑛 (0, ∞)
Sketch a possible graph of 𝑓.
Introduce the idea of the graphing strategy
Class Drill 20 Using the Graphing Strategy to Graph a Polynomial
.
Day 34
X Section 4.5 Absolute Extrema
discuss the Definition (page 293)
discuss Theorem 2 (Locating Absolute Extrema) (page 294)
discuss Theorem 1 (Extreme Value Theorem) (page 294)
Introduce The “Closed Interval Method” (page 295)
Example: 𝑓(𝑥) = 𝑥 4 − 6𝑥 2 + 5. Find extrema on [−3,2].
Some details useful for the solution:
𝑓 ′ (𝑥) = 4𝑥 3 − 12𝑥 = 4𝑥(𝑥 2 − 3) = 4𝑥(𝑥 + √3)(𝑥 − √3).
Started: Step 1: confirmed that interval is closed and f is continuous on that interval.
Step 2: Found critical values in the interval. (discussed incorrect method versus correct method
(using factoring) to find critical values. Made list of important x-values.
Step 3: get list of corresponding y-values
Steps 4,5: write conclusion.
Illustrate with computer graph.
Repeat the example with interval [-1,2]
Observations:
Changing the interval changes the result.
Simply computing y-values at all integer x-values in the interval is not enough.
Day 35
X continuing Section 4.5 Absolute Extrema
today: What happens when extreme value theorem cannot be used?
Graphical examples: do Class Drill 21 Local and Absolute Extrema as a class
Analytical examples where extreme value thm can’t be used
Example 1) Same function as Wed Oct 21: 𝑓(𝑥) = 𝑥 4 − 6𝑥 2 + 5.
Find absolute extrema on (−∞, ∞). (function continuous, but interval not closed)
Some details useful for the solution:
𝑓 ′ (𝑥) = 4𝑥 3 − 12𝑥 = 4𝑥(𝑥 2 − 3) = 4𝑥(𝑥 + √3)(𝑥 − √3).
𝑓 ′′ (𝑥) = 12𝑥 2 − 12 = 12(𝑥 2 − 1).
Illustrate with computer graph.
Another where the Closed Interval Method cannot be used.
250
Example #2 similar to assigned & suggested problems:𝑓(𝑥) = 20 − 4𝑥 − 𝑥 2 .
Find all absolute extrema on (0, ∞). (interval not closed)
Solution: solve using two methods:
 1st derivative information
 2nd derivative information
Illustrate with computer graph. (did not do this)
Useful information:
250
𝑓(𝑥) = 20 − 4𝑥 − 2
𝑥
500
′ (𝑥)
𝑓
= −4 + 3 𝑝𝑎𝑟𝑡𝑖𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 0,5
𝑥
1500
′′ (𝑥)
𝑓
= − 4 𝑝𝑎𝑟𝑡𝑖𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 0
𝑥
Results:local max at (x,y) = (5,-10). absolute max on (0,∞) is y = -10. no absolute min
Quiz 7
.
Day 36
X Section 4.6 Optimization
Today: Area and Perimeter
Discuss general issues:
Optimization problems are simply Max/Min problems, but they may have complications
May be presented as word problems.
May have domains that are not closed intervals
You will probably have to figure out the domain
May involve more than one variable
Example Similar to suggested exercise 4.6#9,17
Find pos numbers x, y such that
 The sum 2x+y = 900.
 The product maximized.
(Step 1) Identify Equation I 2x + y = 900
(Step 2) Write equation II involving x and y and the letter P for the product. Result P=xy. We
want to maximize P.
(Step 3) Solve equation I for y in terms of x. New equation I: y = 900-x.
(Step 4) Substitute Equation I into Equation II and simplify to get a new equation that gives the
product A as a function of just one variable x. Call this function P(x).Result: P(x) = x(900-2x).
(Step 5) Using Calculus, find value of x that maximizes P(x). Result: x = 225
(Step 6) Find corresponding values of y and the product. Result: y=450, product = 101,250
Example similar to suggested exercise 4.6#34,35, using numbers from previous example.
Fence problem: Farmer needs to build a fence to make a rectangular corral next to an adjacent
pasture. He needs only needs to fence three sides, because the fourth side has already been fenced.
He has 900 feet of fencing. What are the dimensions of the pasture that will enclose the largest
possible area? 225, y=450, Area is A = 101,250 square feet
.
Day 37
X Continuing Section 4.6 Optimization Area and Perimeter
Example #1
Similar to suggested problems 4.6#13,15
Find positive numbers x, y such that
 The product is 9000.
 The sum 10x + 25y is minimized.
Solution:
(step 1) Write an equation I involving x and y expressing the fact that the product is 9000: Result
xy = 9000.
