Notes I typed using Microsoft Word for pre-calculus and calculus exams. Most of the images were also created by me. I shared them with other students in my class to increase their chance of success as well. Upon completion of the courses I donated them to the math center to help other math students.

1. Math 141 Midterm Notes (Winter 2016)
|Instructor: Russel Sherif | Author: William Faber |
Adding complex numbers:
(−6 + 5i) + (−6 − 8i) = −6 + 5i −6 − 8i = -12-3i
Subtracting complex numbers:
(−7 + 5i) - (−6 − 8i) = -7 + 5i +6 + 8i =
-1 + 13i
Multiplying complex numbers:
“F.O.I.L” (−4+3i)(2−3i)  -8+12i+6i-9i2

“
i2
= -1”  -8+12i+6i+9 = 1+18i
Rooting negative numbers:
√−49  √(49)(−1) = 7i
√−27
3
 √(−3)(−3)(−3)
3
= -3
Find the vertex: [Formula = a(x-h)2
+k]
f(x) = −2(x − 6)2
− 4  a(x-(-6))2
+ (-4) 
Vertex = (6,-4)
X2
= y
X2
= 1  √𝑥2 = √1  x = ±1
Quadratic formula:
−b ± √b2 − 4(a)(c)
2(a)
Object dropped @ height of 270 feet falls:
[h(t)=270−16t2
]
t is seconds & h(t) is feet.
Height of the object after 0.5 seconds
h(t)=270−16(0.5)2
= 266
Time for object to hit the ground
0 = 270-16t2
 16𝑡2
= 270 
16𝑡2
16
=
270
16

√𝑡2= √
270
16
= 4.11 rounded
Find the zeros on a graph:
Crosses x axis @ (-2, 3)
List the zeros: y = x(2x+3)
(X = 0)(2x+3 = 0) 
2𝑥
2
=
−3
2
 x = -
3
2
X= (-
3
2
, 0)
List the zeros: f(x) = x3
+ 4x2
+ 3x
x(x2
+ 4x + 3)  (x=0)(x2
+ 4x + 3 = 0)
x =
−(4)±√42−4(1)(3)
2(1)
 x = 0,1,3
Degree 3 polynomial having zeros -1, 1 &
6. Coefficient of x3
equal 1:
“F.O.I.L” (x+1)(x-1)(x-6) x2
-x+x-1(x-6) 
(x2
-1)(x-6)  x3
-6x2
–x +6
Factor completely given that one of the
roots is 2:
y3
+ 2y2
− 68y + 120
2 1 2 -68 120
2 8 -120
1 4 -60 0
(y-2)(y2
+4y-60)  Quadratic formula 
(y-2)(y+10)(y-6)
Function & one to one function:
Function = passes the vertical line test
(crosses only once)
One to one function = passes the vertical
and horizontal line test (crosses only once)
𝒇(𝒙) = 𝒙 𝟐
− 𝟏𝟐𝒙 & 𝒈(𝒙) = 𝒙 + 𝟏𝟎:
𝑓 + 𝑔 = (𝑥2
− 12𝑥) + (𝑥 + 10) =
𝑥2
− 11𝑥 + 10
𝑓 – 𝑔 = (𝑥2
− 12𝑥) − ( 𝑥 + 10) =
𝑥2
− 13𝑥 − 10
𝑓𝑔 = (𝑥2
− 12𝑥)(𝑥 + 10) =
𝑥3
− 2x2
− 120x
𝑓
𝑔
=
𝑥2−12𝑥
𝑥 + 10
Use the graphs of 𝒇 and 𝒈 to evaluate
(𝒇 ∘ 𝒈)(𝟐) and (𝒈 ∘ 𝒇)(𝟐):
f◦g(2) or “f(g(2))” g(2) = 4, so f(4) = 3
g◦f(2) or “g(f(2))”  f(2) = -2, so g(-2) = 1
Find 𝒇(𝒈(𝒙)) & 𝒈(𝒇(𝒙)) :
𝒇(𝒙) = 𝟐𝒙 + 𝟒 & 𝒈 (𝒙) = 𝒙 𝟐
+ 𝟔𝒙 + 𝟕
f(g(x)) = 2(𝑥2
+ 6𝑥 + 7) + 4 =
2x2
+12x +18
g(f(x)) = (2𝑥 + 4 ) 2
+ 6(2𝑥 + 4 ) + 7 
(2𝑥 + 4 )(2𝑥 + 4 ) + 6(2𝑥 + 4 ) + 7 =
4x2 + 28x + 47
Functions & Invers functions using
graph:
𝑓(0) = 4
If 𝑓(𝑥) = 0 , 𝑡ℎ𝑒𝑛 𝑥 = 2
𝑓 −1 (0) = 2
If 𝑓 −1 (𝑥) = 0, 𝑡ℎ𝑒𝑛 𝑥 = 4
If 𝒇(𝒙) = 𝟑𝒙 + 𝟏𝟓 , then find
𝒇−𝟏(𝒙) :
𝑥 = 3𝑦 + 15  𝑥 − 15 = 3𝑦 
𝑥−15
3
=
3𝑦
3
 𝒇−𝟏
(𝑥) =
𝑥−15
3
Horizontal asymptotes
𝑓(𝑥) =
−5𝑥 + 4𝑥3
− 1
3𝑥3 + 5𝑥2 − 3
Coefficient of the highest degree over the
coefficient of the same highest degree.
Top degree < Bottom degree then y = 0
Bottom degree < Top degree, then there
are no horizontal asymptotes.
The y-intercept, Roots at A & B, Vertical
asymptotes
𝑓(𝑥) =
2𝑥2
+ 1𝑥 − 1
2𝑥2 + 1𝑥 − 3
Y-intercept in this case is
−1
−3
 0,
1
3
Roots at A & B (Factor the function in the
numerator to find the zeros)
Vertical asymptotes (Factor the function in
the denominator to find the zeros)

2. Math 141 Final Exam Notes (Winter 2016)
|Instructor: Russel Sherif | Author: William Faber |
Logarithmic to exponential:
“Base to the answer power = the number”
log525 = 2  52
=25
log42 =
1
2
 41/2
=2
loguV=Z  Uz
=V
Exponential to logarithmic:
“Reverse process as above”
29
=512  log2512=9
1001/2
=10  log10010 =
1
2
ab
= c  logac = b
Evaluate logarithms:
log327  3x
= 27  X = 3
log648  64x
= 8  X =
1
2
Expressions to single Logarithm:
log4(5) + log4(13)  log4(5∙13) 
log4(65)
log5(63) – log5(7)  log5(
63
7
)  log5(9)
log7(2) + log7(8) + log7(a+3) 
log7(2∙8∙[a+3])  log7(16a + 48)
Expand single logarithm:
“Reverse process as above”
log5(u∙v)  log5(u) + log5(v)
log5(25∙x)  log5(25) + log5(x)
log3(
𝑥
𝑦
) log3(x) – log3(y)
log7√ 𝑥
4
 log7x1/4

1
4
log7(x)
Solve for x in exponential equations:
5x
= 27  log5x
= log27  x log5 = log27
 x =
𝑙𝑜𝑔27
𝑙𝑜𝑔5
 x = 2.048
ex+2
= 100  lnex+2
= ln100  (x+2) lne =
ln100  x =
𝑙𝑛100
𝑙𝑛𝑒
− 2 x = 2.605
20 + 27(9x
) = 55  27(9x
) = 35  9x
=
35
27
x log9 = log
35
27
 x =
log 35/27
log 9
 x =
0.118
Solve for x in logarithmic equations:
ln(x) = 8  loge(x) = 8  x = e8
 x =
2980.958
log2(25-x) = 3  25-x = 23
 25-x = 8  x =
25-8  x = 17
log2(x) + log2(x-12) = 6  log2(x2
-12x) = 6 
26
= x2
-12x  64 = x2
– 12x  x2
– 12x – 64 =
0  (x+4)(x-16) = 0  x= 16 (x must be
greater than zero)
-6 ∙ log8(7x) + 5 = 3  log8(7x) =
3−5
−6

log8(7x) =
1
3
 7x = 81/3
 7x = 2  x =
2
7
Investment of $10,000 earns 14% annually &
compounds continuously. What is investment
value in 20 years?
Formula is: A = P ∙ ert
A = 10,000 (e)(.14)(20)
 A = $164,446.47
Equation of circle centered at (3, -5) & diameter
of 16:
Divide diameter by 2 to get radius then change to
formula: (x-h)2
+ (y-k)2
= r2
(x-3)2
+ (y-(-5))2
= 82
 (x-3)2
+ (y+5)2
= 64
Equation of circle in standard form:
Formula is: (x-h)2
+ (y-k)2
= r2
(x-3)2
+ (y+5)2
= 16
Find center & radius of circle with standard form
of:
(x+7)2
+ (x-3)2
= 576  h = -7, k = 3, r = √576 = 24
Find center and radius of circle with equation of:
x2
+ y2
– 10x – 14y - 26 = 0  (x2
– 10x) + (y2
–
14y) = 26  (x2
– 10x +25) + (y2
– 14y +49) = 26
+25 +49  (x- 5)2
+ (y- 7)2
= 100  center = (5, 7)
& radius = 10
Vertex of Parabola:
f(x) = -3(x-10)2
+ 8  vertex: (10, 8)
Equation of parabola in vertex form: y-k = (x-h)2
:
y = x2
- 12x – 27  1[(x2
- 12x +36) -36 – 27]  (x-
6)2
- 63  y + 63 = (x-6)2
Find the vertex of:
f(x) = -3x2
-18x -21  f(x) = -3(x2
+ 6x + 7)  f(x) =
-3[(x2
+ 6x +9) – 9 +7]  f(x) = -3[(x+3)2
– 2] 
f(x) = -3(x+3)2
+ 6  Vertex = (-3,6)
Standard form equation from ellipse (horizontal
major axis): Formula is:
(𝒙−𝒉) 𝟐
𝒂 𝟐 +
(𝒚−𝒌) 𝟐
𝒃 𝟐 = 𝟏
(𝒙−𝟎) 𝟐
𝟓 𝟐 +
(𝒚−𝟎) 𝟐
𝟐 𝟐 = 𝟏 
𝒙 𝟐
𝟐𝟓
+
𝒚 𝟐
𝟒
= 𝟏
Standard form equation from ellipse (vertical
major axis): Formula is:
(𝒙−𝒉) 𝟐
𝒃 𝟐 +
(𝒚−𝒌) 𝟐
𝒂 𝟐 = 𝟏
(𝒙−𝟎) 𝟐
𝟑 𝟐 +
(𝒚−𝟎) 𝟐
𝟒 𝟐 = 𝟏 
𝒙 𝟐
𝟗
+
𝒚 𝟐
𝟏𝟔
= 𝟏
Equation of ellipse in standard form:
16x2
+ 49y2
= 784 
𝟏𝟔𝒙 𝟐
𝟕𝟖𝟒
+
𝟒𝟗𝒚 𝟐
𝟕𝟖𝟒
=
𝟕𝟖𝟒
𝟕𝟖𝟒

𝒙 𝟐
𝟒𝟗
+
𝒚 𝟐
𝟏𝟔
= 𝟏

3. Math 142 Midterm Notes (Spring 2016)
|Instructor: Will Webber | Author: William Faber |
The Unit circle:
Basic Trig Identities:
Sin =
𝑶𝒑𝒑
𝑯𝒚𝒑
=
𝒄𝒐𝒔
𝒄𝒐𝒕
= cos tan
Cos =
𝑨𝒅𝒋
𝑯𝒚𝒑
=
𝒔𝒊𝒏
𝒕𝒂𝒏
= sin cot
Tan =
𝒐𝒑𝒑
𝒂𝒅𝒋
=
𝒔𝒊𝒏
𝒄𝒐𝒔
Cot =
𝑨𝒅𝒋
𝒐𝒑𝒑
=
𝒄𝒐𝒔
𝒔𝒊𝒏
=
𝟏
𝒕𝒂𝒏
Sec =
𝒉𝒚𝒑
𝒂𝒅𝒋
=
𝒕𝒂𝒏
𝒔𝒊𝒏
=
𝟏
𝒄𝒐𝒔
Csc =
𝒉𝒚𝒑
𝒐𝒑𝒑
=
𝒄𝒐𝒕
𝒄𝒐𝒔
=
𝟏
𝒔𝒊𝒏
Sin csc = 1 Cos sec = 1 Tan cot = 1
𝒔𝒊𝒏 𝟐
+ 𝒄𝒐𝒔 𝟐
= 𝟏
𝒕𝒂𝒏 𝟐
+ 𝟏 = 𝒔𝒆𝒄 𝟐
𝟏 + 𝒄𝒐𝒕 𝟐
= 𝒄𝒔𝒄 𝟐
𝟏 − 𝒄𝒐𝒔 𝟐
= 𝒔𝒊𝒏 𝟐
𝟏 − 𝒔𝒊𝒏 𝟐
= 𝒄𝒐𝒔 𝟐
𝒔𝒆𝒄 𝟐
− 𝟏 = 𝒕𝒂𝒏 𝟐
𝒔𝒆𝒄 𝟐
− 𝒕𝒂𝒏 𝟐
= 𝟏
𝒄𝒔𝒄 𝟐
− 𝟏 + 𝒄𝒐𝒕 𝟐
𝒄𝒔𝒄 𝟐
− 𝒄𝒐𝒕 𝟐
= 𝟏
Angles:
Complementary: Either of two angles whose sum is 90°
Supplementary: Either of two angles whose sum is 180°
Acute: Angle smaller than 𝟗𝟎○
Obtuse: Angle larger than 𝟗𝟎○
Formulas:
Degrees to Radians:
𝒙○ 𝝅
𝟏𝟖𝟎○
Radians to Degrees: 𝐫𝐚𝐝 ∗
𝟏𝟖𝟎
𝝅
Arc Length formula (angles in radians) 
s= arc length & r = radius  𝒔 = Ɵ𝒓
Revolutions to Angular speed:
Rev/min to radians/min 
𝒙 𝒓𝒆𝒗
𝒎𝒊𝒏
∗
𝟐𝝅 𝒓𝒂𝒅
𝒓𝒆𝒗
Angular speed to linear speed:
𝒙𝝅 𝒓𝒂𝒅
𝒎𝒊𝒏
∗
𝒓 𝒊𝒏
𝒓𝒂𝒅
Linear speed to angular speed:
𝒙𝝅 𝒊𝒏
𝒎𝒊𝒏
∗
𝒓𝒂𝒅
𝒙 𝒊𝒏
Area:
𝒃∗𝒉
𝟐
or
𝟏
𝟐
𝒂𝒃𝒔𝒊𝒏𝑪
Law of sines:
𝒂
𝒔𝒊𝒏 𝑨
=
𝒃
𝒔𝒊𝒏 𝑩
=
𝒄
𝒔𝒊𝒏 𝑪
Law of cosines:
𝒄 𝟐
= 𝒂 𝟐
+ 𝒃 𝟐
− 𝟐𝒂𝒃𝒄𝒐𝒔𝑪 (always use to find the big angle)
Graphing trig functions:
1.) 𝑨𝒔𝒊𝒏(𝑩(𝒙 − 𝑪)) + 𝑫
2.) 𝑨𝒄𝒐𝒔(𝑩(𝒙 − 𝑪)) + 𝑫
Definition of One Radian:
One radian is the angle made at the center of a circle by an arc
whose length is equal to the radius of the circle.
Basic Identities problem (simplify):
𝒄𝒐𝒕(𝒕)
𝒄𝒔𝒄(𝒕) − 𝒔𝒊𝒏(𝒕)
=
𝒄𝒐𝒔(𝒕)
𝒔𝒊𝒏(𝒕)
𝟏
𝒔𝒊𝒏(𝒕)
−
𝒔𝒊𝒏(𝒕)
𝟏
=
𝒄𝒐𝒔(𝒕)
𝒔𝒊𝒏(𝒕)
𝟏 − 𝒔𝒊𝒏 𝟐(𝒕)
𝒔𝒊𝒏(𝒕)
=
𝒄𝒐𝒔(𝒕)
𝒔𝒊𝒏(𝒕)
𝒄𝒐𝒔 𝟐(𝒕)
𝒔𝒊𝒏(𝒕)
=
𝒄𝒐𝒔(𝒕)
𝒔𝒊𝒏(𝒕)
⋅
𝒔𝒊𝒏(𝒕)
𝒄𝒐𝒔 𝟐(𝒕)
=
𝟏
𝒄𝒐𝒔(𝒕)
= 𝒔𝒆𝒄(𝒕)
Function graphs (Identification):
𝒚 = 𝒕𝒂𝒏(𝒙) 𝒚 = 𝒄𝒔𝒄(𝒙)
𝒚 = 𝒄𝒐𝒕(𝒙) 𝒚 = 𝒔𝒆𝒄(𝒙)
Function graphs (examples):
𝒚 = 𝟐𝒔𝒊𝒏(𝟑𝒙) 𝒚 = 𝟐𝒔𝒊𝒏(−𝟐𝒙) + 𝟐
𝒚 = 𝟑𝒄𝒐𝒔(−𝟐𝒙) − 𝟐 𝒚 = 𝟒𝒔𝒊𝒏(−𝒙) − 𝟏
𝒚 = 𝟒𝒔𝒊𝒏(𝟐𝒙) + 𝟐 𝒚 = 𝟒𝒄𝒐𝒔(𝒙) − 𝟐
The curve below is the graph of a sinusoidal function.
𝑭(𝒙) = −𝟐𝒔𝒊𝒏 (
𝝅
𝟑
(𝒙 − 𝟐))
It goes through the points (-4,0) and (2,0).
Law of sines (Ambiguous Case):
If the information given is side-side-angle than this happens…
𝑺𝒊𝒏𝑪
𝟕
=
𝒔𝒊𝒏(𝟓𝟐 𝒐
)
𝟔
→ 𝒔𝒊𝒏𝑪 =
𝟕𝒔𝒊𝒏(𝟓𝟐 𝒐
)
𝟔
→ 𝒔𝒊𝒏𝑪 =. 𝟗𝟏𝟗…
Case #1
𝒔𝒊𝒏−𝟏(. 𝟗𝟏𝟗…) →
𝑪 = 𝟔𝟔. 𝟖...𝒐
Case #2
𝟏𝟖𝟎 − 𝟔𝟔. 𝟖….
→
𝑪 = 𝟏𝟏𝟑. 𝟏𝟕 𝒐
𝑩 = 𝟏𝟖𝟎 − 𝟓𝟐 − 𝟔𝟔. 𝟖…
= 𝟔𝟏. 𝟏𝟕 𝒐
→
𝒃
𝒔𝒊𝒏(𝟔𝟏. 𝟏𝟕 𝒐)
=
𝟔
𝒔𝒊𝒏(𝟓𝟐 𝒐)
→ 𝒃 = 𝟔. 𝟔𝟕
𝑩 = 𝟏𝟖𝟎 − 𝟓𝟐 − 𝟏𝟏𝟑. 𝟏𝟕
= 𝟏𝟒. 𝟖𝟑 𝒐
→
𝒃
𝒔𝒊𝒏(𝟏𝟒. 𝟖𝟑 𝒐)
=
𝟔
𝒔𝒊𝒏(𝟓𝟐 𝒐)
→ 𝒃 = 𝟏. 𝟗𝟒𝟖𝟖
Law of cosines (example of finding sides):
1.) 𝜽 = 𝟏𝟖𝟎 𝒐
− 𝟕𝟒 𝒐
= 𝟏𝟎𝟔 𝒐
2.) 𝒄 = 𝟑𝟒𝟎𝟎𝒄𝒔𝒄(𝟕𝟒 𝒐)
=
𝟑𝟒𝟎𝟎
𝒔𝒊𝒏(𝟕𝟒 𝒐)
= 𝟑𝟓𝟑𝟕. 𝟎𝟏𝟖...
3.) 𝒂 = √𝟖𝟏𝟎 𝟐 + 𝟑𝟓𝟑𝟕. 𝟎𝟏𝟖…𝟐 − 𝟐(𝟖𝟏𝟎)(𝟑𝟓𝟑𝟕. 𝟎𝟏𝟖…)𝒄𝒐𝒔(𝟏𝟎𝟔 𝒐)
= 𝟑𝟖𝟒𝟎. 𝟎𝟓𝟎…
Height of a mountain problem
Solve both equations for h, then make them equal to each
other and solve for x.
#1) 𝒕𝒂𝒏(𝟐𝟕 𝒐) =
𝒉
𝟐𝟎𝟎𝟎+𝒙
→
𝒉 = (𝟐𝟎𝟎𝟎 + 𝒙)𝒕𝒂𝒏(𝟐𝟕 𝒐
)
#2) 𝒕𝒂𝒏𝟑𝟎 𝒐
=
𝒉
𝒙
→
𝒉 = 𝒙 𝒕𝒂𝒏(𝟑𝟎 𝒐
)
(𝟐𝟎𝟎𝟎 + 𝒙)𝒕𝒂𝒏(𝟐𝟕 𝒐) = 𝒙 𝒕𝒂𝒏(𝟑𝟎 𝒐) →
𝟐𝟎𝟎𝟎 𝒕𝒂𝒏(𝟐𝟕 𝒐) = 𝒙(𝒕𝒂𝒏𝟑𝟎 𝒐
− 𝒕𝒂𝒏𝟐𝟕 𝒐) →
𝒙 =
𝟐𝟎𝟎𝟎 𝒕𝒂𝒏(𝟐𝟕 𝒐)
(𝒕𝒂𝒏𝟑𝟎 𝒐 − 𝒕𝒂𝒏𝟐𝟕 𝒐)
→ 𝒙 = 𝟏𝟓𝟎𝟐𝟒. 𝟕𝟒𝟗𝟏
𝒕𝒂𝒏(𝟑𝟎 𝒐) =
𝒉
𝟏𝟓𝟎𝟐𝟒. 𝟕𝟒𝟗𝟏
→
𝒉 = (𝟏𝟓𝟎𝟐𝟒. 𝟕𝟒𝟗𝟏)𝒕𝒂𝒏(𝟑𝟎 𝒐) →
𝒉 = 𝟖𝟔𝟕𝟒. 𝟓𝟒 ft.
Solving big nice angles (without a calculator):
𝟏𝟔𝒄𝒐𝒔 𝟐(−𝟏𝟓𝟎)
𝟏
+
𝟔
𝒔𝒊𝒏 𝟐(−𝟑𝟎𝟎)
→
𝟏𝟔(−
√𝟑
𝟐
)
𝟐
𝟏
+
𝟔
(−
√𝟑
𝟐
)
𝟐 →
(𝟏𝟔(−
√𝟑
𝟐
)
𝟐
+𝟔)(−
√𝟑
𝟐
)
𝟐
+𝟔
(−
√𝟑
𝟐
)
𝟐 =
(𝟏𝟔(
𝟑
𝟒
))(
𝟑
𝟒
)+𝟔
𝟑
𝟒
=
𝟏𝟐(
𝟑
𝟒
)+𝟔
𝟑
𝟒
=
𝟗+𝟔
𝟑
𝟒
=
𝟏𝟓
𝟑
𝟒
=
𝟏𝟓
𝟏
⋅
𝟒
𝟑
=
𝟔𝟎
𝟑
= 𝟐𝟎
Finding the area between the circles (using law of cosines):
1.) Find all sides of the triangle.
2.) Find the area of the triangle.
𝟏𝟏. 𝟒 𝟐
= 𝟏𝟑. 𝟔 𝟐
+ 𝟏𝟒. 𝟖 𝟐
− 𝟐(𝟏𝟑. 𝟔)(𝟏𝟒. 𝟖)𝒄𝒐𝒔(𝑪) →
𝒄𝒐𝒔(𝑪) =
𝟏𝟑.𝟔 𝟐+𝟏𝟒.𝟖 𝟐−𝟏𝟏.𝟒 𝟐
𝟐(𝟏𝟑.𝟔)(𝟏𝟒.𝟖)
→
𝑪 = 𝒄𝒐𝒔−𝟏
(
𝟏𝟑.𝟔 𝟐+𝟏𝟒.𝟖 𝟐−𝟏𝟏.𝟒 𝟐
𝟐(𝟏𝟑.𝟔)(𝟏𝟒.𝟖)
) = 𝟒𝟕. 𝟎𝟗…𝒐
𝑨𝒓𝒆𝒂 =
𝟏
𝟐
(𝟑. 𝟔)(𝟏𝟒. 𝟖)𝒔𝒊𝒏(𝟒𝟕. 𝟎𝟗𝟖𝟐…𝒐 ) = 𝟕𝟑. 𝟕𝟐…
3.) Find the remaining angles.
𝒄𝒐𝒔(𝑨) =
𝟏𝟏.𝟒 𝟐+𝟏𝟒.𝟖 𝟐−𝟏𝟑.𝟔 𝟐
𝟐(𝟏𝟏.𝟒)(𝟏𝟒.𝟖)
→
𝑨 = 𝒄𝒐𝒔−𝟏
(
𝟏𝟏.𝟒 𝟐+𝟏𝟒.𝟖 𝟐−𝟏𝟑.𝟔 𝟐
𝟐(𝟏𝟏.𝟒)(𝟏𝟒.𝟖)
) = 𝟔𝟎. 𝟗𝟏…𝒐
𝑩 = 𝟏𝟖𝟎 𝒐
− 𝟔𝟎. 𝟗𝟏…𝒐
− 𝟒𝟕. 𝟎𝟗𝟖𝟐…𝒐
= 𝟕𝟏. 𝟗𝟖…𝒐
4.) Fin the area of each sector, add them together, then
subtract from the triangle.
𝝅(𝟔. 𝟑) 𝟐
(
𝟔𝟎.𝟗𝟏…𝒐
𝟑𝟔𝟎 𝒐
) + 𝝅(𝟓. 𝟏) 𝟐
(
𝟕𝟏.𝟗𝟖…𝒐
𝟑𝟔𝟎 𝒐
) + 𝝅(𝟖. 𝟓) 𝟐
(
𝟒𝟕.𝟎𝟗…𝒐
𝟑𝟔𝟎 𝒐
)
= 𝟔𝟕. 𝟏𝟑…
→
5.) 𝑨𝒓𝒆𝒂 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒄𝒊𝒓𝒄𝒍𝒆𝒔 = (𝟕𝟑. 𝟕𝟐…
) − (𝟔𝟕. 𝟏𝟑…
)
= 𝟔. 𝟓𝟖…

