Calculus AB – Semester Notes Condensed
Chapter 1 - Limits
1) graphically
 2) numerically

Look at all problems 4 ways: 
3) symbolically (algebraically)
 4) verbally
Limits – Graphically – as x is getting closer to c, what is y getting closer to?
Numerically - substitution or set up a table
Symbolically – if substitution does not work, - then algebraically manipulate
ZOOM IN!
3

ex.: lim
1) approach  or -
x 0 x

1

3 cases where limits DNE -  2) oscillates
ex.: lim sin
x 0
x


x
3) approaches diffferent values ex. lim
x 0 x

Examples:
lim
x 2
lim
x 3
lim
x 0
x2
x2
1
1
 lim
 lim

2
x

2
x

2
x 4
x2 4
 x  2 x  2
1 x  2
1 x  4
x  3
1
1
 lim
 lim
 lim

x

3
x

3
x

3
x3
1 x  2 4
 x  3 1  x  2
 x  3 1  x  2



4   x  4
1
1
x
1

 lim
 lim

x

0
x

0
x  x  4 4x
4x  x  4
4 x  x  4  16
Special Trig Limits:
sin x
1
x 0
x
1) lim
1  cos x
0
x 0
x
2) lim
Example:
tan 2 x
sin 2 x
sin x sin x sin 0 0
lim 
 lim
 lim

 1 0
2
x 0
x

0
x

0
x
cos x x
cos 2 x x
cos2 0 1
One Sided Limits
lim
x 4
x4
x4
 1

Continuity – a function is continuous at c if 3 conditions are met (i.e. No Breaks):
f(x) exists
 lim
 x c
 f(c) is defined
 lim f(x)=f(c)
 x c
Removable vs. Non-Removable:
Examples:
Where is it discontinuous? Is it removable or non-removable?
f ( x) 
x 3
x 3
1


2
x  9  x  3 x  3 x  3
Discontinuous at x=-3 non-removable, x=3 removable
2 x  3 x  1 lim  1
individual functions are continuous and continuous at x=1 so continuous
f ( x)   2
 x x  1 lim  1
Intermediate value Theorem: if f(x) is continuous in [a, b] and if c is in [a, b], then there is a K value between f (a) & f (b)
such that f(c) =K.
Example:
Find c guaranteed by IVT
f ( x)  x 2  6 x  8 [0, 3] and f(c) =0
f (c )  0  x 2  6 x  8
F(b)
 ( x  4)( x  2)
F(c)
x  4, x  2
F(a)
solution is only x=2 because in interval
a c
b
Limits to 
Example:
 1
lim  1   Vertical Asymptote is  or  so… then
x 0 
x
1
f ( x)  1 
 1  very large neg num=-
very small neg number
Vertical Asymptotes
Example:
 x  2  x  2  Hole at x=-2 and then there is no vertical asymptote
x2  4
f ( x) 

 x  2   x 2  1  x  2   x 2  1
Chapter 2 – Derivatives

1) f is continuous at c (no hole or V.A.)
A function is differentiable at a point c if: 
2) lim f '( x)  lim f '( x) (no sharp turns)


Example:
x c
x c
Is f(x) differentiable at x=2?
 check #1 lim  lim  2 – all ok
x 2
x 2

Yes f(x) is differentiable at 2

1
check
#2
f'(x)=for
both
=
all
ok


2
1
 x  1, x  2
f ( x)   2
 2x , x  2

Definition of Derivative: lim
h 0
f ( x  h)  f ( x )
h
1
1
f ( x)  2
x
Example:
 x  h
f '( x)  lim
h 0
2

1
x2
h
 lim
h 0
 lim
h 0
2 x  h
x2  x  h 
2

x2   x  h 
hx 2  x  h 
2
2
 lim
h 0
x 2  x 2  2 xh  h 2
hx 2  x  h 
2
x3
Power down Shortcuts:
Trig Derivatives:
Example:
f ( x)  sin x  f '( x)  cos x
f ( x)  sec x  f '( x )  sec x tan x
f ( x)  cos x  f '( x)   sin x
f ( x)  cot x  f '( x )   csc 2 x
f ( x)  tan x  f '( x)  sec 2 x
f ( x)  csc x  f '( x)   csc x cot x

Write the equation of the tangent line to y  x 2  2 x
y  x3  3x 2  2 x
slope  y '  3x 2  6 x  2
m @(1, 6)  3(1) 2  6(1)  2  6
( y  6)  6( x  1)
Horizontal Tangent: set f’(x) =0
Average Rate of Change:
y f ( x2 )  f ( x1 )

x
x2  x1
  x  1 @(1, 6)
2
Instantaneous Rate of Change: f’(x) at x
Example:
Find the average rate of change over the interval & the instantaneous rate of change at the endpoints of
f ( x)  3x 2  4,[1, 2.5]
the interval
f (2.5)  f (1)
 10.5
2.5  1
Inst.: f '( x)  6 x @1  6
Ave :
2.5  15
s(t )  16t 2  v0t  s0 v0  initial velocity s0  initial height
Position Function under Gravity:
Product Rule:
if h( x)  f ( x) g ( x)  h '( x)  f ( x) g '( x)  f '( x) g ( x)
Quotient Rule: if h( x) 
Examples:
f ( x)
g ( x) f '( x)  f ( x) g '( x)
 h '( x) 
g ( x)
( g ( x)) 2
Write the equation of the tangent line to f ( x) 
m  f '( x) 
x 1  1 
@  2, 
x 1  3 
( x  1)(1)  ( x  1)(1)
2
2
 1