(step 2) Write an equation II involving x and y and the letter S for sum: S=10x + 25y.
(step 3) Solve Equation I for y in terms of x. New Equation I. y=9000/x
(step 4) Substitute Equation I into Equation II and simplify to get a new equation that gives the
sum S as a function of just one variable x. Call this function S(x). S(x)=10x + 25(9000/x)
(step 5) Using calculus, find the value of x that minimizes S(x). x = 150
(step 6)) Find the corresponding values of y and sum, S. y = 60, S=3000
Example #2 similar to suggested exercise 4.6#33, using numbers from previous example.
Fence problem: Farmer needs to build a rectangular fenced chicken yard. He wants an area of
9000ft^2. West side of fence must be wood, which costs $20/foot. East, North, South sides can be
wire that costs $5/foot. What are the dimensions of the cheapest fence? How much does it cost?
Solution:
(step 1) Write an equation I involving x and y that expresses the fact that the area is $9000.
Result: xy = 9000
(step 2) Write an equation II involving x and y and the cost C. C=10x + 25y.
At this point, realize that the math matches the math of Example #1.
Conclude that the best dimensions are x=150 and y=60. Cost is C=$3000
Maximizing Revenue and Profit
Similar to 4.6#19 A company manufactures and sells x cameras per week.
The weekly price-demand and cost equations are, respectively
p = 300 - x/30
C(x) = 90,000 + 30x.
(A) If the goal is to maximize the weekly revenue, what price should the company charge for the
cameras, and how many cameras should be produced per week? Use both calculus and noncalculus methods.Result: x=4500 cameras per week at p=$150 per camera.
(B) What is the maximum possible weekly revenue? Result: R(4500)=675,000
(C) If the goal is to maximize the weekly profit, what price should the company charge for the
cameras, and how many cameras should be produced per week? Use calculus methods.
Result x=4050 cameras per week at p=$165 per camera.
(D) What is the maximum possible weekly profit? Result: P(4050)=456,750
(E) Confirm with graph on GraphSketch.com
Day 38
X Continuing Section 4.6 Maximizing Revenue and Profit
Book Exercise 4.6#26, similar to suggested Exercise 4.6#25
(A) Coffee shop sells 1600 cups of coffee per day when the price is $2.40 per cup
Market survey predicts that for every $0.05 price reduction, 50 more cups of coffee will be sold.
How much should the coffee shop charge per cup in order to max revenue? ($2.00)
(B) A different market survey predicts that for every $0.05 price reduction, only 30 more cups of
coffee will be sold. Who much should coffee shop charge to max revenue?
(keep price $2.40, or in fact raise the price.)
Class Drill 22 Maximizing Revenue. Similar to above example, but includes costs, and asks
student to maximize revenue and then to maximize profit.
Day 39 is In-Class Exam 3 on Chapter 4
Day 40
X Section 5.1 Antiderivatives
Antiderivatives
Definition of “Antiderivative”
Words: “F is an antiderivative of f.”
Meaning: f is the derivative of F. That is, f = F’.
Arrow diagram: F  take derivative  f
Example #1 F(x) = x^3/3 is an antiderivative of f(x) = x^2. Check
Example #2 G(x) = x^3/3 +17 is an antiderivative of f(x) = x^2. Check
So there is more than one antiderivative of f(x). That is, given one function F(x) that is known to
be an antiderivative of f(x), we can make lots of other antiderivatives. Any function of the form
F(x)+C (where C is a constant that can be any real number) will also be an antiderivative of f(x).
Draw family of all antiderivatives of f(x) = x^2, showing them as vertical translations of F(x). By
contrast, draw family of horizontal translations of F(x) are not antiderivatives.
Theorem 1: These are the ONLY antiderivatives of f(x). That is, if one antiderivative of f(x) is
F(x), all other antiderivatives of f(x) are of the form F(x)+C.
(5𝑥+7)3
Example like 5.1#29 Is 𝐹(𝑥) = 3 an antiderivative of 𝑓(𝑥) = (5𝑥 + 7)2 ? NO
Example like 5.1#27 Is 𝐹(𝑥) = 𝑥 ln(𝑥) − 𝑥 + 5 an antiderivative of 𝑓(𝑥) = ln(𝑥)? YES
2
2
Example like 5.1#31 Is 𝐹(𝑥) = 𝑒 (𝑥 ) an antiderivative of 𝑓(𝑥) = 𝑒 (𝑥 ) ? No.
𝑥3
)
3
(
2
Is 𝐹(𝑥) = 𝑒
an antiderivative of 𝑓(𝑥) = 𝑒 (𝑥 ) ? No.
2
Discuss fact that 𝑓(𝑥) = 𝑒 (𝑥 ) does not have an antiderivative that can be written in a normal way
as a function 𝐹(𝑥).