4. Math 142 Final Exam Notes (Spring 2016)
|Instructor: Will Webber | Author: William Faber |
Basic Trig Identity problems
𝟏+𝒄𝒐𝒔𝒚
𝟏+𝒔𝒆𝒄𝒚

𝟏+𝒄𝒐𝒔𝒚
𝒄𝒐𝒔𝒚
𝒄𝒐𝒔𝒚
+
𝟏
𝒄𝒐𝒔𝒚

𝟏+𝒄𝒐𝒔𝒚
𝒄𝒐𝒔𝒚+𝟏
𝒄𝒐𝒔𝒚

𝒄𝒐𝒔𝒚(𝒄𝒐𝒔𝒚+𝟏)
𝒄𝒐𝒔𝒚+𝟏
𝒄𝒐𝒔(𝒚)
𝐭𝐚𝐧 𝟐
𝒕 − 𝐬𝐢𝐧 𝟐
𝒕 =
𝐬𝐢𝐧 𝒂
𝒕
𝐜𝐨𝐬 𝒃 𝒕
−→ 𝒇𝒊𝒏𝒅 𝒂 & 𝒃
𝒕𝒂𝒏 𝟐
𝒕 − 𝒔𝒊𝒏 𝟐
𝒕 
𝒔𝒊𝒏 𝟐 𝒕
𝒄𝒐𝒔 𝟐 𝒕
−
𝒔𝒊𝒏 𝟐 𝒕
𝟏

𝒔𝒊𝒏 𝟐 𝒕−𝒔𝒊𝒏 𝟐 𝒕 𝒄𝒐𝒔 𝟐 𝒕
𝒄𝒐𝒔 𝟐 𝒕

𝒔𝒊𝒏 𝟐 𝒕 (𝟏−𝒄𝒐𝒔 𝟐 𝒕)
𝒄𝒐𝒔 𝟐 𝒕

𝒔𝒊𝒏 𝟒 𝒕
𝒄𝒐𝒔 𝟐 𝒕
(𝒕𝒂𝒏𝒙 + 𝒔𝒆𝒄𝒙) 𝟐
=
𝑨+𝒔𝒊𝒏𝒙
𝑩−𝒔𝒊𝒏𝒙

(
𝒔𝒊𝒏𝒙
𝒄𝒐𝒔𝒙
+
𝟏
𝒄𝒐𝒔𝒙
) (
𝒔𝒊𝒏𝒙
𝒄𝒐𝒔𝒙
+
𝟏
𝒄𝒐𝒔𝒙
) 
(𝒔𝒊𝒏𝒙+𝟏) 𝟐
𝐜𝐨𝐬 𝟐 𝒙

(𝒔𝒊𝒏𝒙+𝟏) 𝟐
𝟏−𝐬𝐢𝐧 𝟐 𝒙

(𝟏+𝒔𝒊𝒏𝒙) 𝟐
(𝟏−𝒔𝒊𝒏𝒙)(𝟏+𝒔𝒊𝒏𝒙)

𝟏+𝒔𝒊𝒏𝒙
𝟏−𝒔𝒊𝒏𝒙
𝐜𝐨𝐭(−𝒙) 𝐜𝐨𝐬(−𝒙) + 𝐬𝐢𝐧(−𝒙) = −
𝟏
𝒇(𝒙)
− 𝐜𝐨𝐭(𝒙) ∗ 𝐜𝐨𝐬(𝒙) − 𝐬𝐢𝐧(𝒙) 
−𝐜𝐨𝐬(𝒙)
𝐬𝐢𝐧(𝒙)
∗
𝐜𝐨𝐬(𝒙)
𝟏
−
𝐬𝐢𝐧(𝒙)
𝟏

− 𝐜𝐨𝐬 𝟐(𝒙)
𝐬𝐢𝐧 𝟐(𝒙)
−
𝐬𝐢𝐧 𝟐(𝒙)
𝐬𝐢𝐧(𝒙)

−𝟏(𝐜𝐨𝐬 𝟐(𝒙)+𝐬𝐢𝐧 𝟐(𝒙))
𝐬𝐢𝐧(𝒙)

−𝟏
𝐬𝐢𝐧(𝒙)
√𝐬𝐞𝐜 𝟐 𝒖 − 𝟏  √𝐭𝐚𝐧 𝟐 𝒖  𝐭𝐚𝐧(𝒖)
Sum & Difference Problems
𝒄𝒐𝒔
𝟐𝝅
𝟕
𝒄𝒐𝒔
𝟐𝝅
𝟐𝟏
+ 𝒔𝒊𝒏
𝟑𝝅
𝟕
𝒔𝒊𝒏
𝟐𝝅
𝟐𝟏
= 𝒄𝒐𝒔
𝝅
𝒂
=
𝒃
𝟐
𝒄𝒐𝒔 (
𝟑𝝅
𝟕
−
𝟐𝝅
𝟐𝟏
) 𝒄𝒐𝒔 (
𝟗𝝅
𝟐𝟏
−
𝟐𝝅
𝟐𝟏
) 𝒄𝒐𝒔 (
𝟕𝝅
𝟐𝟏
)  𝒄𝒐𝒔 (
𝝅
𝟑
) =
𝟏
𝟐
𝒕𝒂𝒏𝟕𝟏°−𝒕𝒂𝒏𝟏𝟏°
𝟏+𝒕𝒂𝒏𝟕𝟏° 𝒕𝒂𝒏𝟏𝟏°
= 𝒕𝒂𝒏𝑨°
= √𝑩
𝐭𝐚𝐧(𝟕𝟏°
− 𝟏𝟏°
)  𝐭𝐚𝐧(𝟔𝟎°
) 
√𝟑
𝟐
𝟏
𝟐
√𝟑
𝐜𝐨𝐬 (𝒙 +
𝝅
𝟔
) + 𝐬𝐢𝐧 (𝒙 −
𝝅
𝟑
) = 𝑨
𝒄𝒐𝒔𝒙 ∗ 𝒄𝒐𝒔
𝝅
𝟔
− 𝒔𝒊𝒏𝒙𝒔𝒊𝒏
𝝅
𝟔
 𝒄𝒐𝒔𝒙 ∗
√𝟑
𝟐
− 𝒔𝒊𝒏𝒙 ∗
𝟏
𝟐
𝒔𝒊𝒏𝒙𝒄𝒐𝒔
𝝅
𝟑
− 𝒄𝒐𝒔𝒙𝒔𝒊𝒏
𝝅
𝟑
 𝒔𝒊𝒏𝒙 ∗
𝟏
𝟐
− 𝒄𝒐𝒔𝒙 ∗
√𝟑
𝟐
(
√𝟑
𝟐
𝒄𝒐𝒔𝒙 −
𝟏
𝟐
𝒔𝒊𝒏𝒙) + (
𝟏
𝟐
𝒔𝒊𝒏𝒙 −
√𝟑
𝟐
𝒄𝒐𝒔𝒙) = 𝟎
(problems using only variables have the same principle.)
𝒔𝒊𝒏𝟏𝟔𝟓°
=
√𝑨(√𝑩 − 𝟏)
𝟒
𝐬𝐢𝐧(𝟑𝟎°
+ 𝟏𝟑𝟓°) 𝒔𝒊𝒏𝟑𝟎°
𝒄𝒐𝒔𝟏𝟑𝟓°
+ 𝒄𝒐𝒔𝟑𝟎°
𝒔𝒊𝒏𝟏𝟑𝟓°

(
𝟏
𝟐
∗
−√𝟐
𝟐
) + (
√𝟑
𝟐
∗
√𝟐
𝟐
)
√𝟔−√𝟐
𝟒

√𝟐(√𝟑−𝟏)
𝟒
(Problems using radians have the same principle.)
𝒄𝒐𝒔 ∝ =. 𝟕𝟑𝟓 & 𝒔𝒊𝒏𝜷 = . 𝟑𝟖𝟗 𝒃𝒐𝒕𝒉 𝒓𝒂𝒚𝒔 𝒊𝒏 𝑸𝟏.
𝒇𝒊𝒏𝒅 𝒄𝒐𝒔(∝ +𝜷) & 𝒔𝒊𝒏(∝ +𝜷)
Draw two triangles and keep as decimals.
Use Pythagoras to find all sides. (hypotenuse is 1 for both in this case)
Double Angle problems
𝒄𝒐𝒔 𝟐(𝟑𝟓°) − 𝒔𝒊𝒏 𝟐(𝟑𝟓°) = 𝒄𝒐𝒔(𝑨°
)
𝒄𝒐𝒔(𝟐(𝟑𝟓°)) 𝒄𝒐𝒔(𝟕𝟎°
)
𝒄𝒐𝒔 𝟐(𝟗𝒙) − 𝒔𝒊𝒏 𝟐(𝟗𝒙) = 𝒄𝒐𝒔(𝑩)
𝒄𝒐𝒔(𝟐(𝟗𝒙))𝒄𝒐𝒔(𝟏𝟖𝒙)
𝒄𝒐𝒔𝒙 =
𝟒
𝟓
, 𝒄𝒔𝒄𝒙 < 𝟎 (𝒎𝒆𝒂𝒏𝒔 𝒊𝒕𝒔 𝒊𝒏 𝑸𝟒)
𝒔𝒊𝒏𝟐𝒙𝟐𝒔𝒊𝒏𝒙 ∗ 𝒄𝒐𝒔𝒙 −
𝟐𝟒
𝟐𝟓
𝒄𝒐𝒔𝟐𝒙𝒄𝒐𝒔 𝟐
𝒙 − 𝒔𝒊𝒏 𝟐𝒙

𝟕
𝟐𝟓
𝒕𝒂𝒏𝟐𝒙
𝟐𝒕𝒂𝒏𝒙
𝟏 − 𝒕𝒂𝒏 𝟐 𝒙
 −
𝟐𝟒
𝟕
𝒄𝒐𝒔 𝟐(𝟐𝒙) − 𝒔𝒊𝒏 𝟐(𝟐𝒙) = 𝒄𝒐𝒔(𝒇(𝒙))
𝒄𝒐𝒔(𝟐(𝟐𝒙)) = 𝒄𝒐𝒔(𝟒𝒙)
𝒄𝒐𝒔 𝟒(𝟐𝒙) − 𝒔𝒊𝒏 𝟒(𝟐𝒙) = 𝒄𝒐𝒔(𝒈(𝒙))
(𝒄𝒐𝒔 𝟐(𝟐𝒙) − 𝒔𝒊𝒏 𝟐(𝟐𝒙))(𝒄𝒐𝒔 𝟐(𝟐𝒙) + 𝒔𝒊𝒏 𝟐(𝟐𝒙))
 𝒄𝒐𝒔𝟒𝒙(𝒄𝒐𝒔 𝟐(𝟐𝒙) + 𝒔𝒊𝒏 𝟐(𝟐𝒙)) 𝒄𝒐𝒔𝟒𝒙
Power Reducing formulas (hard problem)
𝒄𝒐𝒔 𝟐
𝒙 𝒔𝒊𝒏 𝟒
𝒙 = 𝑨 + 𝑩𝒄𝒐𝒔𝟐𝒙 + 𝑪𝒄𝒐𝒔𝟒𝒙 + 𝑫𝒄𝒐𝒔𝟐𝒙 ∗ 𝒄𝒐𝒔𝟒𝒙
𝟏
𝟐
(𝟏 + 𝒄𝒐𝒔𝟐𝒙) ∗
𝟏
𝟐
(𝟏 − 𝒄𝒐𝒔𝟐𝒙) 𝟐

