MAT1330A_Ver1_82copies.pdf
LastyearMidterm2VerB.pdf
(Replacing a question about continuity)
Find the limit of f(x) as x goes to 0 from the left, justifying all steps fully and carefully. In particular, if you apply
any theorems from class, you must verify that the hypothesis is met.
Then answer: for what value of a are the two one-sided limits equal, if any?
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MAT1330 DGD WORKBOOK - ANSWER KEY TO SELECT QUESTIONS
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LEC 1 – High School Review.
1.3 # 20 Find the domain of f(x) =
√
2x− 7. Ans: [72 ,∞)
like 1.4 # 17 Find the formulas for the composition f ◦ g and g ◦ f and the product of the functions f
and g; simplify where possible.
f(x) =
x− 1
x+ 1
g(x) = 1/(2x)
Ans: f ◦ g(x) = 1−2x1+2x g ◦ f(x) =
x+1
2x−2 fg(x) =
x−1
2x2+2x
1.4 #33 Sketch the graph, state the domain and range, decide if the function has an inverse and if
so, find it: f(x) = 3
√
x+ 4.
Ans: D : (−∞,∞) R : (−∞,∞) f−1(x) = (x− 4)3
2.2 # 24 Solve 4e2x+1 = 20. Ans: x = 12(ln(5)− 1)
2.2 #26 Solve 4e2x+3 = 7e3x−2. Ans: x = ln 4 + 5− ln 7
2.2 #32 Express y = 0.27x in base e. Ans: y = e(ln(0.27)x
2.2 #42 Solve ln(ln(x)) = 0. Ans: x = e
LEC 2 – High School Review.
2.3 # 68 Graph the function. Give the average, max, min, amplitude, period and phase and mark
them on the graph:
f(x) = 3 + 4 cos
(
2π
(
x− 1
5
))
.
Ans: period P = 5 phase φ = 1 amplitude A = 4 mean M = 3
min M −A = −1 max M +A = 7
Inequalities
Find all solutions to the inequality:
x+ 2
2x− 1
1
x+ 7
. Ans: all x ∈ (−7, 12)
LEC 3 – Intro to DTDS.
3.1 #2 Write the updating function f associated with the following DTDS, and evaluate it at the
given arguments. Is it a linear DTDS?
mt+1 =
m2t
mt + 2
; Evaluate f at mt = 0,mt = 8,mt = 20.
Ans: up. fun. f(x) = x
2
x XXXXXXXXXX
200
11
1Last updated: July 22, 2021.
1
3.1 #12 Write the updating function f associated with the following DTDS. Is it a linear DTDS?
Mt+1 = 0.75Mt + 2
Determine the backward DTDS associated to the above DTDS. Use the backward DTDS to
find the value M0 (the value at the previous time step) given that M1 = 16.
Ans: up. fun. f(x) = 0.75x + 2 (linear) f−1(x) = 43x −
8
3 (up. fun. for backward DTDS)
M0 =
56
3
3.1 #18 & 22 Graph some values of the following DTDS, starting with the given initial condition:
`t+1 = `t − 1.7 with initial value `0 = 13.1 cm
Write down a formula for the general solution and sketch its graph. Sketch the graph of
the updating function. Label the axes for each graph!
Ans: general solution lt = 13.1− 1.7t graphical answers omitted
3.1 #19 & 23 Graph some values of the following DTDS, starting with the given initial condition:
nt+1 = 0.5nt with initial value n0 = 1200
Write down a formula for the general solution and sketch its graph. Sketch also the graph
of the updating function.
Ans: up. fun. f(x) = 0.5x general solution xt = (0.5t XXXXXXXXXXgraphical answers omitted
LEC 4 – Fixed Points and Cobwe
ing.
3.2 like #6 Given the DTDS governing the daily dose of a drug, Mt+1 = 0.75Mt+2, do three iterations
of a cobweb starting at M0 = 16 mg/L. Then plot the solution you found this way on a
graph of t vs Mt. Compare with the general solution formula we proved in class.
Ans: omitted.
3.2 #8 & 26 Graph the updating function underlying the DTDS zt+1 = 0.9zt + 1. Then cobweb fou
steps, starting from z0 = 3. Label the axes!
Next: solve for all fixed points, and classify their stability using cobweb diagrams.
Ans: fixed point: x∗ = 10 (stable) graphical answers omitted.
Ex. Graph the updating function underlying the DTDS zt+1 = 1.1zt − 1. Then cobweb fou
steps, starting from z0 = 3. Label the axes!
Next: solve for all fixed points, and classify their stability using cobweb diagrams.
Ans: fixed point: x∗ = 10 (unstable) graphical answers omitted.
3.2 #12 & 30 Using the graph of the updating function underlying the DTDS xt+1 =
xt
xt − 1
, cobweb fou
steps, starting from x0 = 3. (Restrict your updating function to the domain x > 1.) Label
the axes!
Next: solve for all fixed points, and classify their stability using cobweb diagrams.
Ans: one fixed point: x∗ = 2 (keep in mind: domain is restricted to x > 1)
Based on cobweb, solutions that start near x∗ = 2 do not approach x∗ = 2, so we conside
this an unstable fixed point.
Note: textbook calls this “stable” because nea
y solutions are not moving away from
x∗ = 2, but in MAT1330, we defined “stable” as “all nea
y solutions must move closer to
the fixed point”.
2
Ex. Consider the DTDS xt+1 = −15x
2
t + 2xt. Cobweb this DTDS starting at x0 = 2.