Indefinite Integrals
Definition of Indefinite Integral
words: the indefinite integral of f
symbol: (indefinite integral symbol)
meaning: the symbol represents the family of all possible antiderivatives of f(x).
Remark: Since we know that given one antiderivativ F(x), we can get all other antiderivativs by
adding constants to F(x), we can represent the whole family by writing F(x) + C, where C is a
constant that can be any real number. That is, If 𝐹 ′ (𝑥) = 𝑓(𝑥) then ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶.
Arrow diagram showing F(x)  take derivative  f(x),
but reverse arrow landing in a larger set F(x) + C  take indefinite integral  f(x)
Indefinite integrals of Basic functions
Rule 1: The power Rule in Reverse indef(x^n) = (x^(n+1)/(n+1)) + C. Check it!
Example: indefinite integral of x^8. Check
Example: indefinite integral of x. Check
Example: indefinite integral of 1. Check
Example: indefinite integral of 1/x^4. Check
Example: indefinite integral of 1/x. Gives undefined result. What went wrong? Remember that the
formula is only allowed when n is not equal to -1. So how do we do it?
.
Day 41
X Continuing Section 5.1 Antiderivatives
Do Class Drill #23 Conceptual questions about antiderivatives
[1] The constant function 𝑓(𝑥) = 𝜋 is an antiderivative of the constant function 𝑘(𝑥) = 0. T F
[2] The constant function 𝑘(𝑥) = 0 is an antiderivative of the constant function 𝑓(𝑥) = 𝜋. T F
[3] The constant function 𝑘(𝑥) = 0 is an antiderivative of itself. T F
[4] The function 𝑔(𝑥) = 5𝑒 (𝑥) is an antiderivative of itself. T F
[5] The function is ℎ(𝑥) = 5𝑒 𝜋 an antiderivative of itself. T F
𝑥 𝑛+1
[6] If 𝑛 is an integer, then 𝑛+1 is an antiderivative of 𝑥 𝑛 . T F
Observe from [4] Obvious Rule 2: indefinite integral of e^x
What’s up with [6] being false?!? Review Yesterday Example: indefinite integral of 1/x. Gives
undefined result. What went wrong? Remember the Power Rule is only allowed when 𝑛 ≠ −1. so
𝑥 𝑛+1
when 𝑛 is the integer 𝑛 = −1, then 𝑛+1 is NOT an antiderivative of 𝑥 𝑛 .
Then how DO we find an antiderivative of 1/x?
Rule 3: indefinite integral of 1/x
point out that the derivative of ln(x) is 1/x. This equation is true for x>0
reviewed graph
new fact: the derivative of ln|x| is 1/x. This equation is true for all x except 0.
reviewed graphs
New rule: indefinite integral of 1/x is ln|x| + C This equation is true for x not equal to zero
Obvious question: What is ∫ ln(𝑥) 𝑑𝑥?
Rule not presented as a numbered rule in the book: ∫ ln(𝑥) 𝑑𝑥 = 𝑥 ln(𝑥) − 𝑥.
Constant multiple rule
Sum Rule
Example: indefinite integral of 5e^x. Discuss how constant 5C is replace by the constant C.
Example: Indefinite integral of 2/(x^1/3) – 5x^1/3, using radical notation. Check
.
Day 42
X Continuing Section 5.1 Antiderivatives
Do Class Drill #24 Good and Bad Indefinite Integral Solutions
today: More difficult indefinite integral problems
trick problem: indefinite integral of (1-x^2)/3x.
Different language: Find the antiderivative of the derivative 𝐶 ′ (𝑥) = 12𝑥 2 − 10𝑥.
Similar to 5.1#55. Find the particular antiderivative of the derivative 𝐶 ′ (𝑥) = 12𝑥 2 − 10𝑥
that satisfies 𝐶(0) = 4,000.
𝑑𝑓
Sim to 5.1#61. Find particular antideriv of derivative 𝑑𝑥 = 5𝑒 (𝑥) − 4 that satisfies 𝑓(0) = 3.
Now change variable to 𝑡 and function name to 𝑥(𝑡).
𝑑𝑥
Find particular antideriv of the derivative 𝑑𝑡 = 5𝑒 (𝑡) − 4 that satisfies 𝑥(0) = 3.
.
Day 43
X Section 5.2 Integration by substitution
Reviewed list of indefinite integrals in course packet
consider Reversing the chain rule
Remember that the chain rule for Derivatives is used for taking the derivative of nested functions:
𝑑
Chain Rule for Derivatives: 𝑑𝑥 𝑜𝑢𝑡𝑒𝑟(𝑖𝑛𝑛𝑒𝑟(𝑥)) = 𝑜𝑢𝑡𝑒𝑟′(𝑖𝑛𝑛𝑒𝑟(𝑥)) ∙ 𝑖𝑛𝑛𝑒𝑟′(𝑥).