𝟏
𝟐
(𝟏 + 𝒄𝒐𝒔𝟐𝒙) ∗
𝟏
𝟐
(𝟏 + 𝒄𝒐𝒔𝟐𝒙) ∗
𝟏
𝟐
(𝟏 + 𝒄𝒐𝒔𝟐𝒙)
𝟏
𝟖
(𝟏 + 𝒄𝒐𝒔𝟐𝒙)(𝟏 − 𝒄𝒐𝒔𝟐𝒙)(𝟏 − 𝒄𝒐𝒔𝟐𝒙)
𝟏
𝟖
(𝟏 − 𝒄𝒐𝒔 𝟐
𝟐𝒙)(𝟏 − 𝒄𝒐𝒔𝟐𝒙)
𝟏
𝟖
(𝒔𝒊𝒏 𝟐
𝟐𝒙)(𝟏 − 𝒄𝒐𝒔𝟐𝒙)
𝟏
𝟖
(
𝟏−𝒄𝒐𝒔𝟒𝒙
𝟐
)(𝟏 − 𝒄𝒐𝒔𝟐𝒙)
𝟏
𝟏𝟔
(𝟏 − 𝒄𝒐𝒔𝟒𝒙)(𝟏 − 𝒄𝒐𝒔𝟐𝒙)
𝟏
𝟏𝟔
(𝟏 + 𝒄𝒐𝒔𝟐𝒙 + 𝒄𝒐𝒔𝟒𝒙 + 𝒄𝒐𝒔𝟐𝒙 ∗ 𝒄𝒐𝒔𝟒𝒙) 
𝟏
𝟏𝟔
+
𝟏
𝟏𝟔
𝒄𝒐𝒔𝟐𝒙 +
𝟏
𝟏𝟔
𝒄𝒐𝒔𝟒𝒙 +
𝟏
𝟏𝟔
𝒄𝒐𝒔𝟐𝒙 ∗ 𝒄𝒐𝒔𝟒𝒙
Half Angle Formula Problems
𝑺𝒊𝒏𝟏𝟓°
=
𝟏
𝟐
√ 𝑨 − √𝑩
(multiply degrees by 2 and find all sides)
𝒔𝒊𝒏
𝟑𝟎°
𝟐
= √
𝟏−𝒄𝒐𝒔𝟑𝟎°
𝟐
√ 𝟏−
√𝟑
𝟐
𝟐
√ 𝟏−√𝟑
𝟒

√𝟐−√𝟑
𝟐

𝟏
𝟐
√ 𝟐 − √𝟑
𝒄𝒔𝒄𝒙 = 𝟒, {𝟗𝟎°
< 𝒙 < 𝟏𝟖𝟎°
}divide by two to find quadrant
𝒔𝒊𝒏
𝒙
𝟐
=  ±√ 𝟏−
−√𝟏𝟓
𝟒
𝟐
√ 𝟒+√𝟏𝟓
𝟖
𝒄𝒐𝒔
𝒙
𝟐
=  ±
√ 𝟏 +
−√𝟏𝟓
𝟒
𝟐
√
𝟒 − √𝟏𝟓
𝟖
𝒕𝒂𝒏
𝒙
𝟐
= 
𝟏−
−√𝟏𝟓
𝟒
𝟏
𝟒

𝟒+√𝟏𝟓
𝟒
𝟏
𝟒

𝟏𝟔+𝟒√𝟏𝟓
𝟒
 𝟒 + √𝟏𝟓
Product to sum Problems
𝒄𝒐𝒔𝟑𝟕. 𝟓°
𝒔𝒊𝒏𝟕. 𝟓° √𝑨−𝑩
𝟒

𝟏
𝟐
[𝒔𝒊𝒏(𝟑𝟕. 𝟓 + 𝟕. 𝟓) − 𝒔𝒊𝒏(𝟑𝟕. 𝟓 − 𝟕. 𝟓)]
𝟏
𝟐
[𝒔𝒊𝒏(𝟒𝟓°) − 𝒔𝒊𝒏(𝟑𝟎°)]
𝟏
𝟐
(
√𝟐
𝟐
−
𝟏
𝟐
)
𝟏
𝟒
(√𝟐 − 𝟏)
√𝟐−𝟏
𝟒
Sum to Product Problems
𝒔𝒊𝒏𝟒𝟏°
+ 𝒔𝒊𝒏𝟏𝟗°
= 𝒔𝒊𝒏𝑨°
𝒊𝒏 𝑸𝟏
𝟐𝒔𝒊𝒏 (
𝟒𝟏+𝟏𝟗
𝟐
) 𝒄𝒐𝒔 (
𝟒𝟏−𝟏𝟗
𝟐
) 𝟐𝒔𝒊𝒏(𝟑𝟎°) ∗ 𝒄𝒐𝒔(𝟏𝟏°
) 𝟐 (
𝟏
𝟐
) 𝒄𝒐𝒔(𝟏𝟏°) 
𝒄𝒐𝒔(𝟏𝟏°). 𝟗𝟖𝟏𝟔𝟐𝟕……
𝒔𝒊𝒏−𝟏(. 𝟗𝟖𝟏𝟔𝟐𝟕……) = 𝟕𝟗°
 𝒔𝒊𝒏(𝟕𝟗°
)
Trig equation problems
𝑭𝒊𝒏𝒅 𝒂𝒍𝒍 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔 𝒐𝒇 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒔𝒊𝒏𝒙 − 𝟏 = 𝟎
𝑨𝒏𝒔𝒘𝒆𝒓 𝒊𝒏 [𝑨 + 𝑩𝒌 𝝅]𝒇𝒐𝒓𝒎 | 𝟎 < 𝑨 < 𝝅 | 𝒌 𝒊𝒔 𝒂𝒏𝒚 𝒊𝒏𝒕𝒆𝒈𝒆𝒓 .
𝒔𝒊𝒏
𝝅
𝟐
= 𝟏  𝒔𝒊𝒏
𝝅
𝟐
− 𝟏 = 𝟎  consult the unit circle
𝝅
𝟐
+ 𝟐𝒌𝝅
𝑭𝒊𝒏𝒅 𝒂𝒍𝒍 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔 𝒐𝒇 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒕𝒂𝒏 𝟓
𝒙 − 𝟗𝒕𝒂𝒏𝒙 = 𝟎
𝑨𝒔𝒏𝒘𝒆𝒓𝒆 𝒊𝒏 [𝑨𝒌𝝅] | 𝒌 𝒊𝒔 𝒂𝒏𝒚 𝒊𝒏𝒕𝒆𝒈𝒆𝒓.
𝒕𝒂𝒏𝒙((𝒕𝒂𝒏 𝟒
𝒙 − 𝟗) = 𝟎 𝒕𝒂𝒏𝒙(𝒕𝒂𝒏 𝟐
𝒙 + 𝟑)(𝒕𝒂𝒏 𝟐
𝒙 − 𝟑) = 𝟎𝒕𝒂𝒏𝒙 =
𝟎, ±√−𝟑, ±√𝟑  𝒕𝒂𝒏−𝟏
(±√𝟑) = ±
𝝅
𝟑

𝝅
𝟑
𝒌 
𝟏
𝟑
𝒌𝝅
𝑨𝒍𝒍 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔 𝒐𝒇 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟒𝒔𝒊𝒏𝒙𝒄𝒐𝒔𝒙 + 𝟐𝒔𝒊𝒏𝒙 − 𝟐𝒄𝒐𝒔𝒙 − 𝟏 = 𝟎
𝑨𝒏𝒔𝒘𝒆𝒓𝒆 𝒊𝒏 𝑨 + 𝟐𝒌𝝅, 𝑩 + 𝟐𝒌𝝅, 𝑪 + 𝟐𝒌𝝅, 𝒂𝒏𝒅 𝑫 + 𝟐𝒌𝝅
𝒌 𝒊𝒔 𝒂𝒏𝒚 𝒊𝒏𝒕𝒆𝒈𝒆𝒓 𝒂𝒏𝒅 𝟎 < 𝑨 < 𝑩 < 𝑪 < 𝑫 < 𝟐𝝅
(𝟒𝒔𝒊𝒏𝒙 ∗ 𝒄𝒐𝒔𝒙 + 𝟐𝒔𝒊𝒏𝒙) − 𝟐𝒄𝒐𝒔𝒙 − 𝟏 = 𝟎 
𝟐𝒔𝒊𝒏𝒙(𝟐𝒄𝒐𝒔𝒙 + 𝟏) − 𝟐𝒄𝒐𝒔𝒙 − 𝟏 = 𝟎
𝟐𝒔𝒊𝒏𝒙(𝟐𝒄𝒐𝒙 + 𝟏) − 𝟏(𝟐𝒄𝒐𝒙 + 𝟏) = 𝟎(𝟐𝒔𝒊𝒏𝒙 − 𝟏)(𝟐𝒄𝒐𝒔𝒙 + 𝟏) = 𝟎
𝒔𝒊𝒏𝒙 =
𝟏
𝟐

𝝅
𝟔
,
𝟓𝝅
𝟔
𝒄𝒐𝒔𝒙 = −
𝟏
𝟐

𝟐𝝅
𝟑
,
𝟒𝝅
𝟑
𝝅
𝟔
+ 𝟐𝒌𝝅,
𝟐𝝅
𝟑
+ 𝟐𝒌𝝅,
𝟓𝝅
𝟔
+ 𝟐𝒌𝝅, &
𝟒𝝅
𝟑
+ 𝟐𝒌𝝅
𝑨𝒍𝒍 𝒔𝒐𝒖𝒕𝒊𝒐𝒏𝒔 𝒇𝒐𝒓 𝟐𝒔𝒊𝒏𝒙 + √𝟑 = 𝟎 | 𝒂𝒏𝒔𝒘𝒓𝒆 𝒊𝒏 𝑨 + 𝑩𝒌𝝅 & 𝑪 + 𝑫𝒌𝝅
𝒌 𝒊𝒔 𝒂𝒏𝒚 𝒊𝒏𝒕𝒆𝒈𝒆𝒓 | 𝟎 < 𝑨 < 𝑪 < 𝟐𝝅 |
𝟐𝒔𝒊𝒏𝒙 = −√𝟑 𝒔𝒊𝒏𝒙 = −
√𝟑
𝟐
 𝒙 =
𝟒𝝅
𝟑
,
𝟓𝝅
𝟑

𝟒𝝅
𝟑
+ 𝟐𝒌𝝅 &
𝟓𝝅
𝟑
+ 𝟐𝒌𝝅
𝑭𝒊𝒏𝒅 𝒂𝒍𝒍 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔 𝒐𝒇 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒔𝒆𝒄𝟒𝒙 − 𝟐 = 𝟎
𝑨𝒏𝒔𝒘𝒆𝒓𝒆 𝒊𝒔 𝑨 +
𝒌
𝟐
𝝅 𝒂𝒏𝒅 𝑩 +
𝒌
𝟐
𝝅 | 𝒌 𝒊𝒔 𝒂𝒏𝒚 𝒊𝒏𝒕𝒆𝒈𝒆𝒓 | 𝟎 < 𝑨 < 𝑩 <
𝝅
𝟐
𝒔𝒆𝒄𝟒𝒙 =
𝟐
𝟏
𝒄𝒐𝒔𝟒𝒙 =
𝟏
𝟐
𝒄𝒐𝒔−𝟏 𝟏
𝟐
= 𝟒𝒙
𝝅
𝟑
&
𝟓𝝅
𝟑
= 𝟒𝒙𝒙 =
𝝅
𝟏𝟐
&
𝟓𝝅
𝟏𝟐

𝝅
𝟏𝟐
+
𝒌
𝟐
𝝅 &
𝟓𝝅
𝟏𝟐
+
𝒌
𝟐
𝑭𝒊𝒏𝒅 𝒂𝒍𝒍 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔 𝒐𝒇 𝟐𝒄𝒐𝒔𝟑𝒙 = 𝟏 𝒊𝒏 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 [𝟎, 𝝅)
𝟐𝒄𝒐𝒔𝟑𝒙 = 𝟏𝒄𝒐𝒔𝟑𝒙 =
𝟏
𝟐
𝒄𝒐𝒔−𝟏
(
𝟏
𝟐
) = 𝟑𝒙rotate round the unit circle
to find possibilities that will stay inside the interval after dividing by 3
𝟑𝒙 =
𝝅
𝟑
,
𝟓𝝅
𝟑
,
𝟕𝝅
𝟑
𝒙 =
𝝅
𝟗
,
𝟓𝝅
𝟗
,
𝟕𝝅
𝟗
𝑫𝒐𝒖𝒃𝒍𝒆 − 𝒉𝒂𝒍𝒇 𝒂𝒏𝒈𝒍𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔 𝒐𝒇 𝟐𝒙 + 𝒄𝒐𝒔𝒙 = 𝟎
𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍[𝟎, 𝟐𝝅)
𝟐𝒙 + 𝒄𝒐𝒔𝒙 = 𝟎𝟐𝒔𝒊𝒏𝒙𝒄𝒐𝒔𝒙 + 𝒄𝒐𝒔𝒙 = 𝟎𝒄𝒐𝒔𝒙(𝟐𝒔𝒊𝒏𝒙 + 𝟏) = 𝟎 
𝟎(𝟐(±𝟏) + 𝟏) = 𝟎 𝒐𝒓
√𝟑
𝟐
(𝟐 (−
𝟏
𝟐
) + 𝟏) = 𝟎 𝒙 =
𝝅
𝟐
,
𝟕𝝅
𝟔
,
𝟑𝝅
𝟐
,
𝟏𝟏𝝅
𝟔
𝑭𝒊𝒏𝒅 𝒂𝒍𝒍 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔 𝒕𝒐 𝒄𝒐𝒔(𝟗𝜶) − 𝒄𝒐𝒔(𝟑𝜶) = 𝒔𝒊𝒏(𝟔𝜶)𝒐𝒏 𝟎 ≤ 𝜶 <
𝟐𝝅
𝟑
−𝟐𝒔𝒊𝒏 (
𝟗𝜶+𝟐𝜶
𝟐
) 𝒔𝒊𝒏 (
𝟗𝜶−𝟑𝜶
𝟐
) = 𝒔𝒊𝒏𝟔𝜶 −𝟐𝒔𝒊𝒏(𝟔𝜶)𝒔𝒊𝒏(𝟑𝜶) = 𝒔𝒊𝒏𝟔𝜶
−𝟐𝒔𝒊𝒏(𝟔𝜶)𝒔𝒊𝒏(𝟑𝜶) − 𝒔𝒊𝒏𝟔𝜶 = 𝟎
−𝒔𝒊𝒏(𝟔𝜶)[𝟐𝒔𝒊𝒏(𝟑𝜶) + 𝟏] = 𝟎𝒔𝒊𝒏(𝟔𝜶) = 𝟎 𝒐𝒓 𝒔𝒊𝒏(𝟑𝜶) = −
𝟏
𝟐

𝑭𝒊𝒏𝒅 𝒂𝒍𝒍 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔 𝒕𝒐 𝒄𝒐𝒔(𝟐𝜽) = −
√𝟐
𝟐
| 𝒂𝒏𝒔𝒘𝒆𝒓𝒔 𝒊𝒏 𝑨 + 𝑩𝒌 & 𝑪 + 𝑫𝒌
𝒌 𝒊𝒏 𝒂𝒏𝒚 𝒊𝒏𝒕𝒆𝒈𝒆𝒓, 𝒂𝒏𝒅 𝟎 ≤ 𝑨 < 𝑪 < 𝟐𝝅
𝒄𝒐𝒔−𝟏
(−
√𝟐
𝟐
) = 𝟐𝜽  𝟐𝜽 =
𝟑𝝅
𝟒
,
𝟒𝝅
𝟒
 𝟐𝜽 =
𝟑𝝅
𝟒
+ 𝟐𝝅𝒌,
𝟒𝝅
𝟒
+ 𝟐𝝅𝒌 
𝜽 =
𝟑𝝅
𝟖
+ 𝝅𝒌,
𝟒𝝅
𝟖
+ 𝝅𝒌
Polar Coordinates Problems
𝒄𝒐𝒏𝒗𝒆𝒓𝒕 𝒕𝒉𝒆 𝒑𝒐𝒍𝒂𝒓 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆 (𝟐,
𝟐𝝅
𝟑
) 𝒕𝒐 𝒄𝒂𝒓𝒕𝒆𝒔𝒊𝒂𝒏 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔.
𝒙 = 𝟐 ∗ 𝒄𝒐𝒔 (
𝟐𝝅
𝟑
) = −𝟏 𝒚 = 𝟐 ∗ 𝒔𝒊𝒏 (
𝟐𝝅
𝟑
) = 𝟏. 𝟕𝟑𝟐….
𝑪𝒐𝒏𝒗𝒆𝒓𝒕 𝒕𝒉𝒆 𝑪𝒂𝒓𝒕𝒆𝒔𝒊𝒂𝒏 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆 (−𝟒, −𝟓)𝒕𝒐 𝒑𝒐𝒍𝒂𝒓 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔,
𝟎 ≤ 𝜽 < 𝟐𝝅
𝒓 = √(−𝟒 𝟐) + (−𝟓 𝟐) = √𝟒𝟏 𝜽 = 𝒕𝒂𝒏−𝟏
(
−𝟓
−𝟒
) + 𝝅 = 𝟒. 𝟎𝟑𝟕…
𝑹𝒆𝒘𝒓𝒊𝒕𝒆 𝒕𝒉𝒆 𝑪𝒂𝒓𝒕𝒆𝒔𝒊𝒂𝒏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒚 = 𝟔𝒙 𝟐
𝒂𝒔 𝒂 𝒑𝒐𝒍𝒂𝒓 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏.
𝒓𝒔𝒊𝒏𝜽 = 𝟔(𝒓𝒄𝒐𝒔𝜽) 𝟐
 𝒓𝒔𝒊𝒏𝜽 = 𝟔 ∗ 𝒓 𝟐
𝒄𝒐𝒔 𝟐
𝜽𝒓 =
𝒔𝒊𝒏𝜽
𝟔𝒄𝒐𝒔 𝟐 𝜽
𝒓 =
𝟒
𝟓 − 𝟒𝒔𝒊𝒏𝜽
| 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒙 𝒂𝒏𝒅 𝒚 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕(𝒔)𝒂𝒏𝒅 𝒇𝒐𝒄𝒊
𝒓 =
𝟒
𝟓−𝟒𝒔𝒊𝒏𝜽
 𝒓 =
𝟒
𝟓
𝟏−
𝟒
𝟓
𝒔𝒊𝒏𝜽