Next: solve for all fixed points, and classify them according to their stability.
Ans: two fixed points: x∗ = 0 (unstable) x∗ = 5 (stable)
LEC 5 – Stability of Fixed Points.
3.2 like #14 (or 3.4 #16) Given the following graph of f , which is the updating function of a DTDS,
determine the number of fixed points of the DTDS and determine their stability using
cobwe
ing. Write you conclusions in full sentences.
Ans: two fixed points x∗1 ≈ 1.3 (stable) x∗2 ≈ 5.5 (unstable)
3.2 like #30 Find the equili
ia of the following DTDS. Use cobwe
ing to check each equili
ium fo
stability.
xt+1 =
2xt
xt − 1
(x > 1).
Ans: fixed point x∗ = 3 (stable)
3.3 #28 In 1990 there were about 5000 southern mountain caribou in BC. In 2009, only about 1900
emained. Assume the annual per capita decline is constant. How long until the popula-
tion falls below m = 500 (which is a level, below which it is expected the species will go
extinct)?
Ans: when t > 19 ln(1/10)ln(19/50) years have passed since XXXXXXXXXXis the year for which t = 0)
3.4 #14 Find all nonnegative equili
ia of the following DTDS, where a is some real positive pa-
ameter:
xt+1 =
xt
a+ xt
.
(Extra biza
e: what happens if a = 0?)
Ans: fixed point(s) x∗ = 0 and x∗ = 1−a. Since a > 0, we would need a ≤ 1 in order for x∗ = 1−a
to be non-negative
If a = 0, then f(x) would degenerate to f(x) = 1 for all x.
LEC 6 – Limits and Continuity.
4.2 (typical question) Sketch the graph and decide if the left-hand and right-hand limits at
a = 0 exist, and then decide if the limit at a = 0 exists.
f(x) =
{
x+ 1 if x ≤ 1
x2 if x > 1
Ans: limx→ 0−f(x) = 1 limx→ 0+f(x) = 1 limx→ 0f(x) = 1
Ans: limx→ 1−f(x) = 2 limx→ 1+f(x) = 1 limx→ 0f(x)DNE
4.2 like #13 Sketch the graph and decide if the left-hand and right-hand limits at a = 0 exist, and then
decide if the limit exists.
f(x) =
{
|x+ 1|+ 1 if x ≤ 0
|x− 2| if x > 0
Ans: limx→ 0−f(x) = 2 limx→ 0+f(x) = 2 limx→ 0f(x) = 2
3
Evaluate each of the following limits, showing all your steps:
4.2 #48 lim
x→0
x2 − 3x
x3 − 9x
Ans: 13
4.2 #48 lim
x→4
√
x− 2
4− x
Ans: −14
4.4 #36 Sketch the graph and discuss continuity of
f(x) =
x2 − 4
x− 2
if x 6= 2
0 if x = 2
Ans: f is discontinuous at x = 2 because lim
x→2
f(x) = 4 but f(2) = 0.
LEC 7 – Infinite Limits & Limits at Infinity. Evaluate each of the following limits, showing all of
your steps:
4.3 #12 lim
x→1+
x
x2 − 1
Ans: +∞ (DNE)
4.3 #14 lim
x→−7+
√
1
x+ 7
Ans: +∞ (DNE)
4.3 #36 lim
x→∞
0.7x Ans: 0
4.3 #43 lim
x→∞
x3 − 6x+ 4
3− x3
Ans: −1
4.3 like #45 lim
x→∞
(x− 1)(x− 3)(x− 5)
x2 − 4
Ans: ∞ (DNE)
4.3 #50 lim
x→−∞
ln(3− x3) Ans: ∞ (DNE)
4.4 # 6 Find a formula for a function g(x) that makes the composition sin(g(x)) discontinuous at
x = π.
Ans: many answers possible. One possibility: g(x) =
{
0 if x > π
π/2 if x ≤ π
Ex. (like Course Guide Lecture 7 question 7) Let f(x) =
x2 − 4x+ 3
(x− 1)3
if x 6= ±1
x+ b if x = ±1
Find the limit of f as x → 1 and as x → −1. Is there a value of b that makes f continuous
at x = 1? Is there a value of b that makes f continuous at x = −1?
Ans: lim
x→1−
f(x) = −∞ so lim
x→1
f(x) DNE. f(1) = 1 + b. There is no b that makes f continuous at x = 1.
lim
x→−1−
f(x) = −1 lim
x→−1+
f(x) = −1 lim
x→−1
f(x) = −1 f(−1) = −1 +
If b = 0, then f would be continuous at x = −1.
LEC 8 – The Derivative: Definition & Basic Rules.
Let f be a function defined on a interval around x.
The derivative of f at x is, by definition,
4
Ans: f ′(x) = lim
h→0
f(x+ h)− f(x)
h
if this limit exists.
The derivative of f(x) at a point a is, by definition,
Ans: f ′(a) = lim
h→0
f(a+ h)− f(a)
h
if this limit exists.
Using the definition, compute the derivative of each of the following functions:
a. f(x) = 2 +
√
3x+ 1 Ans: f ′(x) = 3
2
√
3x+1
. g(x) =
3
4 + 2x
Ans: g′(x) = − 6
(4+2x)2
c. h(x) =
√
x2 + 1 Ans: h′(x) = x√
x2+1
Using the rules of differentiation and simplifications where appropriate, compute the de-
ivative