Present Method from Course Packet Reference 7 The Substitution Method
Show them how different the Course Packet method is from the book’s method.
basic examples (No leftover constants)
Example: 5.2#12 (similar to #11) Find ∫(𝑥 6 + 1)4 (6𝑥 5 )𝑑𝑥. Check by differentiating.
1
Example: 5.2#18 (similar to #17) Find ∫ 5𝑥−7 (5)𝑑𝑥. Check by differentiating.
Examples where there are leftover constants
𝑡2
Example: 5.2#34 (similar to #33) Find ∫ (𝑡 3 −2)5 𝑑𝑡. Check by differentiating.
Class Drill #25 The Substitution Method
Quiz 8
.
Day 44
X Continuing Section 5.2 Integration by substitution
Other Examples
Find ∫ 𝑒 (𝑘𝑥) 𝑑𝑥. Check by differentiating.
Notice Exponential Function Rules for Derivatives and Indefinite Integrals (Reference 4)
𝑑
Derivative Rule #2 𝑑𝑥 𝑒 (𝑘𝑥) = 𝑘𝑒 (𝑘𝑥) .
𝑒 (𝑘𝑥)
Indefinite Integral Rule #2 ∫ 𝑒 (𝑘𝑥) 𝑑𝑥 = 𝑘 + 𝐶.
Applied Examples using this rule
Similar to 5.2#85 Yeast culture growing at rate 𝑊′(𝑡) = 0.3𝑒 (−.1𝑡) grams per hour.
Starting culture weighs 3 grams.
What will be the weight of the culture after 𝑡 hours? After 10 hours?
Question for class: What is the variable? What is the function?
Similar to 5.2#81 Car company is going to introduce a new model of car.
It is estimated that sales (in millions of dollars) will increase at the monthly rate of
𝑆′(𝑡) = 7 − 7𝑒 (−.1𝑡) , 0 ≤ 𝑡 ≤ 60
at 𝑡 months after the campaign has started.
What will be the total sales 𝑆(𝑡) at time 𝑡 months after the new model is introduced?
What will be estimated total sales during the first year?
Antiderivative Questions from Section 5.2
sim to 5.2#51 is 𝐹(𝑥) = 𝑥 3 𝑒 (𝑥) an antiderivative of 𝑓(𝑥) = 3𝑥 2 𝑒 (𝑥) ? Explain.
1
5.2#52 is 𝐹(𝑥) = 𝑥 an antiderivative of 𝑓(𝑥) = ln(𝑥)? Explain.
sim to 5.2#53 is 𝐹(𝑥) = (𝑥 2 + 7)5 an antiderivative of 𝑓(𝑥) = 10𝑥(𝑥 2 + 7)4 ? Explain.
sim to 5.2#55 is 𝐹(𝑥) = 𝑒 (3𝑥) + 5 an antiderivative of 𝑓(𝑥) = 𝑒 (3𝑥) ? Explain.
sim to 5.2#56 is 𝐹(𝑥) = −0.5𝑒 (−2𝑥) + 7 an antiderivative of 𝑓(𝑥) = 𝑒 (−2𝑥) ? Explain.
sim to 5.2#57 is 𝐹(𝑥) = 0.5(ln(𝑥))2 + 7 an antiderivative of 𝑓(𝑥) =
.
ln(𝑥)
𝑥
? Explain.
Day 45
X Section 5.4 The Definite Integral
Today: approximating areas by Left & Right Sums
New Idea: Area under the graph of a function from x=a to x=b.
Introduce Unsigned Area (USA)
Introduce Signed Area (SA).
for the signed area introduce name (the “Definite Integral of f from x=a to x=b”) and symbol:
𝑥=𝑏
𝑆𝑖𝑔𝑛𝑒𝑑𝐴𝑟𝑒𝑎 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥. This is an “informal definition”.
Start students on Class Drill 26: Definite Integrals for a Simple Graph. Do the first one. Let
them finish.
The class drill introduces us to the additive properties of the definite Integral.
point out that integral from x=a to x=a must be zero.
What about the integral from x=b to x=a? Define it to be the negative integral from x=a to x=b.
New Question: what do we do if graph of f is not a simple shape? How do we find the area
𝑥=𝑏
𝑆𝐴 = ∫ 𝑓(𝑥)𝑑𝑥
𝑥=𝑎
if we are not given it? For example, what would be the definite integral for curvy graph shown.
Show graph of an increasing function, with left & right endpoints shown.
We start by thinking about approximations of the area.