𝒙 − 𝒊𝒏𝒕 (
𝟒
𝟓
, 𝟎) (−
𝟒
𝟓
, 𝟎)
𝒚 − 𝒊𝒏𝒕(𝟒, 𝟎) (−
𝟒
𝟗
, 𝟎)
𝑭𝒐𝒄𝒊 𝟏(𝟎, 𝟎)
𝟏
𝟐
(𝟒 + (−
𝟒
𝟗
)) =
𝟏𝟔
𝟗
 𝒄 = (𝟎,
𝟏𝟔
𝟗
)
𝑭𝒐𝒄𝒊 𝟐 = (𝟎,
𝟏𝟔
𝟗
+
𝟏𝟔
𝟗
) −→ 𝑭 𝟐 = (𝟎,
𝟑𝟐
𝟗
)
Parametric Problems
𝒙 𝟐
𝟒 𝟐
+
𝒚 𝟐
𝟓 𝟐
= 𝟏 𝒄𝒂𝒏 𝒃𝒆 𝒅𝒓𝒂𝒘𝒏 𝒘𝒊𝒕𝒉 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒓𝒊𝒄 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔.
𝒊𝒇 𝒙 = 𝒓𝒄𝒐𝒔(𝒕), 𝒕𝒉𝒆𝒏 𝒓 = ? 𝒂𝒏𝒅 𝒚 = ?
𝒙 𝟐
𝟒 𝟐
= 𝒄𝒐𝒔 𝟐
𝒕
𝒙
𝟒
= 𝒄𝒐𝒔(𝒕) 𝒙 = 𝟒𝒄𝒐𝒔(𝒕) 𝒓 = 𝟒
𝒚 𝟐
𝟓 𝟐
= 𝒔𝒊𝒏 𝟐
𝒕
𝒚
𝟓
= 𝒔𝒊𝒏(𝒕)𝒚 = 𝟓𝒔𝒊𝒏(𝒕)
𝑬𝒍𝒊𝒎𝒊𝒏𝒂𝒕𝒆 𝒕𝒉𝒆 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒆𝒓 𝒕 𝒕𝒐 𝒇𝒊𝒏𝒅 𝒂 𝑪𝒂𝒓𝒕𝒆𝒔𝒊𝒂𝒏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒇𝒐𝒓:
𝒙 = 𝒕 𝟐
& 𝒚 = 𝟑 + 𝟓𝒕 | 𝒕𝒉𝒆 𝒓𝒆𝒔𝒖𝒍𝒕𝒊𝒏𝒈 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝒊𝒔 𝒙 = 𝑨𝒚 𝟐
+ 𝑩𝒚 + 𝑪
Find values of t
𝒙 = 𝒕 𝟐
 𝒕 = √𝒙 | 𝒚 = 𝟑 + 𝟓𝒕 𝒕 =
𝒚−𝟑
𝟓
Make them equal to each other
√𝒙 =
𝒚−𝟑
𝟓
 𝒙 =
(𝒚−𝟑) 𝟐
𝟓 𝟐
 𝒙 =
𝟏
𝟐𝟓
𝒚 𝟐
−
𝟔
𝟐𝟓
𝒚 +
𝟗
𝟐𝟓
𝑷𝒂𝒓𝒂𝒎𝒆𝒕𝒓𝒊𝒄 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒍𝒊𝒏𝒆 𝒔𝒆𝒈𝒎𝒆𝒏𝒕 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 (𝟏, 𝟒)& (𝟖, 𝟏)
𝒉𝒂𝒗𝒆 𝒕𝒉𝒆 𝒇𝒐𝒓𝒎: 𝒙(𝒕) = 𝒂 + 𝒃𝒕 & 𝒚(𝒕) = 𝒄 + 𝒅(𝒕)
𝒄𝒖𝒓𝒗𝒆 𝒔𝒕𝒂𝒓𝒕𝒔 𝒂𝒕 (𝟏, 𝟒)𝒘𝒉𝒆𝒏 𝒕 = 𝟎 𝒂𝒏𝒅 𝒆𝒏𝒅𝒔 (𝟖, 𝟏)𝒂𝒕 𝒕 = 𝟏
𝒙 = 𝒂 + 𝒃(𝟎)𝒙 = 𝒂 = 𝟏
𝒚 = 𝒄 + 𝒅(𝟎)𝒚 = 𝒃 = 𝟒
𝒙 = 𝒂 + 𝒃(𝟏)𝒙 = 𝟏 + 𝒃 = 𝟖𝒃 = 𝟕
𝒚 = 𝒄 + 𝒅(𝟏)𝒚 = 𝟒 + 𝒃 = 𝟏𝒅 = −𝟑
Basic Polar Graphs Basic Parametric graphs
𝒓 = 𝒄𝒐𝒔(𝟕𝜽)
𝒓 = 𝒄𝒐𝒔(𝟑𝜽)
𝒓 = 𝒔𝒊𝒏(𝟔𝜽)
𝒓 = 𝒄𝒐𝒔(𝟔𝜽)
𝒓 = 𝒄𝒐𝒔(𝟒𝜽)
𝑿 = 𝟕𝒔𝒊𝒏(𝒕) + 𝟑𝒔𝒊𝒏(𝟓𝒕)
𝒀 = 𝟕𝒄𝒐𝒔(𝒕) + 𝟑𝒄𝒐𝒔(𝟓𝒕)
𝑿 = 𝟕𝒔𝒊𝒏(𝒕) − 𝟑𝒔𝒊𝒏(𝟕𝒕)
𝒀 = 𝟕𝒄𝒐𝒔(𝒕) + 𝟑𝒄𝒐𝒔(𝟕𝒕)
𝑿 = 𝒔𝒊𝒏(𝒕) + 𝟐𝒔𝒊𝒏(𝟓𝒕)
𝒀 = 𝒄𝒐𝒔(𝒕) + 𝟐𝒄𝒐𝒔(𝟓𝒕)
𝑿 = 𝒔𝒊𝒏(𝒕) + 𝒔𝒊𝒏(𝟐𝒕)
𝒀 = 𝒄𝒐𝒔(𝒕) − 𝒄𝒐𝒔(𝟐𝒕)
𝑿 = 𝟑𝒄𝒐𝒔(𝒕) + 𝟐𝒄𝒐𝒔(𝟔𝒕)
𝒀 = −𝟔𝒔𝒊𝒏(𝒕) + 𝟐𝒔𝒊𝒏(𝟑𝒕)
𝒔𝒊𝒏(𝟔𝜶) = 𝟎
𝒔𝒊𝒏−𝟏(𝟎) = 𝟔𝜶
𝟔𝜶 = 𝟎, 𝝅, 𝟐𝝅, , 𝟑𝝅, …
𝜶 = 𝟎,
𝝅
𝟔
,
𝝅
𝟑
,
𝝅
𝟐
𝒔𝒊𝒏(𝟑𝜶) = −
𝟏
𝟐

𝒔𝒊𝒏−𝟏
(−
𝟏
𝟐
) = 𝟑𝜶
𝟑𝜶 =
𝟕𝝅
𝟔
,
𝟏𝟏𝝅
𝟔
…
𝜶 =
𝟕𝝅
𝟏𝟖
,
𝟏𝟏𝝅
𝟏𝟖
𝜽 𝒓
𝟎 𝟒
𝟓
𝝅 𝟒
𝟓
𝝅
𝟐
𝟒
𝟑𝝅
𝟐
𝟒
𝟗

5. Math 142 Final Exam Notes (Spring 2016)
|Instructor: Will Webber | Author: William Faber |
Basic Identities
Sin =
𝑶𝒑𝒑
𝑯𝒚𝒑
=
𝒄𝒐𝒔
𝒄𝒐𝒕
= cos tan
Cos =
𝑨𝒅𝒋
𝑯𝒚𝒑
=
𝒔𝒊𝒏
𝒕𝒂𝒏
= sin cot
Tan =
𝒐𝒑𝒑
𝒂𝒅𝒋
=
𝒔𝒊𝒏
𝒄𝒐𝒔
Cot =
𝑨𝒅𝒋
𝒐𝒑𝒑
=
𝒄𝒐𝒔
𝒔𝒊𝒏
=
𝟏
𝒕𝒂𝒏
Sec =
𝒉𝒚𝒑
𝒂𝒅𝒋
=
𝒕𝒂𝒏
𝒔𝒊𝒏
=
𝟏
𝒄𝒐𝒔
Csc =
𝒉𝒚𝒑
𝒐𝒑𝒑
=
𝒄𝒐𝒕
𝒄𝒐𝒔
=
𝟏
𝒔𝒊𝒏
Sin csc = 1 Cos sec = 1 Tan cot = 1
𝒔𝒊𝒏 𝟐
+ 𝒄𝒐𝒔 𝟐
= 𝟏
𝒕𝒂𝒏 𝟐
+ 𝟏 = 𝒔𝒆𝒄 𝟐
𝟏 + 𝒄𝒐𝒕 𝟐
= 𝒄𝒔𝒄 𝟐
𝟏 − 𝒄𝒐𝒔 𝟐
= 𝒔𝒊𝒏 𝟐
𝟏 − 𝒔𝒊𝒏 𝟐
= 𝒄𝒐𝒔 𝟐
𝒔𝒆𝒄 𝟐
− 𝟏 = 𝒕𝒂𝒏 𝟐
𝒔𝒆𝒄 𝟐
− 𝒕𝒂𝒏 𝟐
= 𝟏
𝒄𝒔𝒄 𝟐
− 𝟏 + 𝒄𝒐𝒕 𝟐
𝒄𝒔𝒄 𝟐
− 𝒄𝒐𝒕 𝟐
= 𝟏
Conversions (degrees & Radians)
𝑫𝒆𝒈𝒓𝒆𝒆𝒔 𝒕𝒐 𝑹𝒂𝒅𝒊𝒂𝒏𝒔 →
𝒙○
𝝅
𝟏𝟖𝟎○
𝑹𝒂𝒅𝒊𝒂𝒏𝒔 𝒕𝒐 𝑫𝒆𝒈𝒓𝒆𝒆𝒔 → 𝒓𝒂𝒅 ∗
𝟏𝟖𝟎
𝝅
Arc Length Formula
𝑨𝒓𝒄 𝑳𝒆𝒏𝒈𝒕𝒉 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 (𝒂𝒏𝒈𝒍𝒆𝒔 𝒊𝒏 𝒓𝒂𝒅𝒊𝒂𝒏𝒔)
𝒔 = 𝒂𝒓𝒄 𝒍𝒆𝒏𝒈𝒕𝒉 & 𝒓 = 𝒓𝒂𝒅𝒊𝒖𝒔  𝒔 = Ɵ𝒓
Cofunction Identities
𝒔𝒊𝒏 (
𝝅
𝟐
− 𝒖) = 𝒄𝒐𝒔 𝒖
𝒄𝒐𝒔 (
𝝅
𝟐
− 𝒖) = 𝒔𝒊𝒏 𝒖
𝒕𝒂𝒏 (
𝝅
𝟐
− 𝒖) = 𝒄𝒐𝒕 𝒖
𝒄𝒐𝒕 (
𝝅
𝟐
− 𝒖) = 𝒕𝒂𝒏 𝒖
𝒔𝒆𝒄 (
𝝅
𝟐
− 𝒖) = 𝒄𝒔𝒄 𝒖
𝒔𝒊𝒏 (
𝝅
𝟐
− 𝒖) = 𝒔𝒆𝒄 𝒖
Even/Odd Identities
𝒔𝒊𝒏(−𝒖) = −𝒔𝒊𝒏 𝒖
𝒄𝒐𝒔(−𝒖) = 𝒄𝒐𝒔 𝒖
𝒕𝒂𝒏(−𝒖) = −𝒕𝒂𝒏 𝒖
𝒄𝒐𝒕(−𝒖) = −𝒄𝒐𝒕 𝒖
𝒔𝒆𝒄(−𝒖) = 𝒔𝒆𝒄 𝒖
𝒄𝒔𝒄(−𝒖) = −𝒄𝒔𝒄 𝒖
Double-Angle Formulas
𝒔𝒊𝒏(𝟐𝒖) = 𝟐 𝒔𝒊𝒏 𝒖 𝒄𝒐𝒔 𝒖
𝒄𝒐𝒔(𝟐𝒖) = 𝒄𝒐𝒔 𝟐
𝒖 − 𝒔𝒊𝒏 𝟐
𝒖 = 𝟐 𝒄𝒐𝒔 𝟐
𝒖 − 𝟏
= 𝟏 − 𝟐 𝒔𝒊𝒏 𝟐 𝒖
𝒕𝒂𝒏(𝟐𝒖) =
𝟐 𝒕𝒂𝒏 𝒖
𝟏 − 𝒕𝒂𝒏 𝟐 𝒖
Sum & Difference Formulas
𝒔𝒊𝒏(𝒖 ± 𝒗) = 𝒔𝒊𝒏 𝒖 𝒄𝒐𝒔 𝒗 ± 𝒄𝒐𝒔 𝒖 𝒔𝒊𝒏 𝒗
𝒄𝒐𝒔(𝒖 ± 𝒗) = 𝒄𝒐𝒔 𝒖 𝒄𝒐𝒔 𝒗 ∓ 𝒔𝒊𝒏 𝒖 𝒔𝒊𝒏 𝒗
𝒕𝒂𝒏(𝒖 ± 𝒗) =
𝒕𝒂𝒏 𝒖 ± 𝒕𝒂𝒏 𝒗
𝟏 ∓ 𝒕𝒂𝒏 𝒖 𝒕𝒂𝒏 𝒗
Product-to-Sum Formulas
𝒔𝒊𝒏 𝒖 𝒔𝒊𝒏 𝒗 =
𝟏
𝟐
[𝒄𝒐𝒔(𝒖 − 𝒗) − 𝒄𝒐𝒔(𝒖 + 𝒗)]
𝒄𝒐𝒔 𝒖 𝒄𝒐𝒔 𝒗 =
𝟏
𝟐
[𝒄𝒐𝒔(𝒖 − 𝒗) + 𝒄𝒐𝒔(𝒖 + 𝒗)]
𝒔𝒊𝒏 𝒖 𝒄𝒐𝒔 𝒗 =
𝟏
𝟐
[𝒔𝒊𝒏(𝒖 + 𝒗) + 𝒔𝒊𝒏(𝒖 − 𝒗)]
𝒄𝒐𝒔 𝒖 𝒔𝒊𝒏 𝒗 =
𝟏
𝟐
[𝒔𝒊𝒏(𝒖 + 𝒗) − 𝒔𝒊𝒏(𝒖 − 𝒗)]
Sum-to-Product Formulas
𝒔𝒊𝒏 𝒖 + 𝒔𝒊𝒏 𝒗 = 𝟐 𝒔𝒊𝒏 (
𝒖 + 𝒗
𝟐
) 𝒄𝒐𝒔 (
𝒖 − 𝒗
𝟐
)
𝒔𝒊𝒏 𝒖 − 𝒔𝒊𝒏 𝒗 = 𝟐 𝒄𝒐𝒔(
𝒖 + 𝒗
𝟐
) 𝒔𝒊𝒏 (
𝒖 − 𝒗
𝟐
)
𝒄𝒐𝒔 𝒖 + 𝒄𝒐𝒔 𝒗 = 𝟐 𝒄𝒐𝒔 (
𝒖 + 𝒗
𝟐
) 𝒄𝒐𝒔 (
𝒖 − 𝒗
𝟐
)
𝒄𝒐𝒔 𝒖 − 𝒄𝒐𝒔 𝒗 = −𝟐 𝒔𝒊𝒏 (
𝒖 + 𝒗
𝟐
) 𝒔𝒊𝒏 (
𝒖 − 𝒗
𝟐
)
Half-Angle Formulas
𝒔𝒊𝒏 𝑨 = ±√
𝟏 − 𝒄𝒐𝒔𝟐𝑨
𝟐
𝒄𝒐𝒔 𝑨 = ±√
𝟏 + 𝒄𝒐𝒔𝟐𝑨
𝟐
𝒕𝒂𝒏 𝑨 = ±√
𝟏 − 𝒄𝒐𝒔𝟐𝑨
𝟏 + 𝒄𝒐𝒔𝟐𝑨
𝒔𝒊𝒏
𝒙
𝟐
= ±√
𝟏 − 𝒄𝒐𝒔 𝒙
𝟐
𝒄𝒐𝒔
𝒙
𝟐
= ±√
𝟏 + 𝒄𝒐𝒔 𝒙
𝟐
𝒕𝒂𝒏
𝒙
𝟐
= ±√
𝟏 − 𝒄𝒐𝒔 𝒙
𝟏 + 𝒄𝒐𝒔
𝒕𝒂𝒏
𝒙
𝟐
=
𝟏 − 𝒄𝒐𝒔 𝒙
𝒔𝒊𝒏 𝒙
Power-Reducing Formulas
𝒔𝒊𝒏 𝟐
𝒖 =
𝟏 − 𝒄𝒐𝒔 𝟐𝒖
𝟐
𝒄𝒐𝒔 𝟐
𝒖 =
𝟏 + 𝒄𝒐𝒔 𝟐𝒖
𝟐
𝒕𝒂𝒏 𝟐
𝒖 =
𝟏 − 𝒄𝒐𝒔 𝟐𝒖
𝟏 + 𝒄𝒐𝒔 𝟐𝒖
Alternate Notation
𝒔𝒊𝒏−𝟏
𝒙 = 𝒂𝒓𝒄𝒔𝒊𝒏 𝒙
𝒄𝒐𝒔−𝟏
𝒙 = 𝒂𝒓𝒄𝒄𝒐𝒔 𝒙
𝒕𝒂𝒏−𝟏
𝒙 = 𝒂𝒓𝒄𝒕𝒂𝒏 𝒙
Domain and Range
Function
𝒚 = 𝒔𝒊𝒏−𝟏
𝒙
Domain
−𝟏 ≤ 𝒙 ≤ 𝟏
Range
−
𝝅
𝟐
≤ 𝒚 ≤
𝝅
𝟐
𝒚 = 𝒄𝒐𝒔−𝟏
𝒙 −𝟏 ≤ 𝒙 ≤ 𝟏 −𝟎 ≤ 𝒚 ≤ 𝝅
𝒚 = 𝒕𝒂𝒏−𝟏
𝒙 −∞ ≤ 𝒙 ≤ ∞ −
𝝅
𝟐
≤ 𝒚 ≤
𝝅
𝟐
Polar Coordinates (conversions)
Convert Polar to Rectangular Coordinates
𝒙 = 𝒓 𝒄𝒐𝒔 𝜽 𝒚 = 𝒓 𝒔𝒊𝒏 𝜽
Convert Rectangular to Polar Coordinates
𝑰𝒇 𝒙 = 𝒚 = 𝟎, 𝒕𝒉𝒆𝒏 𝒓 = 𝟎, 𝜽𝒄𝒂𝒏 𝒉𝒂𝒗𝒆 𝒂𝒏𝒚 𝒗𝒂𝒍𝒖𝒆
𝒆𝒍𝒔𝒆 𝒓 = √𝒙 𝟐 + 𝒚 𝟐
𝜽 =
{
𝒕𝒂𝒏−𝟏
(
𝒚
𝒙
)
𝒕𝒂𝒏−𝟏
(
𝒚
𝒙
) + 𝝅
𝝅
𝟐
−
𝝅
𝟐
QI or QIV
QII or QIII
𝒙 = 𝟎, 𝒚 > 𝟎
𝒙 = 𝟎, 𝒚 < 𝟎
Implicit & Parametric forms
Implicit Form Parametric Form
Circle 𝒙 𝟐
+ 𝒚 𝟐
− 𝒓 𝟐
= 𝟎 𝒙(𝒕) = 𝒓
𝟏 − 𝒕 𝟐
𝟏 + 𝒕 𝟐
𝒚(𝒕) = 𝒓
𝟐𝒕
𝟏 + 𝒕 𝟐
Ellipse 𝒙 𝟐
𝒂 𝟐
+
𝒚 𝟐
𝒃 𝟐
− 𝟏 = 𝟎 𝒙(𝒕) = 𝒂
𝟏 − 𝒕 𝟐
𝟏 + 𝒕 𝟐
𝒚(𝒕) = 𝒃
𝟐𝒕
𝟏 + 𝒕 𝟐
Hyperbola 𝒙 𝟐
𝒂 𝟐
−
𝒚 𝟐
𝒃 𝟐
− 𝟏 = 𝟎 𝒙(𝒕) = 𝒂
𝟏 + 𝒕 𝟐
𝟏 − 𝒕 𝟐
𝒚(𝒕) = 𝒃
𝟐𝒕
𝟏 − 𝒕 𝟐
Parabola 𝒚 𝟐
− 𝟐𝒑𝒙 = 𝟎
𝒙(𝒕) = 𝒂
𝒕 𝟐
𝟐𝒑
𝒚(𝒕) = 𝒕
Eliminate the parameter
𝒚 = 𝟑𝒙 − 𝟑 𝒙 = 𝒕 + 𝟏
𝒚 = 𝒙 − 𝟏
𝒔𝒐𝒍𝒗𝒊𝒏𝒈 𝒙 = 𝒕 + 𝟏 𝒇𝒐𝒓 𝒕 𝒙 = 𝒕 + 𝟏
𝒕 = 𝒙 − 𝟏
𝑺𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒊𝒏𝒈 𝒕𝒉𝒂𝒕 𝒊𝒏𝒕𝒐
𝒚 = 𝟑𝒕
𝒚 = 𝟑(𝒙 − 𝟏)
𝒚 = 𝟑𝒙 − 𝟑
The Unit Circle