@  2,   m 
2
2
( x  1)
( x  1)
9
 3
1 2

so...  y     x  2 
3 9

Find the 2nd derivative:
f ( x)  sec x
f '( x)  sec x tan x
f ''( x)  sec x sec2 x  tan x sec x tan x  sec3 x  tan 2 x sec x
s (t )  position function
Position Function:
s'(t)=v(t)=veloctiy function
s''(t)=v'(t)=a(t)=acceleration function
Chain Rule:
Examples:




if h( x)  f g  x   h '( x)  f ' g  x  g '  x 
f ( x)   3x 2  1  f '( x)  4  3x 2  1  6 x   24 x  3x 2  1
4
3
3
f ( x)  5cos2  x  5  cos  x    f '( x)  5 2  cos  x   sin  x      10 cos  x   sin  x  
2
Implicit Differentiation:
x 2 y ' y 2 x  y 2 1  x  2 yy '   0
Example:
x 2 y  y 2 x  2
x 2 y ' 2 xyy '  2 xy  y 2
2 xy  y 2
y' 2
x  2 xy
Normal Line:
Horizontal Tangent Line:
Vertical Tangent Line:
Perpendicular to tangent line, slopes are opposite reciprocals
let m=0, numerator =0
Let denominator = 0
Related Rate Word Problems:
always derive with respect to time so
Geometry Formulas:
Pythagorean Theorem:
dy
dx
dV
 y ',  x ',
 V ', etc.
dt
dt
dt
4
Vsphere   r 3  V '  4 r 2 r '
3
x2  y 2  z 2
2 xx ' 0  2 zz '
6
constant
10
Chapter 3 – Extrema, Concavity, Point of Inflection
Critical Numbers:
1) f is defined at c
2) f '(c)=0 or f '(c) is DNE
c is a critical number of f if: 
1) set f '(x)=0 and solve
To find critical numbers: 2) where is f '(x) undefined
3) Be sure f(c) is defined
Find the critical numbers of f ( x)  5 x 4  3x3
Example:
f '( x)  20 x 3  9 x 2  0
x 2  20 x  9   0
x0
x
9
20
function is not undefined
 9 
check to see if f(0) & f   exist
 20 
Absolute Extrema:
on a closed interval, find the critical numbers. Then check critical numbers and endpoints in the
function to see which gives the maximum and minimum values.
Rolles Theorem:
If f(x) is continuous on [a, b] and differentiable on (a, b) and if f (a) =f (b) then there is at least
one c value in (a, b) that gives f’(c) =0
Example:
f ( x)  x2  3x  2 in 1, 2
Does Rolles Theorem apply?
is f(x) continuous? - yes
is f(x) differentiable? - yes
does f(1)=f(2)? -  f 1  0, f (2)  0  - yes
3
and it is in 1, 2 
Yes it applies and when f '( x)  2 x  3  0  x 
2
Mean Value Theorem: if f(x) is continuous on [a, b] and is differentiable on (a, b), then there is a c value in (a, b) such
that f '(c) 
f (b)  f (a)
(this guarantees a secant line parallel to a tangent line at c)
ba
c is where theaverage rate of change = instantaneous rate of change
a
Example:
c
b
Find a c value where the mean value theorem would apply
f (6)  f (2) 2  0 1


62
4
2
is it continuous? - yes
is it differentiable? - yes
1
So…
f ( x)   x  2 
2( x  2)
 x  2
1
1
2

2
1
1
2
x  2 1
x 3c 3
1
f '( x) 
2
 2, 6
1
1
 x  2 2
2
f(x)
f’(x)
Increase
+
1st Derivative Test:
decrease
To find the maximum/minimum in any interval
1) find critical numbers (f '(x)=0 or undefined)

2) Is f(x) defined at critical points?
3) Does the f(x) change from incr to decr or decr to incr?