Define 𝐿5 = 𝑙𝑒𝑓𝑡 𝑠𝑢𝑚 𝑤𝑖𝑡ℎ 5 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒𝑠.
Discuss partition of the interval. Divide the interval 𝑎 ≤ 𝑥 ≤ 𝑏 into 5 equal subintervals. Name
the points x0 = a, … , x5 = b. Also show Δ𝑥.
𝑤𝑖𝑑𝑡ℎ_𝑜𝑓_𝑤ℎ𝑜𝑙𝑒_𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
𝑏−𝑎
Point out that the width of each subinterval is Δ𝑥 = 𝑛𝑢𝑚𝑏𝑒𝑟_𝑜𝑓_𝑠𝑢𝑏𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠 = 5
Park a rectangle on top of each subinterval
Introduce terminology of “Left Rectangle”
Find area of the individual rectangles and add them up. The resulting number is called 𝐿5 .
Generalize to more rectangles and other kinds of sums
Ln = left sum (using “left rectangles”) with n subintervals
Rn = right sum (using “right rectangles”) with n subintervals
Did not do these general formulas
define Ln to be the sum of the area of n left rectangles put on the interval 𝑎 ≤ 𝑥 ≤ 𝑏.
That is, 𝐿𝑛 = Δ𝑥 ⋅ 𝑓(𝑥0 ) + Δ𝑥 ⋅ 𝑓(𝑥1 ) + ⋯ + Δ𝑥 ⋅ 𝑓(𝑥𝑛−2 ) + Δ𝑥 ⋅ 𝑓(𝑥𝑛−1 ).
Factor out the Δ𝑥 to get 𝐿𝑛 = Δ𝑥 ⋅ (𝑓(𝑥0 ) + 𝑓(𝑥1 ) + ⋯ + 𝑓(𝑥𝑛−2 ) + 𝑓(𝑥𝑛−1 )).
And define Rn to be the sum of the area of n right rectangles put on interval 𝑎 ≤ 𝑥 ≤ 𝑏.
That is, 𝑅𝑛 = Δ𝑥 ⋅ 𝑓(𝑥1 ) + Δ𝑥 ⋅ 𝑓(𝑥2 ) + ⋯ + Δ𝑥 ⋅ 𝑓(𝑥𝑛−1 ) + Δ𝑥 ⋅ 𝑓(𝑥𝑛 ).
Factor out the Δ𝑥 to get 𝐿𝑛 = Δ𝑥 ⋅ (𝑓(𝑥1 ) + 𝑓(𝑥2 ) + ⋯ + 𝑓(𝑥𝑛−1 ) + 𝑓(𝑥𝑛 )).
Now do Class Drill 27: Estimating the Area Under a Graph Using Riemann Sums.
𝑥=𝑏
observe sandwich: If f is increasing function, then 𝐿𝑛 < 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 < 𝑅𝑛 .
𝑥=𝑏
Of course, If f is decreasing function, then 𝑅𝑛 < 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 < 𝐿𝑛 .
And in general, if 𝑓 is a mix of increasing & decreasing, we can’t predict.
.
Day 46
X Continuing Section 5.4 The Definite Integral
Today: The Definite Integral as a Limit of Sums
Review of Definite Integrals & Riem Sums from Tuesday: We were interested in finding values
𝑥=𝑏
for definite integrals:𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 the signed area under graph of f from x=a to x=b.
We introduced Riemann sums Ln and Rn that are approximations of the actual signed area. We
𝑥=𝑏
noticed in particular that if f is increasing function, then 𝐿𝑛 < 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 < 𝑅𝑛 .
Now consider area problems where the function f is given by a formula
𝑥2
𝑥=12
Let 𝑓(𝑥) = 5 + 10. We are interested in the definite integral 𝑆𝐴 = ∫𝑥=2 𝑓(𝑥)𝑑𝑥.
Compute L5 by hand. Result: 𝐿5 = 94
Confirm with Wolfram
Similary, find R5.. Used Wolfram Result: 𝑅5 = 122. NEXT TIME DO BY HAND.
𝑥=12
Observe that L5 < Actual Area A < R5. That is,94 < 𝐴 = ∫𝑥=2 𝑓(𝑥)𝑑𝑥 < 122.
𝑥=12
Observe that L5 < Actual Area A < R5. That is,94 < 𝐴 = ∫𝑥=2 𝑓(𝑥)𝑑𝑥 < 122
now consider using n=10, 100, 1000
We notice that as 𝑛 → ∞, the values of Ln & Rn seemed to be getting closer & closer together.
Inspired by that, we define the signed area SA in the following way:
Big definition:
𝑥=𝑏
𝑆𝐴 = ∫ 𝑓(𝑥)𝑑𝑥 ≝ lim 𝐿𝑛 = lim 𝑅𝑛
𝑛→∞
𝑛→∞
𝑥=𝑎
Big fact: these limits do exist and they are the same number.