6. Math 151 Final Notes (Summer 2016)
|Instructor: Julian Trujillo | Author: William Faber |
Basic Trig Identities:
Sin =
𝑶𝒑𝒑
𝑯𝒚𝒑
=
𝒄𝒐𝒔
𝒄𝒐𝒕
= cos tan
Cos =
𝑨𝒅𝒋
𝑯𝒚𝒑
=
𝒔𝒊𝒏
𝒕𝒂𝒏
= sin cot
Tan =
𝒐𝒑𝒑
𝒂𝒅𝒋
=
𝒔𝒊𝒏
𝒄𝒐𝒔
Cot =
𝑨𝒅𝒋
𝒐𝒑𝒑
=
𝒄𝒐𝒔
𝒔𝒊𝒏
=
𝟏
𝒕𝒂𝒏
Sec =
𝒉𝒚𝒑
𝒂𝒅𝒋
=
𝒕𝒂𝒏
𝒔𝒊𝒏
=
𝟏
𝒄𝒐𝒔
Csc =
𝒉𝒚𝒑
𝒐𝒑𝒑
=
𝒄𝒐𝒕
𝒄𝒐𝒔
=
𝟏
𝒔𝒊𝒏
Sin csc = 1
Cos sec = 1
Tan cot = 1
𝒔𝒊𝒏 𝟐
+ 𝒄𝒐𝒔 𝟐
= 𝟏
𝒕𝒂𝒏 𝟐
+ 𝟏 = 𝒔𝒆𝒄 𝟐
𝟏 + 𝒄𝒐𝒕 𝟐
= 𝒄𝒔𝒄 𝟐
𝟏 − 𝒄𝒐𝒔 𝟐
= 𝒔𝒊𝒏 𝟐
𝟏 − 𝒔𝒊𝒏 𝟐
= 𝒄𝒐𝒔 𝟐
𝒔𝒆𝒄 𝟐
− 𝟏 = 𝒕𝒂𝒏 𝟐
𝒔𝒆𝒄 𝟐
− 𝒕𝒂𝒏 𝟐
= 𝟏
𝒄𝒔𝒄 𝟐
− 𝟏 + 𝒄𝒐𝒕 𝟐
𝒄𝒔𝒄 𝟐
− 𝒄𝒐𝒕 𝟐
= 𝟏
The Unit Circle: 𝑻𝒂𝒏−𝟏(𝒙) + 𝒄𝒐𝒕−𝟏(𝒙) =
𝝅
𝟐
Slope Formulas:
Slope: Tangent Line Equation:
𝑺𝒍𝒐𝒑𝒆 =
𝒓𝒊𝒔𝒆
𝒓𝒖𝒏
= 𝒎 =
𝒚 𝟐−𝒚 𝟏
𝒙 𝟐−𝒙 𝟏
→ 𝒚 − 𝒚 𝟏 = 𝒎(𝒙 − 𝒙 𝟏)
Or solve algebraically for
slope intercept form:
𝒚 = 𝒎𝒙 + (𝒃)
Slope-intercept:
𝒚 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕: (𝒃) →
Limits:
Precise Definition:
𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙) = 𝑳 if for every 𝜺 > 𝟎 𝒕𝒉𝒆𝒓𝒆 𝒊𝒔 𝒂 𝜹 > 𝟎 such that
whenever 𝟎 < |𝒙 − 𝒂| < 𝜹, then |𝒇(𝒙) − 𝑳| < 𝜺.
Working Definition:
𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙) = 𝑳 if we can make 𝒇(𝒙) as close to L as we want by
taking 𝒙 sufficiently close to a (on either side of 𝒂)
without letting 𝒙 = 𝒂
Left hand limit definition: Right hand limit definition:
𝐥𝐢𝐦
𝒙→𝒂−
𝒇(𝒙) 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒔 𝒙 < 𝒂 𝐥𝐢𝐦
𝒙→𝒂+
𝒇(𝒙) 𝒓𝒆𝒒𝒊𝒓𝒆𝒔 𝒙 > 𝒂
Relationship between the limit and one-sided limits:
 𝐥𝐢𝐦
𝒙→ 𝒂
𝒇(𝒙) = 𝑳 ↔ 𝐥𝐢𝐦
𝒙→ 𝒂+
𝒇(𝒙) = 𝐥𝐢𝐦
𝒙→ 𝒂−
𝒇(𝒙) = 𝑳
 𝐥𝐢𝐦
𝒙→ 𝒂+
𝒇(𝒙) ≠ 𝐥𝐢𝐦
𝒙→ 𝒂−
𝒇(𝒙) → 𝐥𝐢𝐦
𝒙→ 𝒂
𝒇(𝒙) 𝑫𝒐𝒆𝒔 𝑵𝒐𝒕 𝑬𝒙𝒊𝒔𝒕
Limit Laws:
Assume 𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙) 𝒂𝒏𝒅 𝐥𝐢𝐦
𝒙→𝒂
𝒈(𝒙) both exist and c is any number, then
𝐥𝐢𝐦
𝒙→𝒂
𝒄 = 𝒄 𝐥𝐢𝐦
𝒙→𝒂
𝒙 = 𝒂 𝐥𝐢𝐦
𝒙→𝒂
𝒙 𝒏
= 𝒂 𝒏
 𝐥𝐢𝐦
𝒙→𝒂
√𝒙
𝒏
= √𝒂
𝒏 𝐥𝐢𝐦
𝒙→𝒂
[𝒄𝒇(𝒙)] = 𝒄 𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙)
𝐥𝐢𝐦
𝒙→𝒂
[𝒇(𝒙) ± 𝒈(𝒙)] = 𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙) ± 𝐥𝐢𝐦
𝒙→𝒂
𝒈(𝒙)
𝒍𝒊𝒎
𝒙→𝒂
[𝒇(𝒙)𝒈(𝒙)]
= 𝒍𝒊𝒎
𝒙→𝒂
𝒇(𝒙) 𝒍𝒊𝒎
𝒙→𝒂
𝒈(𝒙)
𝐥𝐢𝐦
𝒙→𝒂
[𝒇(𝒙)] 𝒏
= [𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙)]
𝒏
 𝐥𝐢𝐦
𝒙→𝒂
[
𝒇(𝒙)
𝒈(𝒙)
] =
𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙)
𝐥𝐢𝐦
𝒙→𝒂
𝒈(𝒙)
 𝐥𝐢𝐦
𝒙→𝒂
[√𝒇(𝒙)
𝒏
] = √ 𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙)𝒏
 Note: A “0” in the denominator or an even root of a negative number
Does Not Exist (
𝟎
𝟎
𝒐𝒓
±∞
±∞
, 𝒓𝒆𝒇𝒆𝒓 𝒕𝒐 𝑳′
𝑯𝒐̂𝒑𝒊𝒕𝒂𝒍′
𝒔 𝑹𝒖𝒍𝒆).
Basic Limit Evaluations at ± ∞:
Note: 𝒔𝒈𝒏(𝒂) = 𝟏 if 𝒂 > 𝟎 and 𝒔𝒈𝒏(𝒂) = −𝟏 if 𝒂 < 𝟎
𝐥𝐢𝐦
𝒙→∞
𝒆 𝒙
= ∞ & 𝐥𝐢𝐦
𝒙→−∞
𝒆 𝒙
= 𝟎 𝒏 𝒆𝒗𝒆𝒏: 𝐥𝐢𝐦
𝒙→±∞
𝒙 𝒏
= ∞
𝐥𝐢𝐦
𝒙→∞
𝒍𝒏(𝒙) = ∞ & 𝐥𝐢𝐦
𝒙→𝟎−
𝒍𝒏(𝒙) = −∞ 𝒏 𝒐𝒅𝒅: 𝐥𝐢𝐦
𝒙→±∞
𝒙 𝒏
= ±∞
If 𝒓 > 𝟎 then 𝐥𝐢𝐦
𝒙→∞
𝒃
𝒙 𝒓
= 𝟎 𝒏 𝒆𝒗𝒆𝒏:
𝐥𝐢𝐦
𝒙→±∞
𝒂𝒙 𝒏
+ ⋯ + 𝒃𝒙 + 𝒄 = 𝒔𝒈𝒏(𝒂)∞
If 𝒓 > 𝟎 and 𝒙 𝒓
is real for
−𝒙, then 𝐥𝐢𝐦
𝒙→−∞
𝒃
𝒙 𝒓
= 𝟎
𝒏 𝒐𝒅𝒅:
𝐥𝐢𝐦
𝒙→±∞
𝒂𝒙 𝒏
+ ⋯ + 𝒃𝒙 + 𝒄 = ±𝒔𝒈𝒏(𝒂)∞
Continuous Functions:
𝑰𝒇 𝒇(𝒙) is continuous at a, then 𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙) = 𝒇(𝒂)
Continuous Functions and Composition:
If 𝒇(𝒙) is continuous at 𝒃 and 𝐥𝐢𝐦
𝒙→𝒂
𝒈(𝒙) = 𝒃, then
𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒈(𝒙)) = 𝒇 (𝐥𝐢𝐦
𝒙→𝒂
𝒈(𝒙)) = 𝒇(𝒃)
Factor and cancel
𝒍𝒊𝒎
𝒙→𝟐
𝒙 𝟐+𝟒𝒙−𝟏𝟐
𝒙 𝟐−𝟐𝒙
= 𝒍𝒊𝒎
𝒙→𝟐
(𝒙−𝟐)(𝒙+𝟔)
𝒙(𝒙−𝟐)
= 𝒍𝒊𝒎
𝒙→𝟐
𝒙+𝟔
𝒙
=
𝟖
𝟐
= 𝟒
Rationalize Numerator/Denominator
𝐥𝐢𝐦
𝒙→𝟗
𝟑−√𝒙
𝒙 𝟐−𝟖𝟏
= 𝐥𝐢𝐦
𝒙→𝟗
𝟑−√𝒙
𝒙 𝟐−𝟖𝟏
∙
𝟑+√𝒙
𝟑+√𝒙
= 𝐥𝐢𝐦
𝒙→𝟗
𝟗−𝒙
(𝒙 𝟐−𝟖𝟏)(𝟑+√𝒙)
= 𝐥𝐢𝐦
𝒙→𝟗
−𝟏
(𝒙+𝟗)(𝟑+√𝒙)
=
𝐥𝐢𝐦
𝒙→𝟗
−𝟏
(𝟏𝟖)(𝟔)
= 𝐥𝐢𝐦
𝒙→𝟗
−
𝟏
𝟏𝟎𝟖
Combine Rational Expressions
𝐥𝐢𝐦
𝒉→𝟎
𝟏
𝒉
(
𝟏
𝒙+𝒉
−
𝟏
𝒙
) = 𝐥𝐢𝐦
𝒉→𝟎
𝟏
𝒉
(
𝒙−(𝒙+𝒉)
𝒙(𝒙+𝒉)
) = 𝐥𝐢𝐦
𝒉→𝟎
𝟏
𝒉
(
−𝒉
𝒙(𝒙+𝒉)
) = 𝐥𝐢𝐦
𝒉→𝟎
−𝟏
𝒙(𝒙+𝒉)
= −
𝟏
𝒙 𝟐
Polynomials at Infinity
𝒑(𝒙)and 𝒒(𝒙)are polynomials. To compute 𝐥𝐢𝐦
𝒉→±∞
𝒑(𝒙)
𝒒(𝒙)
factor largest power of x
out of both 𝒑(𝒙)𝒂𝒏𝒅 𝒒(𝒙)and then compute limit.
𝐥𝐢𝐦
𝒉→−∞
𝟑𝒙 𝟐−𝟒
𝟓𝒙−𝟐𝒙 𝟐
= 𝐥𝐢𝐦
𝒉→−∞
𝒙 𝟐(𝟑−
𝟒
𝒙 𝟐)
𝒙 𝟐(
𝟓
𝒙
−𝟐)
= 𝐥𝐢𝐦
𝒉→−∞
𝟑−
𝟒
𝒙 𝟐
𝟓
𝒙
−𝟐
= −
𝟑
𝟐
𝑵𝒐𝒕𝒆: (𝑪𝒂𝒏 𝒂𝒍𝒔𝒐 𝒃𝒆 𝒔𝒐𝒍𝒗𝒆𝒅 𝒖𝒔𝒊𝒏𝒈 𝑳′
𝑯𝒐̂𝒑𝒊𝒕𝒂𝒍′
𝒔 𝒓𝒖𝒍𝒆)
Piecewise Function:
𝐥𝐢𝐦
𝒙→−𝟐
𝒈(𝒙) where 𝒈(𝒙) = {
𝒙 𝟐
+ 𝟓, 𝒊𝒇 𝒙 < −𝟐
𝟏 − 𝟑𝒙, 𝒊𝒇 𝒙 ≥ −𝟐
Compute two “one-sided limits”,
[ 𝐥𝐢𝐦
𝒙→−𝟐−
𝒈(𝒙) = 𝐥𝐢𝐦
𝒙→−𝟐−
𝒙 𝟐
+ 𝟓 = 𝟗] and [ 𝐥𝐢𝐦
𝒙→−𝟐+
𝒈(𝒙) = 𝐥𝐢𝐦
𝒙→−𝟐+
𝟏 − 𝟑𝒙 = 𝟕]
The one sided limits are different so 𝐥𝐢𝐦
𝒙→−𝟐
𝒈(𝒙) Does Not Exist. (If the two
"one-sided limits" had been equal then 𝐥𝐢𝐦
𝒙→−𝟐
𝒈(𝒙) would have existed and
had the same value.)
𝑳′
𝑯𝒐̂𝒑𝒊𝒕𝒂𝒍′
𝒔 𝑹𝒖𝒍𝒆:
If 𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙)
𝒈(𝒙)
=
𝟎
𝟎
𝒐𝒓 𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙)
𝒈(𝒙)
=
±∞
±∞
, then 𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙)
𝒈(𝒙)
= 𝐥𝐢𝐦
𝒙→𝒂
𝒇′(𝒙)
𝒈′(𝒙)
𝒂 is a 𝒏𝒖𝒎𝒃𝒆𝒓, ∞, 𝒐𝒓 − ∞
Note: Continue differentiating if outcome is still the same.
List of continuous functions:
A partial list of continuous functions and the values of x for which they are
continuous.
𝑷𝒐𝒍𝒚𝒏𝒐𝒎𝒊𝒂𝒍𝒔 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙 𝒍𝒏 𝒙 𝒇𝒐𝒓 𝒙 > 𝟎
𝑹𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏, 𝒆𝒙𝒄𝒆𝒑𝒕 𝒇𝒐𝒓
𝒙’𝒔 𝒕𝒉𝒂𝒕 𝒈𝒊𝒗𝒆 𝒅𝒊𝒗𝒊𝒔𝒊𝒐𝒏 𝒃𝒚 𝒛𝒆𝒓𝒐
𝒄𝒐𝒔(𝒙) 𝒂𝒏𝒅 𝒔𝒊𝒏(𝒙) 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙
𝒕𝒂𝒏(𝒙)𝒂𝒏𝒅 𝒔𝒆𝒄(𝒙)𝒑𝒓𝒐𝒗𝒊𝒅𝒆𝒅
𝒙 ≠ ⋯ , −
𝟑𝝅
𝟐
, −
𝝅
𝟐
,
𝝅
𝟐
,
𝟑𝝅
𝟐
,⋅⋅⋅√𝒙
𝒏
(𝒏 𝒐𝒅𝒅)𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙
√𝒙
𝒏
(𝒏 𝒆𝒗𝒆𝒏)𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙 ≥ 𝟎 𝒄𝒐𝒕(𝒙)𝒂𝒏𝒅 𝒄𝒔𝒄(𝒙)𝒑𝒓𝒐𝒗𝒊𝒅𝒆𝒅
𝒙 ≠ ⋯ − 𝟐𝝅, −𝝅, 𝟎, 𝝅, 𝟐𝝅,⋅⋅⋅𝒆 𝒙
𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙
The intermediate Value Theorem:
Suppose that 𝒇(𝒙) is continuous on [𝒂, 𝒃]and let 𝑴 be any number between
𝒇(𝒂) and 𝒇(𝒃). Then there exists a number 𝒄 such that 𝒂 < 𝒄 < 𝒃 and
𝒇(𝒄) = 𝑴.
Derivatives:
Definition and Notation
 If 𝒚 = 𝒇(𝒙)then the derivative is defined to be 𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
𝒇(𝒙+𝒉)−𝒇(𝒙)
𝒉
If 𝒚 = 𝒇(𝒙), then all of the following are equivalent notations for the
derivative. 𝒇′(𝒙) = 𝒚′
=
𝒅𝒇
𝒅𝒙
=
𝒅𝒚
𝒅𝒙
=
𝒅
𝒅𝒙
(𝒇(𝒙)) = 𝑫𝒇(𝒙)
If 𝒚 = 𝒇(𝒙), then all of the following are equivalent notations for the
derivative evaluated at 𝒙 = 𝒂. 𝒇′(𝒂) = 𝒚′
| 𝒙=𝒂 =
𝒅𝒇
𝒅𝒙
| 𝒙=𝒂
𝒅𝒚
𝒅𝒙
| 𝒙=𝒂 = 𝑫𝒇(𝒂)
Alternative Formulas:
 The derivative of a function 𝒇 at a number 𝒂, denoted by 𝒇’(𝒂), is
𝒇′(𝒂) = 𝐥𝐢𝐦
𝒉→𝟎
𝒇(𝒂+𝒉)−𝒇(𝒂)
𝒉
at point (𝒂, 𝒇(𝒂))
 If we write 𝒙 = 𝒂 + 𝒉, then we have 𝒉 = 𝒙 − 𝒂 and 𝒉 approaches 𝟎, if and
only if, 𝒙 approaches 𝒂. Therefore, at point (𝒂, 𝒇(𝒂)), 𝒇′(𝒂) = 𝐥𝐢𝐦
𝒙→𝒂
𝒇(𝒙)−𝒇(𝒂)
𝒙−𝒂
 The Instantaneous rate of change of 𝒚 with respect to 𝒙 at 𝒙 = 𝒙 𝟏, which is
interpreted as the slope of the tangent to the curve 𝒚 = 𝒇(𝒙) at
𝑷(𝒙 𝟏, 𝒇(𝒙 𝟏)). Instantaneous rate of change = 𝐥𝐢𝐦
𝚫𝒙→𝟎
𝚫𝒚
𝚫𝒙
= 𝐥𝐢𝐦
𝒙 𝟐→𝒙 𝟏
𝒇(𝒙 𝟐)−𝒇(𝒙 𝟏)
𝒙 𝟐−𝒙 𝟏
Interpretation of the Derivative
If 𝒚 = 𝒇(𝒙) then,
1. 𝒎 = 𝒇′(𝒂)is the slope of the tangent line to 𝒚 = 𝒇(𝒙) 𝒂𝒕 𝒙 = 𝒂 and
the equation of the tangent line at 𝒙 = 𝒂 is given by
𝒚 = 𝒇(𝒂) + 𝒇′(𝒂)(𝒙 − 𝒂) Refer to “Slope Formulas:”
2. 𝒇′(𝒂) is the instantaneous rate of change of 𝒇(𝒙) at 𝒙 = 𝒂.
3. If 𝒇(𝒙)is the position of an object at 𝒙, then 𝒇′(𝒂)is the velocity of the
object at 𝒙 = 𝒂.
Basic Properties and Formulas
If 𝒇(𝒙) and 𝒈(𝒙) are differentiable functions, and (𝒄 and 𝒏) are any real
numbers, then,
(𝒄𝒇)′
= 𝒄𝒇′
Constant multiple
rule
𝒅
𝒅𝒙
(𝒄) = 𝟎
Constant function rule
𝒅
𝒅𝒙
(𝒙) = 𝟏
(𝒇 ± 𝒈)′
= 𝒇′ ± 𝒈′
Sum & Difference rule
𝒅
𝒅𝒙
(𝒙 𝒏) = 𝒏𝒙 𝒏−𝟏
Power rule
(𝒇𝒈)′
= 𝒇′
𝒈 + 𝒇𝒈′
Product rule
𝒅
𝒅𝒙
(𝒇(𝒈(𝒙))) = 𝒇′
(𝒈(𝒙))𝒈′(𝒙)
Chain rule
(
𝒇
𝒈
)
′
=
𝒇′ 𝒈−𝒇𝒈′
𝒈 𝟐
Quotient rule
𝒅
𝒅𝒙
(𝒆 𝒙) = 𝒆 𝒙
𝒅
𝒅𝒙
of a natural exponential function
Common Derivatives:
𝒅
𝒅𝒙
(𝒙) = 𝟏
𝒅
𝒅𝒙
(𝒔𝒊𝒏−𝟏
𝒙) =
𝟏
√ 𝟏−𝒙 𝟐
𝒅
𝒅𝒙
(𝒔𝒊𝒏𝒙) = 𝒄𝒐𝒔𝒙
𝒅
𝒅𝒙
(𝒄𝒐𝒔−𝟏
𝒙) = −
𝟏
√ 𝟏−𝒙 𝟐
𝒅
𝒅𝒙
(𝒄𝒐𝒔𝒙) = −𝒔𝒊𝒏𝒙
𝒅
𝒅𝒙
(𝒕𝒂𝒏−𝟏
𝒙) =
𝟏
𝟏+𝒙 𝟐
𝒅
𝒅𝒙
(𝒕𝒂𝒏𝒙) = 𝒔𝒆𝒄 𝟐
𝒙
𝒅
𝒅𝒙
(𝒂 𝒙) = 𝒂 𝒙
𝒍𝒏(𝒂)
𝒅
𝒅𝒙
(𝒔𝒆𝒄𝒙) = 𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙
𝒅
𝒅𝒙
(𝒆 𝒙) = 𝒆 𝒙
𝒅
𝒅𝒙
(𝒄𝒔𝒄𝒙) = −𝒄𝒔𝒄𝒙 𝒄𝒐𝒕𝒙
𝒅
𝒅𝒙
(𝒍𝒏(𝒙)) =
𝟏
𝒙
, 𝒙 > 𝟎
𝒅
𝒅𝒙
(𝒄𝒐𝒕𝒙) = −𝒄𝒔𝒄 𝟐
𝒙
𝒅
𝒅𝒙
(𝒍𝒏|𝒙|) =
𝟏
𝒙
, 𝒙 ≠ 𝟎
𝒅
𝒅𝒙
(𝐥𝐨𝐠 𝒂(𝒙)) =
𝟏
𝒙𝒍𝒏𝒂
, 𝒙 > 𝟎
The Mean Value Theorem:
If 𝒇(𝒙) is continuous on the closed interval [𝒂, 𝒃] and differentiable on the
open interval (𝒂, 𝒃), then there is a number 𝒂 < 𝒄 < 𝒃 such that
𝒇′(𝒄) =
𝒇(𝒃)−𝒇(𝒂)
𝒃−𝒂
.
The Chain Rule:
The chain rule is a formula for computing the derivative of the composition
of two or more functions.
Notations:
𝟏. ) (𝒇 ∘ 𝒈)′
= (𝒇′
∘ 𝒈) ⋅ 𝒈′ 𝟐. ) 𝒇′(𝒈(𝒙)) = 𝒇′
(𝒈(𝒙))𝒈′(𝒙)
𝟑. )
𝒅𝒛
𝒅𝒙
=
𝒅𝒛
𝒅𝒚
⋅
𝒅𝒚
𝒅𝒙
Consider 𝒛 to be a function of the variable 𝒚, which is
itself a function of 𝒙 (𝒚 and 𝒛 are therefore dependent variables), and
so, 𝒛 becomes a function of 𝒙.
Chain Rule Variants
The chain rule applied to some specific functions
𝒅
𝒅𝒙
([𝒇(𝒙)] 𝒏
= 𝒏[𝒇(𝒙)] 𝒏−𝟏
𝒇′(𝒙)
𝒅
𝒅𝒙
(𝒆 𝒇(𝒙)
) = 𝒇′(𝒙)𝒆 𝒇(𝒙)
𝒅
𝒅𝒙
(𝒔𝒊𝒏[𝒇(𝒙)]) = 𝒇′(𝒙) 𝒄𝒐𝒔[𝒇(𝒙)]
𝒅
𝒅𝒙
(𝒄𝒐𝒔[𝒇(𝒙)] = −𝒇′(𝒙) 𝒔𝒊𝒏[𝒇(𝒙)]
𝒅
𝒅𝒙
(𝒕𝒂𝒏[𝒇(𝒙)] = 𝒇′(𝒙) 𝒔𝒆𝒄 𝟐[𝒇(𝒙)]
𝒅
𝒅𝒙
(𝒔𝒆𝒄[𝒇(𝒙)]) = 𝒇′(𝒙) 𝒔𝒆𝒄[𝒇(𝒙)]𝒕𝒂𝒏[𝒇(𝒙)]
𝒅
𝒅𝒙
(𝒕𝒂𝒏−𝟏[𝒇(𝒙)]) =
𝒇′(𝒙)
𝟏+[𝒇(𝒙)] 𝟐
𝒅
𝒅𝒙
𝒇(𝒙) 𝒈(𝒙)
= 𝒇(𝒙) 𝒈(𝒙)
⋅ [𝒍𝒏𝒇(𝒙) ⋅ 𝒈(𝒙)]′
Use product rule to find [𝒍𝒏𝒇(𝒙) ⋅ 𝒈(𝒙)]′
→
𝒅
𝒅𝒙
(𝒍𝒏[𝒇(𝒙)]) =
𝒇′(𝒙)
𝒇(𝒙)
Implicit Differentiation
Find 𝒚’ if 𝒆 𝟐𝒙−𝟗𝒚
+ 𝒙 𝟑
𝒚 𝟐
= 𝒔𝒊𝒏(𝒚) + 𝟏𝟏𝒙. Remember 𝒚 = 𝒚(𝒙) here, so
products/quotients of 𝒙 and 𝒚 will use the product/quotient rule and
derivatives of 𝒚 will use the chain rule. The “trick” is to differentiate as
normal and every time you differentiate a 𝒚 you tack on a 𝒚’ (from the chain
rule). After differentiating solve for 𝒚’.
𝒆 𝟐𝒙−𝟗𝒚(𝟐 − 𝟗𝒚′) + 𝟑𝒙 𝟐
𝒚 𝟐
+ 𝟐𝒙 𝟑
𝒚𝒚′
= 𝒄𝒐𝒔(𝒚)𝒚′
+ 𝟏𝟏
 𝟐𝒆 𝟐𝒙−𝟗𝒚
− 𝟗𝒚′
𝒆 𝟐𝒙−𝟗𝒚
+ 𝟑𝒙 𝟐
𝒚 𝟐
+ 𝟐𝒙 𝟑
𝒚𝒚′
= 𝒄𝒐𝒔(𝒚)𝒚′
+ 𝟏𝟏
 (𝟐𝒙 𝟑
𝒚 − 𝟗𝒆 𝟐𝒙−𝟗𝒚
− 𝒄𝒐𝒔(𝒚))𝒚′
𝟏𝟏 − 𝟐𝒆 𝟐𝒙−𝟗𝒚
− 𝟑𝒙 𝟐
𝒚 𝟐
 𝒚′
=
𝟏𝟏−𝟐𝒆 𝟐𝒙−𝟗𝒚−𝟑𝒙 𝟐 𝒚 𝟐
𝟐𝒙 𝟑 𝒚−𝟗𝒆 𝟐𝒙−𝟗𝒚−𝒄𝒐𝒔(𝒚)
Increasing/Decreasing – Concave Up/Concave Down:
Critical Points:
𝒙 = 𝒄 is a critical point of 𝒇(𝒙), provided that either,
𝟏. ) 𝒇′ (𝒄) = 𝟎 Or 𝟐. ) 𝒇′(𝒄) 𝑫𝑵𝑬
Increasing / Decreasing:
1.) If 𝒇′(𝒙) > 𝟎 for all 𝒙 in an interval 𝑰, then 𝒇(𝒙)is increasing on the
interval 𝑰.
2.) If 𝒇′(𝒙) < 𝟎 for all 𝒙 in an interval 𝑰, then 𝒇(𝒙)is decreasing on the
interval 𝑰.
3.) If 𝒇′(𝒙) = 𝟎 for all 𝒙 in an interval 𝑰, then 𝒇(𝒙)is constant on the
interval 𝑰.
Concave Up / Concave Down:
1.) If 𝒇′′(𝒙) > 𝟎 for all x in an interval 𝑰, then 𝒇(𝒙) is concave up on the
interval 𝑰.
2.) If 𝒇′′(𝒙) < 𝟎 for all x in an interval 𝑰, then 𝒇(𝒙) is concave down on the
interval 𝑰.
Inflection Points:
𝒙 = 𝒄 is an inflection point of 𝒇(𝒙) if the concavity changes at 𝒙 = 𝒄
Extrema:
Absolute Extrema:
1.) 𝒙 = 𝒄 is an absolute Max. of 𝒇(𝒙) if 𝒇(𝒄) ≥ 𝒇(𝒙)for all 𝒙 in the domain.
2.) 𝒙 = 𝒄 is an absolute Min. of 𝒇(𝒙) if 𝒇(𝒄) ≤ 𝒇(𝒙)for all 𝒙 in the domain.
Relative (local) Extrema:
1.) 𝒙 = 𝒄 is a relative (local) Maximum of 𝒇(𝒙) if 𝒇(𝒄) ≥ 𝒇(𝒙)for all 𝒙 near 𝒄.
2.) 𝒙 = 𝒄 is a relative (local) Minimum of 𝒇(𝒙) if 𝒇(𝒄) ≤ 𝒇(𝒙)for all 𝒙 near 𝒄.
Fermat’s Theorem:
If 𝒇(𝒙) has a relative (local) extrema at 𝒙 = 𝒄 , then 𝒙 = 𝒄 is a critical point of
𝒇(𝒙).
Extreme Value Theorem:
If 𝒇(𝒙) is continuous on the closed interval [𝒂, 𝒃] then there exist numbers 𝒄
and 𝒅 so that,
𝟏. ) 𝒂 ≤ 𝒄 , 𝒅 ≤ 𝒃 𝟐. ) 𝒇(𝒄)𝒊𝒔 𝒕𝒉𝒆 𝒂𝒃𝒔. 𝑴𝒂𝒙. 𝒊𝒏 [𝒂, 𝒃]
𝟑. ) 𝒇(𝒅)𝒊𝒔 𝒕𝒉𝒆 𝒂𝒃𝒔. 𝑴𝒊𝒏. 𝒊𝒏 [𝒂, 𝒃]
Finding Absolute Extrema:
To find the absolute extrema of the continuous function 𝒇(𝒙) on interval
[𝒂, 𝒃], use the following process.
1.) Find all critical points of 𝒇(𝒙) in [𝒂, 𝒃].
2.) Evaluate 𝒇(𝒙) at all points found in Step 1.
3.) Evaluate 𝒇(𝒂) and 𝒇(𝒃).
4.) Identify the abs. Max. (largest function value) and abs. Min. (smallest
function value) from the evaluations in Steps 2 & 3.
1st Derivative Test:
If 𝒙 = 𝒄 is a critical point of 𝒇(𝒙) then 𝒙 = 𝒄 is
1. a rel. Max. of 𝒇(𝒙) if 𝒇′(𝒙) > 𝟎 to the left of 𝒙 = 𝒄 and 𝒇′(𝒙) < 𝟎 to the
right of 𝒙 = 𝒄.
2. a rel. Min. of 𝒇(𝒙) if 𝒇′(𝒙) < 𝟎 to the left of 𝒙 = 𝒄 and 𝒇′(𝒙) > 𝟎 to the
right of 𝒙 = 𝒄
3. not a relative extrema of 𝒇(𝒙) if 𝒇′(𝒙) is the same sign on both sides of
𝒙 = 𝒄.
2nd Derivative Test:
If 𝒙 = 𝒄 is a critical point of 𝒇(𝒙) such that 𝒇′(𝒄) = 𝟎, then 𝒙 = 𝒄
1. is a relative maximum of 𝒇(𝒙) if 𝒇′′(𝒄) < 𝟎.
2. Is a relative minimum of 𝒇(𝒙) if 𝒇′′(𝒄) > 𝟎.
3. May be relative maximum, relative minimum, or neither if 𝒇′′(𝒄) = 𝟎.
Finding Relative Extrema and/or Classify Critical Points:
1. Find all critical points of 𝒇(𝒙).
2. Use 1st
derivative test or 2nd
derivative test on each critical point.

7. Problems (examples):
Limits:
1.) For the function 𝒇 whose graph is shown, state the value of each quantity,
if it exists.
a) 𝐥𝐢𝐦
𝒙→−𝟐
𝒇(𝒙) = DNE
b) 𝐥𝐢𝐦
𝒙→𝟎
𝒇(𝒙) = 𝟑
c) 𝐥𝐢𝐦
𝒙→𝟏
𝒇(𝒙) = 𝟐
d) 𝐥𝐢𝐦
𝒙→𝟐−
𝒇(𝒙) = −𝟏
e) 𝐥𝐢𝐦
𝒙→𝟐+
𝒇(𝒙) = ∞
f) 𝐥𝐢𝐦
𝒙→−∞
𝒇(𝒙) = 𝟏
2.) Function 𝒇 is continuous at intervals: (−∞, −𝟐), [−𝟐, 𝟎), (𝟎, 𝟐], (𝟐, ∞)
1.) Evaluate: 𝐥𝐢𝐦
𝒕→𝟎
𝒕 𝟐−𝟗
𝟐𝒕 𝟐+𝟕𝒕+𝟑
if it exists.
𝐥𝐢𝐦
𝒕→𝟎
𝒕 𝟐−𝟗
𝟐𝒕 𝟐+𝟕𝒕+𝟑
=
𝐥𝐢𝐦
𝒕→𝟎
(𝒕−𝟑)(𝒕+𝟑)
𝐥𝐢𝐦
𝒕→𝟎
(𝟐𝒕−𝟏)(𝒕+𝟑)
=
(𝟎−𝟑)
(𝟐(𝟎)+𝟏)
= −𝟑
3.) Evaluate: 𝐥𝐢𝐦
𝒕→−𝟑
𝒕 𝟐−𝟗
𝟐𝒕 𝟐+𝟕𝒕+𝟑
if it exists.
𝐥𝐢𝐦
𝒕→−𝟑
𝒕 𝟐−𝟗
𝟐𝒕 𝟐+𝟕𝒕+𝟑
=
𝐥𝐢𝐦
𝒕→−𝟑
(𝒕−𝟑)(𝒕+𝟑)
𝐥𝐢𝐦
𝒕→−𝟑
(𝟐𝒕−𝟏)(𝒕+𝟑)
=
((−𝟑)−𝟑)
(𝟐(−𝟑)+𝟏)
=
𝟔
𝟓
2.) Evaluate: 𝐥𝐢𝐦
𝒕→∞
𝒕 𝟐−𝟗
𝟐𝒕 𝟐+𝟕𝒕+𝟑
if it exists.
𝐥𝐢𝐦
𝒕→∞
𝒕 𝟐−𝟗
𝟐𝒕 𝟐+𝟕𝒕+𝟑
=
𝐥𝐢𝐦
𝒕→∞
(𝒕 𝟐)
𝐥𝐢𝐦
𝒕→∞
(𝟐𝒕 𝟐)
=
𝟏
𝟐
Derivatives:
𝑳𝒆𝒕 𝒇(𝒙) = 𝟐𝒙 − 𝟑𝒙 𝟐
a.) Find the derivative 𝒇′(𝒙) Using Formula 𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
𝒇(𝒙+𝒉)−𝒇(𝒙)
𝒉

𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
[𝟐(𝒙+𝒉)−𝟑(𝒙+𝒉) 𝟐]−(𝟐𝒙−𝟑𝒙 𝟐)
𝒉
 Simplify 
𝐥𝐢𝐦
𝒉→𝟎
[(𝟐𝒙+𝟐𝒉)−𝟑(𝒙 𝟐+𝒙𝒉+𝒉 𝟐)]−(𝟐𝒙−𝟑𝒙 𝟐)
𝒉
 Simplify  𝐥𝐢𝐦
𝒉→𝟎
𝟐𝒉−𝟔𝒙𝒉−𝟑𝒉 𝟐
𝒉