Another big fact: for graphs made of simple geometric shapes, this gives the same answer as
doing the definite integral using geometric area formulas.
But this is still a little puzzling: We have only seen examples where we have used a computer to
find Ln & Rn for particular values of n, and we guessed at a rough value for the limit.
Obvious questions:
Question (1) Can we do the Riemann Sums analytically? Answer: yes, but tedious.
Question (2) Can we figure out the limit lim 𝐿𝑛 analytically? Answer: yes, but it involves a lot of
𝑛→∞
difficult math that is not taught in a class like this. (The difficulty is comparable to the level of
difficulty you encountered when computing the derivative using the limit definition.)
𝑥=𝑏
So we want an analytic way to compute 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 ≝ lim 𝐿𝑛 = lim 𝑅𝑛 but the
𝑛→∞
𝑛→∞
computation of those Riemann Sums is tedious and their limit is beyond the level of this class.
Question: What do we do?
𝑥=𝑏
Is there another way to find the value of 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 ≝ lim 𝐿𝑛 = lim 𝑅𝑛 ?
𝑛→∞
Answer that question tomorrow
𝑛→∞
Day 47
X Today: Section 5.5 The Fundamental Theorem of Calculus
Review problem from Friday:
𝑥=𝑏
We want an analytic way to compute 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 ≝ lim 𝐿𝑛 = lim 𝑅𝑛 .
𝑛→∞
𝑛→∞
Wish list of three examples where we would like to find signed area but we are unable to:
𝑥=12
𝑥2

Wish List Example #1: Find 𝑆𝐴 = ∫𝑥=2 5 + 10 𝑑𝑥. Draw picture.

Wish List Example #2: Find 𝑆𝐴 = ∫𝑥=0 4𝑒 (𝑥) 𝑑𝑥. Draw picture.

Wish List Example #3: Find 𝑆𝐴 = ∫𝑥=1 𝑥 𝑑𝑥. Draw picture.
𝑥=2
𝑥=5 2
𝑥=𝑏
In each of these three example we are not able to compute 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 because
 the graphs were not simple geometric shapes, so we cannot compute the signed area 𝑆𝐴 =
𝑥=𝑏
∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 using geometry.
 We do not know how to compute the limits of Riemann Sums (it is possible, but beyond
the level of this course), so we cannot compute the signed area using 𝑆𝐴 =
𝑥=𝑏
∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 ≝ lim 𝐿𝑛 = lim 𝑅𝑛 .
𝑛→∞
𝑛→∞
𝑥=𝑏
So what do we do? Is there another way to get value of 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥?
That brings us to Section 5.5 The Fundamental Theorem of Calculus
We have seen two uses of the integral symbol:
(1) The Indefinite Integral, having to do with antiderivative of f.
(2) The Definite Integral, having to do with the signed area under the graph of f.
The two concepts (antiderivative and signed area) seem unrelated. It should seem strange that
such similar-looking symbols would be used to denote such dissimilar mathematical concepts.
But it turns out that there is a very important relationship between the two concepts:
Given a function f(x) and an interval [a,b],
 The indefinite integral can be used to find an antiderivative F(x). This is a function.
 The definite integral represents a number that is the signed area.
𝑥=𝑏
Here is the relationship: 𝑆𝐴 = ∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎).
First test of the Fundamental Theorem: Use it on a function where we actually do know how to
find the signed area. Do Class Drill 28: The Fundamental Theorem of Calculus
𝑥=12
𝑥2
Use Fundamental Theorem for Example #1: Find 𝑆𝐴 = ∫𝑥=2 5 + 10 𝑑𝑥.
Compare result to numerical estimates from last Wednesday and Friday.
Compare to analytic result using Wolfram Alpha.
𝑥=2
Use Fundamental Theorem for Example #2: Find 𝑆𝐴 = ∫𝑥=0 4𝑒 (𝑥) 𝑑𝑥.
Compare to numerical results using Wolfram Mathworld.
Compare to analytic result using Wolfram Alpha.
𝑥=5 2
Use Fundamental Theorem for Example #3: Find 𝑆𝐴 = ∫𝑥=1 𝑥 𝑑𝑥 .
Compare to numerical results using Wolfram Mathworld.
Compare to analytic result using Wolfram Alpha.
Observe that Wolfram site says A=log(25). Why?
Quiz 9
Day 48
X Continuing Section 5.5 The Fundamental Theorem of Calculus
First Applications of the Fundamental Theorem: “Total Change Problems”
We will be thinking about a function, maybe 𝐹(𝑥), and its derivative 𝐹′(𝑥).