𝐥𝐢𝐦
𝒉→𝟎
𝟐 − 𝟔𝒙 − 𝟑𝒉 = 𝐥𝐢𝐦
𝒉→𝟎
𝟐 − 𝟔𝒙 − 𝟑(𝟎) = 𝒇′(𝒙) = 𝟐 − 𝟔𝒙
b.) Find 𝒇′(𝟏) → 𝒇′(𝒙) = 𝟐 − 𝟔(𝟏) = −𝟒
c.) Find the equation of the line tangent to 𝒇(𝒙) at the point (𝟏, −𝟏)
𝒚 − (−𝟏) = −𝟒(𝒙 − 𝟏) → 𝒚 = −𝟒𝒙 + 𝟑
Differentiate 𝒚 = 𝟒𝒙 𝟑
𝒆 𝒙
 𝑷𝒓𝒐𝒅𝒖𝒄𝒕 𝒓𝒖𝒍𝒆 𝒂𝒏𝒅 𝑷𝒐𝒘𝒆𝒓 𝑹𝒖𝒍𝒆 
𝒚′
= (𝟒(𝟑)𝒙 𝟐)(𝒆 𝒙) + 𝟒𝒙 𝟑(𝒆 𝒙) = 𝟏𝟐𝒙 𝟐
𝒆 𝒙
+ 𝟒𝒙 𝟑
𝒆 𝒙
= 𝟒𝒙 𝟐
𝒆 𝒙(𝟑 + 𝒙)
Find the derivative of 𝒚 =
𝟏−𝒔𝒆𝒄𝒙
𝒕𝒂𝒏𝒙
 𝑸𝒖𝒐𝒕𝒊𝒆𝒏𝒕 𝒓𝒖𝒍𝒆 
𝒚′
=
(𝒕𝒂𝒏𝒙)(𝟎−(𝒔𝒆𝒄𝒙𝒕𝒂𝒏𝒙)−(𝟏−𝒔𝒆𝒄𝒙)(𝒔𝒆𝒄 𝟐 𝒙)
𝒕𝒂𝒏 𝟐 𝒙
=
−𝒔𝒆𝒄𝒙𝒕𝒂𝒏 𝟐 𝒙−(𝟏−𝒔𝒆𝒄𝒙)𝒔𝒆𝒄 𝟐 𝒙
𝒕𝒂𝒏 𝟐 𝒙
=
−𝒔𝒆𝒄𝒙𝒕𝒂𝒏 𝟐 𝒙
𝒕𝒂𝒏 𝟐 𝒙
−
(𝟏−𝒔𝒆𝒄𝒙)𝒔𝒆𝒄 𝟐 𝒙
𝒕𝒂𝒏 𝟐 𝒙
= −𝒔𝒆𝒄 −
(𝟏−𝒔𝒆𝒄𝒙)𝒔𝒆𝒄 𝟐 𝒙
𝒕𝒂𝒏 𝟐 𝒙
Find the derivative of 𝒚 = 𝟑𝒆 √𝒙
𝟑
 𝑪𝒉𝒂𝒊𝒏 𝑹𝒖𝒍𝒆  𝒖 = 𝒈(𝒙) = √𝒙
𝟑
= 𝒙
𝟏
𝟑 𝒂𝒏𝒅 𝒚 = 𝒇(𝒖) = 𝟑𝒆 𝒖
→
𝒚′
= (
𝟏
𝟑
𝒙−
𝟐
𝟑) (𝟑𝒆 𝒖) = (
𝟏
𝟑
𝒙−
𝟐
𝟑) (𝟑𝒆 √𝒙
𝟑
) = (
𝟏
𝟑
∙
𝟏
𝒙
𝟐
𝟑
) (𝟑𝒆 √𝒙
𝟑
) =
𝒆 √𝒙
𝟑
√𝒙
𝟑
Implicit Differentiation:
Find
𝒅𝒚
𝒅𝒙
by implicit differentiation.
1.) 𝒚 𝒄𝒐𝒔𝒙 = 𝒙 𝟐
+ 𝒚 𝟐
→
𝒅
𝒅𝒙
(𝒚𝒄𝒐𝒔𝒙) =
𝒅
𝒅𝒙
(𝒙 𝟐
+ 𝒚 𝟐) →
𝒚(−𝒔𝒊𝒏𝒙) + 𝒄𝒐𝒔𝒙 ⋅ 𝒚′
= 𝟐𝒙 + 𝟐𝒚𝒚′
→ 𝒄𝒐𝒔𝒙 ⋅ 𝒚′
− 𝟐𝒚𝒚′
= 𝟐𝒙 + 𝒚𝒔𝒊𝒏𝒙 →
𝒚′(𝒄𝒐𝒔𝒙 − 𝟐𝒚) = 𝟐𝒙 + 𝒚𝒔𝒊𝒏𝒙 → 𝒚′
=
𝟐𝒙+𝒚𝒔𝒊𝒏𝒙
𝒄𝒐𝒔𝒙−𝟐𝒚
2.) 𝒙 𝟒(𝒙 + 𝒚) = 𝒚 𝟐(𝟑𝒙 − 𝒚) →
𝒅
𝒅𝒙
[𝒙 𝟒(𝒙 + 𝒚)] =
𝒅
𝒅𝒙
[𝒚 𝟐(𝟑𝒙 − 𝒚)] →
𝒙 𝟒 (𝟏 + 𝒚′) + (𝒙 + 𝒚) ⋅ 𝟒𝒙 𝟑
= 𝟐𝒚 𝟐(𝟑 − 𝒚′) + (𝟑𝒙 − 𝒚) ⋅ 𝟐𝒚𝒚′
→
𝒙 𝟒
+ 𝒙 𝟒
𝒚′
+ 𝟒𝒙 𝟒
+ 𝟒𝒙 𝟑
𝒚 = 𝟑𝒚 𝟐
− 𝒚 𝟐
𝒚′
+ 𝟔𝒙𝒚 − 𝟐𝒚 𝟐
𝒚′
→
𝒙 𝟒
𝒚′
+ 𝟑𝒚 𝟐
𝒚′
− 𝟔𝒙𝒚′
= 𝟑𝒚 𝟐
− 𝟓𝒙 𝟒
− 𝟒𝒙 𝟑
𝒚 →
(𝒙 𝟒
+ 𝟑𝒚 𝟐
− 𝟔𝒙𝒚)𝒚′
= 𝟑𝒚 𝟐
− 𝟓𝒙 𝟒
− 𝟒𝒙 𝟑
𝒚 → 𝒚′
=
𝟑𝒚 𝟐−𝟓𝒙 𝟒−𝟒𝒙 𝟑 𝒚
𝒙 𝟒+𝟑𝒚 𝟐−𝟔𝒙𝒚
Derivatives of Logarithmic Functions:
Differentiate the function
1.) 𝒇(𝒙) = 𝒍𝒐𝒈 𝟏𝟎(𝒙 𝟑
+ 𝟏) → 𝒇′(𝒙) =
𝟏
(𝒙 𝟑+𝟏)𝒍𝒏𝟏𝟎
𝒅
𝒅𝒙
(𝒙 𝟑
+ 𝟏) =
𝟑𝒙 𝟐
(𝒙 𝟑+𝟏)𝒍𝒏𝟏𝟎
2.) 𝒇(𝒙) = 𝒍𝒏(𝒍𝒏(𝒙)) 𝒈(𝒙) = 𝒍𝒏(𝒙) 𝒂𝒏𝒅 𝒇(𝒙) = 𝒍𝒏(𝒈(𝒙))
𝒇′(𝒙) =
𝒈′(𝒙)
𝒈(𝒙)
=
𝟏
𝒙
÷
𝒍𝒏(𝒙)
𝟏
=
𝟏
𝒙
⋅
𝟏
𝒍𝒏(𝒙)
=
𝟏
𝒙𝒍𝒏(𝒙)
3.) 𝒚 = (𝒄𝒐𝒔𝒙) 𝒙
→ 𝐲′ = (
𝒅
𝒅𝒙
[𝒙] ⋅ 𝒍𝒏(𝒄𝒐𝒔(𝒙)) + 𝒙 ⋅
𝒅
𝒅𝒙
[𝒍𝒏(𝒄𝒐𝒔(𝒙))]) 𝒄𝒐𝒔 𝒙
(𝒙)
= 𝒄𝒐𝒔 𝒙(𝒙)(𝟏 𝒍𝒏(𝒄𝒐𝒔(𝒙)) +
𝟏
𝒄𝒐𝒔(𝒙)
⋅
𝒅
𝒅𝒙
[𝒄𝒐𝒔(𝒙)] ⋅ 𝒙)
= 𝒄𝒐𝒔 𝒙(𝒙)(𝒍𝒏(𝒄𝒐𝒔(𝒙)) +
(−𝒔𝒊𝒏(𝒙))𝒙
𝒄𝒐𝒔(𝒙)
) = 𝒄𝒐𝒔 𝒙(𝒙) (𝒍𝒏(𝒄𝒐𝒔(𝒙)) −
𝒙 𝒔𝒊𝒏(𝒙)
𝒄𝒐𝒔(𝒙)
)
= 𝒄𝒐 𝒔 𝒙
(𝒙)(𝒍𝒏(𝒄𝒐𝒔(𝒙)) − 𝒙 𝒕𝒂𝒏(𝒙))
Rates of Change:
1.) Finding velocity and Acceleration:
The position of a particle along a straight line is given by the function
−𝟏𝟔𝒕 𝟐
+ 𝟐𝟓𝟎𝒕 − 𝟑𝟎 , where 𝒇 is measured in feet and 𝒕 is seconds. Find the
velocity and acceleration of the particle after 4 seconds.
First derivative = Velocity.
𝒗(𝒕) =
𝒅𝒔
𝒅𝒕
= −𝟏𝟔(𝟐)𝐭 + 𝟐𝟓𝟎(𝟏) − 𝟎 = 𝟐𝟓𝟎 − 𝟑𝟐𝐱 →
𝒗(𝟒) = 𝟐𝟓𝟎 − 𝟑𝟐(𝟒) = 𝟏𝟐𝟐 𝒇𝒕/𝒔𝒆𝒄 Second derivative = Acceleration.
𝒂(𝒕) = 𝒅𝒗/𝒅𝒕 = −𝟑𝟐(𝟏) + 𝟐𝟓𝟎(𝟎) = −𝟑𝟐 → 𝒂(𝟒) = −𝟑𝟐 𝒇𝒕 ⁄ 𝒔𝒆𝒄
2.) Change in pressure to change in volume:
Boyle’s Law states that when a sample of gas is compressed at a constant
temperature, the pressure P and volume V satisfy the equation PV=C, where
C is constant. Suppose that at a certain instant the volume is 1000 cm3
, the
pressure is 250 kPa, and the pressure is decreasing at a rate of 15 kPa/min.
At what rate is the volume increasing at this instant?
𝑽 = 𝟏𝟎𝟎𝟎𝒄𝒎 𝟑
𝑷 = 𝟐𝟓𝟎𝒌𝑷𝒂
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑽𝒐𝒍.
=
𝒅𝑽
𝒅𝒕
= 𝑽′
=?
𝒄𝒉𝒂𝒏𝒈 𝒊𝒏 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆
=
𝒅𝑷
𝒅𝒕
= 𝑽′
= −𝟏𝟓
𝒌𝑷𝒂
𝒎𝒊𝒏
𝑷𝑽 = 𝑪 →
𝒅
𝒅𝒕
[𝑷𝑽] =
𝒅
𝒅𝒕
[𝑪] → 𝑷𝒓𝒐𝒅𝒖𝒄𝒕 𝒓𝒖𝒍𝒆 →
𝒅𝑷
𝒅𝒕
⋅ 𝑽 + 𝑷 ⋅
𝒅𝑽
𝒅𝒕
= 𝟎 →
𝑷 ⋅
𝒅𝑽
𝒅𝒕
= −𝑽
𝒅𝑷
𝒅𝒕
→
𝒅𝑽
𝒅𝒕
= −
𝑽
𝑷
⋅
𝒅𝑷
𝒅𝒕
𝒅𝑽
𝒅𝒕
= −
𝟏𝟎𝟎𝟎
𝟐𝟓𝟎
⋅ −𝟏𝟓 = 𝟔𝟎
𝒄𝒎 𝟑
𝒎𝒊𝒏
Related Rates:
 Sketch picture and identify known/unknown quantities. Write down
relating quantities and differentiate with respect to 𝐭 using implicit
differentiation (i.e. add on a derivative every time you differentiate a
function of (𝒕). Plug in known quantities and solve for the unknown
quantity.
1.) Kite string extending:
A kite 100 ft. above the ground moves only horizontally at a speed of 8 ft./s.
At what rate is the angle between the string and the horizontal decreasing
when 200 ft. of string has been let out?
𝒅𝒙
𝒅𝒕
= 𝟖
𝒇𝒕
𝒔𝒆𝒄
→
𝒄𝒐𝒕𝜽 =
𝒙
𝟏𝟎𝟎
→ 𝒙 = 𝟏𝟎𝟎 𝒄𝒐𝒕𝜽 →
𝒅𝒙
𝒅𝒕
= −𝟏𝟎𝟎𝒄𝒔𝒄 𝟐
𝜽
𝒅𝜽
𝒅𝒕
→
𝒅𝜽
𝒅𝒕
= −
𝒔𝒊𝒏 𝟐 𝜽
𝟏𝟎𝟎
⋅ 𝟖. 𝑾𝒉𝒆𝒏 𝒚 = 𝟐𝟎𝟎, 𝒔𝒊𝒏𝜽 =
𝟏𝟎𝟎
𝟐𝟎𝟎
=
𝟏
𝟐
→
𝒅𝜽
𝒅𝒕
= −
(𝟏 𝟐⁄ ) 𝟐
𝟏𝟎𝟎
⋅ 𝟖 = −
𝟏
𝟓𝟎
𝒓𝒂𝒅 𝒔𝒆𝒄⁄
2.) Boat approaching a dock:
A boat is pulled into a dock by a rope attached to the bow of the boat and
passing through a pulley on the dock that is 1 m higher than the bow of the
boat. If the rope is pulled in at a rate of 1m/s, how fast is the boat
approaching the dock when it is 8 m from the dock?
𝒅𝒚
𝒅𝒕
= −𝟏 𝒎 𝒔⁄
𝒅𝒙
𝒅𝒕
=? 𝒘𝒉𝒆𝒏 𝒙 = 𝟖 𝒎
𝒚 𝟐
= 𝒙 𝟐
+ 𝟏 →
𝟐𝒚
𝒅𝒚
𝒅𝒕
= 𝟐𝒙
𝒅𝒙
𝒅𝒕
→
𝒅𝒙
𝒅𝒕
=
𝟐𝒚
𝟐𝒙
𝒅𝒚
𝒅𝒕
= −
𝒚
𝒙
→ 𝑾𝒉𝒆𝒏 𝒙 = 𝟖 𝒚 = √𝟔𝟓 →
𝒅𝒙
𝒅𝒕
= −
√𝟔𝟓
𝟖
→ 𝑨𝒑𝒑𝒓𝒐𝒂𝒄𝒉𝒆𝒔 𝒅𝒐𝒄𝒌 @
√𝟔𝟓
𝟖
≈ 𝟏. 𝟎𝟏 𝒎 𝒔⁄
Linear Approximations and Differentials:
1.) Maximum error in area of disk:
The radius of a circular disk is given as 24 cm with a maximum error in
measurement of 0.2 cm. Use differentials to find the maximum error in the
calculated area of the disk.
𝒂) 𝑨 = 𝝅𝒓 𝟐
→ 𝒅𝑨 = 𝟐𝝅𝒓 ⋅ 𝒅𝒓 → 𝑾𝒉𝒆𝒏 𝒓 = 𝟐𝟒 𝒂𝒏𝒅 𝒅𝒓 = 𝟎. 𝟐,
𝒅𝑨 = 𝟐𝝅(𝟐𝟒)(𝟎. 𝟐) = 𝟗. 𝟔𝝅,
𝑴𝒂𝒙 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒆𝒓𝒓𝒐𝒓 = 𝟗. 𝟔𝝅 ≈ 𝟑𝟎 𝒄𝒎 𝟐
𝒃)𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒆𝒓𝒓𝒐𝒓 =
𝚫𝑨
𝑨
≈
𝒅𝑨
𝑨
=
𝟐𝝅𝒓 𝒅𝒓
𝝅𝒓 𝟐
=
𝟐𝒅𝒓
𝒓
=
𝟐(𝟎.𝟐)
𝟐𝟒
=
𝟎.𝟐
𝟏𝟐
=
𝟏
𝟔𝟎
= 𝟎. 𝟎𝟏𝟔̅ → 𝐏𝐞𝐫𝐜𝐞𝐧𝐭𝐚𝐠𝐞 𝐞𝐫𝐫𝐨𝐫 = 𝟏. 𝟔̅%
2.) Find 𝒅𝒚 and evaluate for given values:
𝒚 = √𝟑 + 𝒙 𝟐 , 𝒙 = 𝟏 , 𝒅𝒙 = −𝟎. 𝟏
𝒂)𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒅𝒚
𝒃)𝑬𝒗𝒂𝒍𝒖𝒂𝒕𝒆 𝒅𝒚 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒈𝒊𝒗𝒆𝒏 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 𝒙 𝒂𝒏𝒅 𝒅𝒙
𝒂) 𝒚 = √𝟑 + 𝒙 𝟐 → 𝒅𝒚 =
𝟏
𝟐
(𝟑 + 𝒙 𝟐)−
𝟏
𝟐(𝟐𝒙)𝒅𝒙 =
𝒙
√ 𝟑+𝒙 𝟐
𝒅𝒙
𝒃) 𝒙 = 𝟏 𝒂𝒏𝒅 𝒅𝒙 = −𝟎. 𝟏 → 𝒅𝒚 =
𝟏
√ 𝟑+𝒙 𝟐
(𝟎. 𝟏) = −𝟎. 𝟎𝟓
3.) Difference of functions:
Find the difference of each function.
𝒂) 𝒚 = 𝒙 𝟐
𝒔𝒊𝒏𝟐𝒙 𝒃) 𝒚 = 𝒍𝒏√𝟏 + 𝒕 𝟐
𝒂) 𝑻𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒅𝒚 𝒊𝒔 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒅𝒙 𝒃𝒚 𝒕𝒉𝒆
𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒅𝒚 = 𝒇’(𝒙)𝒅𝒙.
𝒇𝒐𝒓 𝒚 = 𝒇(𝒙) = 𝒙 𝟐
𝒔𝒊𝒏𝟐𝒙 , 𝒇′(𝒙) = 𝒙 𝟐
𝒄𝒐𝒔𝟐𝒙 ⋅ 𝟐 + 𝒔𝒊𝒏𝟐𝒙 ⋅ 𝟐𝒙
= 𝟐𝒙(𝒙𝒄𝒐𝒔𝟐𝒙 + 𝒔𝒊𝒏𝟐𝒙), 𝒔𝒐 𝒅𝒚 = 𝟐𝒙(𝒙𝒄𝒐𝒔𝟐𝒙 + 𝒔𝒊𝒏𝟐𝒙)𝒅𝒙.
𝒃) For 𝒚 = 𝒇(𝒕) = 𝒍𝒏√𝟏 + 𝒕 𝟐, 𝒇′(𝒕) =
𝟏
𝟐
⋅
𝟏
𝟏+𝒕 𝟐
⋅ 𝟐𝒕 =
𝒕
𝟏+𝒕 𝟐
, 𝒔𝒐
𝒅𝒚 =
𝒕
𝟏+𝒕 𝟐
𝒅𝒕
4.) Find the Linearization 𝑳(𝒙):
Find the linearization 𝑳(𝒙) of the function 𝒇(𝒙) = 𝒙
𝟓
𝟐 at 𝒙 = 𝟒.
𝑺𝒍𝒐𝒑𝒆:
𝒎 = 𝒇′(𝒂)
𝑷𝒐𝒊𝒏𝒕:
(𝒂, 𝒇(𝒂))
𝒚 − 𝒚 𝟏 = 𝒎(𝒙 − 𝒙 𝟏)
𝒚 − 𝒇(𝒂) = 𝒇′(𝒂)(𝒙 − 𝒂)
𝒚 = 𝒇(𝒂) + 𝒇′(𝒂)(𝒙 − 𝒂)
𝑳(𝒙) = (𝟒)
𝟓
𝟐 +
𝟓(𝟒)
𝟑
𝟐
𝟐
(𝒙 − (𝟒))
= 𝟐𝟎𝒙 − 𝟒𝟖
Use answer to estimate 𝟓
𝟓
𝟐 → 𝒇(𝒙) = √𝒙 𝟓 → √𝟓 𝟓 → 𝒇(𝟓) ≈
𝟐𝟎(𝟓) − 𝟒𝟖 = 𝟓𝟐
5.) Verify linear approximation:
Veryfy the given linear approximation at 𝒂 = 𝟎. Then determin the
values of 𝒙 for which the linear approximation is accurate to within 𝟎. 𝟏
𝒍𝒏(𝟏 + 𝒙) ≈ 𝒙
 𝒇(𝒙) = 𝒍𝒏(𝟏 + 𝒙) →
 𝒇′(𝒙) =
𝟏
𝟏+𝒙
→
 𝒇(𝟎) = 𝟎 & 𝒇′(𝟎) = 𝟏
 𝒇(𝒙) ≈ 𝒇(𝟎) + 𝒇′(𝟎)(𝒙 − 𝟎)
= 𝟎 + 𝟏(𝒙) = 𝒙
𝒍𝒏(𝟏 + 𝒙)−. 𝟎𝟏 < 𝒙 < 𝒍𝒏(𝟏 + 𝒙) + 𝟎. 𝟏 𝒘𝒉𝒆𝒏 − 𝟎. 𝟑𝟖𝟑 < 𝒙 < 𝟎. 𝟓𝟏𝟔
2.) Find critical numbers:
Find the critical numbers of the function 𝒇(𝒙) = 𝟐𝒙 𝟑
− 𝟑𝒙 𝟐
= 𝟑𝟔𝒙
𝒇′(𝒙) = 𝟔𝒙 𝟐
− 𝟔𝒙 − 𝟑𝟔 = 𝟔(𝒙 + 𝟐)(𝒙 + 𝟑). 𝒇′
= 𝟎 →
𝒙 = 𝟐, 𝟑 𝐂𝐫𝐢𝐭𝐢𝐜𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐚𝐫𝐞 𝟐 𝒂𝒏𝒅 𝟑
Find the critical numbers of the function
𝒚−𝟏
𝒚 𝟐−𝒚+𝟏
𝒈′(𝒚) =
(𝒚 𝟐−𝒚+𝟏)(𝟏)−(𝒚−𝟏)(𝟐𝒚−𝟏)
(𝒚 𝟐−𝒚+𝟏)
𝟐 =
𝒚(𝟐−𝒚)
(𝒚 𝟐−𝒚+𝟏)
𝟐
𝒈′(𝟎) → 𝒚 = 𝟎, 𝟐. The expression 𝒚 𝟐
− 𝒚 + 𝟏 is never equal to 0, so
𝒈′(𝒚) exists for all ℝ. Critical numbers are 𝟎 𝒂𝒏𝒅 𝟐
Find the critical numbers of the function 𝒇(𝜽) = 𝟐𝒄𝒐𝒔(𝜽) + 𝒔𝒊𝒏 𝟐
(𝜽)
𝒇′(𝜽) = −𝟐𝒔𝒊𝒏(𝜽) + 𝟐𝒔𝒊𝒏(𝜽)𝒄𝒐𝒔(𝜽).
𝒇′(𝜽) = 𝟎 → 𝟐𝒔𝒊𝒏(𝜽)(𝒄𝒐𝒔(𝜽) − 𝟏) = 𝟎 → 𝒔𝒊𝒏(𝜽) = 𝟎 𝒐𝒓 𝒄𝒐𝒔(𝜽) = 𝟏
𝜽 = 𝒏𝝅[𝒏 𝒂𝒏 𝒊𝒏𝒕𝒆𝒈𝒆𝒓]𝒐𝒓 𝜽 = 𝟐𝒏𝝅. Solutions 𝜽 = 𝒏𝝅 include
solutions 𝜽 = 𝟐𝒏𝝅, so critical numbers are 𝜽 = 𝒏𝝅
3.) Find Abs Max/Min Values:
Find the abs Max. and abs Min. values of 𝒇 on the given interval.
1.) 𝒇(𝒙) = 𝟏𝟐 + 𝟒𝒙 − 𝒙 𝟐
, [𝟎, 𝟓]
𝒇′(𝒙) = 𝟒 − 𝟐𝒙 = 𝟎 → 𝒙 = 𝟐. 𝒇(𝟎) = 𝟏𝟐, 𝒇(𝟐) = 𝟏𝟔, 𝒂𝒏𝒅 𝒇(𝟓) = 𝟕.
𝑺𝒐 𝒇(𝟐) = 𝟏𝟔 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒃𝒔 𝑴𝒂𝒙 𝒂𝒏𝒅 𝒇(𝟓) = 𝟕 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒃𝒔 𝒎𝒊𝒏
2.) 𝒇(𝒕) = 𝟐𝒄𝒐𝒔𝒕 + 𝒔𝒊𝒏𝟐𝒕 , [𝟎,
𝝅
𝟐
]
𝒇′(𝒕) = −𝟐𝒔𝒊𝒏𝒕 + 𝒄𝒐𝒔𝟐𝒕 ⋅ 𝟐 = −𝟐𝒔𝒊𝒏𝒕 + 𝟐(𝟏 − 𝟐𝒔𝒊𝒏 𝟐
𝒕)
= −𝟐(𝟐𝒔𝒊𝒏 𝟐
𝒕 + 𝒔𝒊𝒏𝒕 − 𝟏) = −𝟐(𝟐𝒔𝒊𝒏𝒕 − 𝟏)(𝒔𝒊𝒏𝒕 + 𝟏).
𝒇′(𝒕) = 𝟎 → 𝒔𝒊𝒏𝒕 =
𝟏
𝟐
𝒐𝒓 𝒔𝒊𝒏𝒕 = −𝟏 → 𝒕 =
𝝅
𝟔
.
𝒇(𝟎) = √𝟑 +
𝟏
𝟐
√𝟑 =
𝟑
𝟐
√𝟑 ≈ 𝟐. 𝟔𝟎, 𝒂𝒏𝒅 𝒇 (
𝝅
𝟐
) = 𝟎
𝑺𝒐 𝒇 (
𝝅
𝟔
) =
𝟑
𝟐
√𝟑 𝒊𝒔 𝒂𝒃𝒔 𝑴𝒂𝒙 𝒂𝒏𝒅 𝒇 (
𝝅
𝟐
) = 𝟎 𝒊𝒔 𝒂𝒃𝒔 𝒎𝒊𝒏
Derivatives and Graphs
1.) List of problems:
let 𝒇(𝒙) =
𝟏𝟔𝒙
𝒙 𝟐+𝟒
→ 𝒇′
=
𝒖′ 𝒗−𝒗′𝒖
𝒗 𝟐
𝒇′(𝒙) =
−𝟏𝟔𝒙 𝟐+𝟔𝟒
(𝒙 𝟐+𝟒)
𝟐 → 𝒇′′
=
𝒖′ 𝒗−𝒗′𝒖
𝒗 𝟐
𝒇′′(𝒙) =
𝟑𝟐𝒙(𝒙 𝟐−𝟏𝟐)
(𝒙 𝟐+𝟒)
𝟑
a) 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒔 𝒐𝒇 𝒇:
𝟏𝟔𝒙
𝒙 𝟐 + 𝟒
= 𝟎 𝒘𝒉𝒆𝒏 𝒙 = 𝟎
𝒔𝒐 𝒙 𝒊𝒏𝒕. 𝒊𝒔 (𝟎, 𝟎)
b) 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝑨𝒔𝒚𝒎𝒑𝒕𝒐𝒕𝒆𝒔 𝒐𝒇 𝒇:
𝐥𝐢𝐦
𝒙→±∞
𝟏𝟔𝒙
𝒙 𝟐+𝟒
= 𝟎
𝒔𝒐 𝒕𝒉𝒆𝒓𝒆 𝒊𝒔 𝒂 𝑯. 𝑨. @ 𝒚 = 𝟎
c) 𝑭𝒊𝒏𝒅 𝒘𝒉𝒆𝒓𝒆 𝒇 𝒊𝒔 𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒊𝒏𝒈/𝒅𝒆𝒄𝒓𝒆𝒂𝒔𝒊𝒏𝒈:
𝒇′(𝒙) = 𝟎 when −𝟏𝟔𝒙 𝟐
+ 𝟔𝟒 = 𝟎 → 𝒙 𝟐
=
−𝟔𝟒
−𝟏𝟔
= 𝟒 → 𝒙 = ±𝟐
Test
𝒇′(−𝟑) < 𝟎
𝒇′(𝟎) 𝟒
𝒇′(𝟑) < 𝟎 𝑰𝒏𝒄. 𝒐𝒏 (−𝟐, 𝟐)
𝑫𝒆𝒄. 𝒐𝒏 (−∞, 𝟐) ∪ (𝟐, ∞)
d) 𝑭𝒊𝒏𝒅 𝒂𝒏𝒚 𝒆𝒙𝒕𝒓𝒆𝒎𝒂 (𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓𝒔) 𝒐𝒇 𝒇:
𝒇′(−𝟐) = 𝟎 & 𝒇′(𝟐) = 𝟎, → 𝒇(−𝟐) = −𝟒 & 𝒇(𝟐) = 𝟒
𝒇 𝒉𝒂𝒔 𝒂 𝑴𝒊𝒏 @(−𝟐, −𝟒) & 𝑴𝒂𝒙 @ (𝟐, 𝟒)
e) 𝑭𝒊𝒏𝒅 𝒘𝒉𝒆𝒓𝒆 𝒇 𝒊𝒔 𝒄𝒐𝒏𝒄𝒂𝒗𝒆 𝒖𝒑/𝒄𝒐𝒏𝒄𝒂𝒗𝒆 𝒅𝒐𝒘𝒏:
𝒇′′
= 𝟎 𝒘𝒉𝒆𝒏 𝟑𝟐𝒙(𝒙 𝟐
− 𝟏𝟐) = 𝟎. →
𝟑𝟐𝒙 = 𝟎 𝒐𝒓 𝒙 𝟐
− 𝟏𝟐 = 𝟎, 𝒙 = 𝟎, 𝒐𝒓 𝒙 = ±𝟐√𝟑
Test
𝒇′′(−𝟒) < 𝟎
𝒇′′(−𝟏) > 𝟎
𝒇′′(𝟏) < 𝟎
𝒇′′(𝟒) > 𝟎
𝑪. 𝑼. (−𝟐√𝟑 , 𝟎) ∪ (𝟐√𝟑 , ∞)
𝑪. 𝑫. (−∞, −𝟐√𝟑) ∪ (𝟎, 𝟐√𝟑 )
f) 𝑭𝒊𝒏𝒅 𝑰𝒏𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒑𝒐𝒊𝒏𝒕𝒔 𝒐𝒇 𝒇:
𝒇′′
(−𝟐√𝟑) = 𝟎 , 𝒇′′
(𝟐√𝟑) = 𝟎, & 𝒇′′(𝟎) = 𝟎
→ 𝒇(−𝟐√𝟑) = −𝟐√𝟑 , 𝒇(𝟐√𝟑) = 𝟐√𝟑 , & 𝒇(𝟎) = 𝟎
𝑰𝒏𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒑𝒐𝒊𝒏𝒕𝒔 𝒂𝒓𝒆 @ (−𝟐√𝟑, −𝟐√𝟑), (𝟐√𝟑, 𝟐√𝟑), & (𝟎, 𝟎)
Indeterminate forms and 𝑳′
𝑯𝒐̂𝒑𝒊𝒕𝒂𝒍′
𝒔 Rule:
Find the Limit:
 𝐥𝐢𝐦
𝒙→𝟏
𝒙 𝟐−𝟏
𝒙 𝟐−𝒙
→ This limit has the form
𝟎
𝟎
. Factor and simplify to evaluate the
limit. → 𝐥𝐢𝐦
𝒙→𝟏
𝒙 𝟐−𝟏
𝒙 𝟐−𝒙
= 𝐥𝐢𝐦
𝒙→𝟏
(𝒙+𝟏)(𝒙−𝟏)
𝒙(𝒙−𝟏)
= 𝐥𝐢𝐦
𝒙→𝟏
𝒙+𝟏
𝒙
=
𝟏+𝟏
𝟏
= 𝟐
 𝐥𝐢𝐦
𝒙→(
𝝅
𝟐
)
+
𝒄𝒐𝒔(𝒙)
𝟏−𝒔𝒊𝒏(𝒙)
→ This limit has the form
𝟎
𝟎
. → 𝐥𝐢𝐦
𝒙→(
𝝅
𝟐
)
+
𝒄𝒐𝒔(𝒙)
𝟏−𝒔𝒊𝒏(𝒙)
→ 𝐥𝐢𝐦
𝒙→(
𝝅
𝟐
)
+
−𝒔𝒊𝒏(𝒙)
−𝒄𝒐𝒔(𝒙)
= 𝐥𝐢𝐦
𝒙→(
𝝅
𝟐
)
+
𝒕𝒂𝒏(𝒙) = −∞
 𝐥𝐢𝐦
𝒙→∞
𝒍𝒏(𝒙)
√𝒙
→ This limit has the form
∞
∞
. → 𝐥𝐢𝐦
𝒙→∞
𝒍𝒏(𝒙)
√𝒙
→ 𝐥𝐢𝐦
𝒙→∞
𝟏 𝒙⁄
𝟏
𝟐
𝒙
−
𝟏
𝟐
= 𝐥𝐢𝐦
𝒙→∞
𝟐
√𝒙
= 𝟎
 𝐥𝐢𝐦
𝜽→(
𝝅
𝟐
)
+
𝟏−𝒔𝒊𝒏 𝜽
𝒄𝒔𝒄 𝜽
=
𝟎
𝟏
= 𝟎. 𝑳′
𝑯𝒐̂𝒑𝒊𝒕𝒂𝒍′
𝒔 Rule does not apply.
Optimization:
 Sketch picture if needed, write down equation to be optimized and
constraint. Solve constraint for one of the two variables and plug into
first equation. Find critical points of equation in range of variables and
verify that they are min/max as needed.
1.) A Farmer has 2400 ft. of fencing and wants to fence off a rectangular field
that borders a straight river. He needs to fence along the river. What are the
dimensions of the field that has the largest area?
Perimeter = 𝟐𝟒𝟎𝟎 ← Constraint
𝑨 = 𝒙𝒚 ← Objective function
← 𝟐𝒙 + 𝒚 = 𝟐𝟒𝟎𝟎 → 𝒚 = 𝟐𝟒𝟎𝟎 − 𝟐𝒙
𝑨 = 𝒙(𝟐𝟒𝟎𝟎 − 𝟐𝒙) = 𝟐𝟒𝟎𝟎𝒙 − 𝟐𝒙 𝟐
𝑨(𝒙) = 𝟐𝟒𝟎𝟎𝒙 − 𝟐𝒙 𝟐
| 𝟎 ≤ 𝒙 ≤ 𝟏𝟐𝟎𝟎
𝑨′(𝒙) = 𝟐𝟒𝟎𝟎 − 𝟒𝒙 → 𝟐𝟒𝟎𝟎 − 𝟒𝒙 = 𝟎
𝒙 = 𝟔𝟎𝟎 →
𝑨(𝟎) = 𝟎, 𝑨(𝟔𝟎𝟎) = 𝟕𝟐𝟎𝟎𝟎, 𝑨(𝟏𝟐𝟎𝟎) = 𝟎
2.) A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that
will minimize the cost of the metal to manufacture the can.
𝑨 = 𝟐(𝝅𝒓 𝟐
) + 𝟐(𝝅𝒓𝒉) →
𝑽𝒐𝒍. = 𝟏𝑳 = 𝟏𝟎𝟎𝟎𝒄𝒎 𝟑
→ 𝝅𝒓 𝟐 𝒉 = 𝟏𝟎𝟎𝟎 →
𝑨 = 𝟐𝝅𝒓 𝟐
+ 𝟐𝝅𝒓 (
𝟏𝟎𝟎𝟎
𝝅𝒓 𝟐
) = 𝟐𝝅𝒓 𝟐
+
𝟐𝟎𝟎𝟎
𝒓
→
𝑨(𝒓) = 𝟐𝝅𝒓 𝟐
+
𝟐𝟎𝟎𝟎
𝒓
| 𝒓 > 𝟎
𝑨′
(𝒓) = 𝟒𝝅𝒓 −
𝟐𝟎𝟎𝟎
𝒓 𝟐
=
𝟒(𝝅𝒓 𝟑−𝟓𝟎𝟎)
𝒓 𝟐
𝟒(𝝅𝒓 𝟑−𝟓𝟎𝟎)
𝒓 𝟐
= 𝟎 → 𝑹 = √𝟓𝟎𝟎/𝝅
𝟑
𝒉 =
𝟏𝟎𝟎𝟎
𝝅𝒓 𝟐
=
𝟏𝟎𝟎𝟎
𝝅(𝟓𝟎𝟎/𝝅) 𝟐/𝟑
= 𝟐√𝟓𝟎𝟎 𝝅⁄𝟑
= 𝟐𝒓
3.) A box with an open top is to be constructed out of a rectangular piece of
cardboard that is 3 ft. long by 5 ft. long by cutting a square out of each corner
then folding up the sides. Find the largest value of such a box.
𝒗 = 𝒍 ⋅ 𝒘 ⋅ 𝒉 → 𝒗 = (𝟓 − 𝟐𝒙)(𝟑 − 𝟐𝒙)𝒙
𝒗 = (𝟏𝟓 − 𝟏𝟔𝒙 + 𝟒𝒙 𝟐)𝒙 = 𝟏𝟓𝒙 − 𝟏𝟔𝒙 𝟐
+ 𝟒𝒙 𝟑
𝒅𝒗
𝒅𝒙
= 𝟏𝟓 − 𝟑𝟐𝒙 + 𝟏𝟐𝒙 𝟐
→ = 𝟎 →Quadratic
formula→ 𝒙 ≈ 𝟐. 𝟎𝟔 𝒂𝒏𝒅 . 𝟔𝟎𝟕
𝒅 𝟐 𝒗
𝒅 𝟐 𝒙
= −𝟑𝟐 + 𝟐𝟒𝒙 →
𝒇′′(𝟐. 𝟎𝟔) > 𝟎 (𝑴𝒊𝒏), 𝒇′′(. 𝟔𝟎𝟕) < 𝟎 (𝑴𝒂𝒙)
𝑽(. 𝟔𝟎𝟕) ≈ 𝟒. 𝟏 𝒇𝒕 𝟑