𝑥=𝑏
Use these symbols to rewrite the Fundamental Theorem: ∫𝑥=𝑎 𝐹′(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎).
“Total Change Problems” involve this idea:
Given 𝐹′(𝑥) (the derivative), find 𝐹(𝑏) − 𝐹(𝑎) = Δ𝐹 the change in F.
Biology Example: Bacterial Culture. t=time in hours. W(t) = weight in grams. Culture is growing
at a rate 𝑊 ′ (𝑡) = .6𝑒 .2𝑡 . How much does the weight change from t=5 hours to t=15 hours?
Illustrate with a graph and check with wolfram.
Company makes phones.
𝑥
Marginal cost is 𝐶 ′ (𝑥) = 50 − 5 for 0 ≤ 𝑥 ≤ 1000.
Find increase in cost going from production level of 200 bikes per month to 300 bikes per month.
Salvage value problem for copier 𝑉 ′ (𝑡) = 15000(𝑡 − 9) for 0 ≤ 𝑡 ≤ 7.
Find total loss in value in first two years. In second two years.
Day 49
X Continuing Section 5.5 The Fundamental Theorem of Calculus
today: 2nd Application of Fundamental Theorem: “Average Value of a Function over an Interval”
Start with a drawing and the question: How high would a rectangle have to be to enclose the same
∫
𝑥=𝑏
𝑓(𝑥)𝑑𝑥
area. Arrive at the formula for the height ℎ = 𝑥=𝑎𝑏−𝑎 .
Definition:
words: “average value of f(x) over the interval 𝑎 ≤ 𝑥 ≤ 𝑏”
𝑥=𝑏
∫𝑥=𝑎 𝑓(𝑥)𝑑𝑥
meaning: the number ℎ =
𝑏−𝑎
graphical interpretation: that the number h is the height of a rectangle sitting on the interval 𝑎 ≤
𝑥 ≤ 𝑏 that would enclose a signed area that is equal to the signed area of the graph of 𝑓 on the
same interval.
Example: Find the average value of 𝑔(𝑡) = 4𝑡 + 3𝑡 2 over the interval −2 ≤ 𝑡 ≤ 5.
Result: h=25.
.2𝑡
Example: Drug concentration after taking pill is 𝐶(𝑡) = 𝑡 2 +1 mg/liter. Find average concentration
over 1st three hours. Illustrate with drawing and check with Wolfram.
Example: Find the average value of 𝑓(𝑥) = 5√𝑥 from 𝑥 = 1 to 𝑥 = 16.
Day 50 is Exam 4 Covering Chapter 5
.
Day 51
X Section 6.1 The Area Between Two Curves
Review concept of unsigned and signed area
The Definite Integral refers to signed area.
The “Area Between Two Curves” refers to UNSIGNED AREA!! Not spelled out clearly in book.
Present Theorem 1 about the area between two curves with drawing of graph.
If top(x) and bottom(x) are continuous and 𝑏𝑜𝑡𝑡𝑜𝑚(𝑥) ≤ 𝑡𝑜𝑝(𝑥) on the interval [a,b], then the
area bounded by y=top and y=bottom for 𝑎 ≤ 𝑥 ≤ 𝑏 is given by the definite integral
𝑥=𝑏
∫ 𝑡𝑜𝑝(𝑥) − 𝑏𝑜𝑡𝑡𝑜𝑚(𝑥)𝑑𝑥
𝑥=𝑎
.
Example similar to 6.1#55 Find the area bounded by the curves 𝑦 = 𝑥 2 + 1 and 𝑦 = 2𝑥 − 2 from
x = -3 to x = 2. Discuss how we make the drawing (using transformations). Then identify top &
bottom curves. Then set up integral. Then do integral.
𝑥=𝑏
[Pause to introduce notation: 𝐻(𝑥)|𝑥=𝑎
means 𝐻(𝑏) − 𝐻(𝑎)]
Result: USA= 95/3 (Confirmed with Wolfram.)
Theorem 1 doesn’t say anything about is the issue of how you have to use the theorem:
In general, you must determine where the graphs of f and g cross, and figure out the intervals
where one graph is above the other. This breaks the domain into pieces. Then the unsigned area
between f and g is found by computing a definite integral on each piece of the domain, with the
correct “top” function..
Example with crossing graphs: similar to 6.1#55
Find area bounded by the curves
𝑦 = 𝑥 2 − 1, 𝑦 = 𝑥 − 1, 𝑥 = −2, 𝑥 = 1
Graphed. Observed that the graphs cross at x=0. So we have to do two integrals.
Result: USA=29/6. Confirmed with Wolfram
Class Drill 29: Finding the Area Bounded by Curves using Two Different Methods.
Day 52
X Continuing Section 6.1 The Area Between Two Curves
today: more difficult examples and Total Change Problems
Have Class do this Example similar to 6.1#67
Find area of region bounded by y=x^5 and y=16x.
Solution: Draw the graphs separately
Determine the x-coordinate of their intersection points. Result: x=-2,0,2
𝑥=0
𝑥=2
Set up the integrals: 𝑈𝑆𝐴 = ∫𝑥=−2 𝑥 5 − 16𝑥𝑑𝑥 + ∫𝑥=0 16𝑥 − 𝑥 5 𝑑𝑥.
Resume example: Evaluate USA, get USA=128/3.
Compare to value of signed area: Signed area = 0. Discuss why that is true.
Have Class do this Example not involving polynomial functions:
Similar to 6.1#57,58 Find area between graphs of 𝑦 = 𝑒 (𝑥) and 𝑦 = −1/𝑥 for 1 ≤ 𝑥 ≤ 2. Make
Drawing, then integrate. Result: 𝑈𝑆𝐴 = 𝑒 2 − 𝑒 − 𝑙𝑛(2).
Total change problems as area problems
like 6.2#95 Yeast culture growing at a rate w’(t)=.5e^(.3t) grams per hour.
find area between graph of w’ and the t-axis over [0,4] and interpret the results.
Day 53
X Section 6.2 Applications in Business and Economics
Not covering Probability Distributions (Show them pages 421,422)
Total Income for a Continuous Income Stream
Total income problem 7-2#22 (similar to #21)
Find the total income produced by a continuous income stream in the first 10 years if the flow rate
is f(t)=3000.
Problem #24 (similar to #23)
Illustrate with graph.
𝑒 (𝑘𝑥)
Recall useful antiderivative: Exponential Function Rule #2:∫ 𝑒 (𝑘𝑥) 𝑑𝑥 = 𝑘 + 𝐶.
Total income problem 7-2#26 (similar to #25)
Find the total income produced by a continuous income stream in the first 2 years if the flow rate
is f(t)=600e^(.06t).
Problem #28 (similar to #27)
Illustrate with graph.
Future value problems
Future Value of a continuous income stream definition & formula from Page 395
Future value problem 6.2#44 (similar to #43)
Find the future value, at 2.95% interest, compounded continuously for 6 years, of the continuous
income stream with rate of flow f(t)=2000e^(.06t)
Future value problem 6.2#42 (similar to #41)
Starting at age 30, you deposit $2000/year into an IRA account.
Treat the deposits as a continuous income stream.
The account earns 6% interest, compounded continously.
(A) How much will be in the account 35 years later, when you retire at age 65?
(B) How much of the final amount is interest?
Future value problem 6.2#48 (similar to #47)
Investment #1: Clothing store. income flow rate is f(t)=12,000/year.
Investment #2: Computer store. income flow rate is g(t) = 10,000e^.05t
Both investments earn 4% compounded continuously.
Which investement will yield a better return at time 10 years?
.
Day 54
X Section 6.2, continued
Consumers’ Surplus
Idea of Consumers’ Surplus
Definition from book
𝑥=𝑥̅
𝐶𝑆 = ∫ [𝐷(𝑥) − 𝑝̅]𝑑𝑥
𝑥=0
.
7-2#44 Find the consumers’ surplus at a price level of 𝑝̅ = 120 for the price demand equation
𝑝 = 𝐷(𝑥) = 200 − 0.02𝑥
7-2#46 Interpret the result of #44 with a graph and a description.
Producers’ Surplus
Idea of price-supply curve
Idea of Producers’ Surplus
Definition from book
𝑥=𝑥̅
𝑃𝑆 = ∫ [𝑝̅ − 𝑆(𝑥)]𝑑𝑥
𝑥=0
.
7-2#48 Find the producers’ surplus at a price level of 𝑝̅ = 55 for the price supply equation
𝑝 = 𝑆(𝑥) = 15 + .1𝑥 + 0.003𝑥 2
7-2#50 Interpret the result of #48 with a graph and a description.
Quiz 10
Day 55
X Section 6.2, continuedConsumers’ Surplus and Producers’ Surplus
Problems involving both Consumers’ & Producers’ Surplus
Equilibrium price and quantity (𝑥̅ , 𝑝̅).
Examples of finding equilibrium point
7-2#51 𝐷(𝑥) = 50 − 0.1𝑥 and 𝑆(𝑥) = 11 + 0.05𝑥.
7-2#52 𝐷(𝑥) = 25 − 0.004𝑥 2 and 𝑆(𝑥) = 5 + 0.004𝑥 2 .
Consumers’ & Producers’ Surplus problems involving finding equilibrium point
Class Drill 30 Finding Equilibrium Price and Consumers’ & Producers’ Surplus
.