Example:
Find the location of maximum/minimum(s) for f ( x)  x 2  8x  10
f '( x)  2 x  8  0
F(x)
F’(x)
x  -4, it is not undefined
f(x) is not undefined
Concavity:
F(x)
F’(x)
F’’(x)
Inflection Points:
Example:
Decrease
-
Increase
+
4
Decreasing to increasing so minimum at x=4
ccup
Increase
+
ccdn
Decrease
-
Point where a graph changes concavity
Find the extrema, points of inflection, and concavity for the function
To test for extrema:
To test for inflection points:
function is continuous and differentiable
f ''( x)  6 x  0
f '( x)  3x 2  0
x0
x  0 - critical number
F(x)
F’(x)
Increase
+
f ( x)  x 3  1
Increase
+
0
No Extrema
ccdn   ,0 ccup   0,  
F(x)
F’(x)
F’’(x)
ccdn
Decr
-
ccup
Incr
+
0
Point of Inflection at x=0 or (0, 1)
2nd Derivative Test:
If c is a critical number of f(x) and
 f’’(c)>0, then there is a relative minimum at c
 f’’(c)<0, then there is a relative maximum at c
 f’’(c)=0, then there is no conclusion and you must go back and do 1st derivative test
F’’(x)
F’(x)
F(x)
Sketching Functions and Derivatives:
Example:
Sketch f’(x) and f’’(x)
a
F(x)
Dec
Inc
Dec
Inc
F’(x)
-
+
-
+
a
b
ccup
ccdn
ccup
F’(x)
inc
dec
inc
F’’(x)
+
-
+
e
x
3x  2
2x 1
Trig Limits to ∞:
F(x)
d
a
if
then H.A. at
b
dn  dd
y0
dn  dd
From last year we learned:
x 
c
e
If lim  # , then it’s a horizontal asymptote.
Horizontal Asymptotes:
lim
b
c
F(x) extrema →zeros for f’(x)
F(x) decreasing →negative values for f’(x)
F(x) increasing → positive values for f’(x)
F’’(x) positive →concave up for f(x)
F’’(x) negative→concave down for f(x)
F’’(x) zero values →points of inflection
Example:
d
2
lim
x 

y
dn  1
3
 so limit is HA =
dd  1
2
sin x
0
x
lim
x 
x  cos x
1
x
Optimization Problems:
 1)Primary Equation? What is to be maximized or minimized?

 2) If Pri. Eq. has more than 1 variable, find secondary equation & sub into Pri. Eq. to reduce variables
3) set f '(x)=0 & undefined to find critical numbers  minimums and maximums must occur here.

Examples:
x
A box is to be made from a rectangular paper with dimensions of 4’x7’ if equal squares are cut from
each corner turning the sides up. Find the dimensions of the box that will maximize the volume.
7
x
x
x
V  lwh
7-2x
x
4
x
x
x
x
4-2x
Pri Eq.
V  (7  2 x)(4  2 x)( x)
V  4 x 3  22 x 2  28 x
V '  12 x 2  44 x  28
x  .819 x  2.847
so .819 x 2.362 x 5.362
is the solution
A rectangle is bound by the x and y axis and the graph of y  
2
x  4 . What will the dimensions of the
3
rectangle be that will maximize the area?
2
is Pri Eq. and y   x  4
3
2
 2

A  x   x  4    x2  4x
3
 3

4
A'   x  4  0
3
x  3 so y  2
A x y
y
x
Differentials:
is sec eq.
dy
 f '( x)  dy  f '( x)dx
dx
Rewriting the derivative equation
Linearization:
Examples:
Find the linearization T(x) for f ( x) 
x2
at (3, 3). Then approximate the error for
3
x  .1 x  .5 x  1
2
f '( x)  x @(3,3)
3
y  3  2( x  3)
x
T(x)
F(x)
error
.1
3.2
3.2033
.0033
.5
4
4.0833
.0833
1
5
5.3333
.3333
f '( x)  2
so….
T ( x)  3  2( x  3)
Use Linear approximations to estimate f (2.1) for the function where f’ (2) =-3 and f (2) =5
Slope
point (2, 5) so…
y  5  3( x  2)
T ( x)  y  5  3( x  2)
at 2.1  y  5  3(2.1)  1.3
Chapter 4 – Antiderivative and Integrals
Power Up:
Example:
1
1
1
x 1
2 3
1
dx    x  1 x  2 dx   x 2  x 2 dx  x 2  2 x 2  c
3
x

Know your Trig Derivatives!!
Solving Differential Equations:
Examples:
Find f(x) or y
Find y for
dy
 3x 4
dx
 dy   3x

4
dx 
y
1
c
x3
Find f(x) if f ''( x)  x 2 , f '(0)  6, f (0)  3
 f ''( x)   x
f '( x) 
1
 f '( x)   3 x
2
1 3
x c
3
f ( x) 
1
6  (0)3  c
3
c6
Riemann Sums:
3
6
1 4
x  6x  c
12
1
(0) 4  6(0)  c
12
c3
f ( x) 
1 4
x  6x  3
12
3
Using rectangles to estimate the area under the curve – Left, Right, and Midpoint Sums along
with Trapezoid sums.
Definite Integrals:

Fundamental Theorem of Calculus:
b
a
f ( x)dx  F (b)  F (a)
anitderivative is F
Example:
2
2
   

 
2
t

cos
t
dt

t

sin
t
|


sin



sin










 2

2
2  2 
2 
2

2

2
2
=
Mean Value Theorem for Integrals:
2
4
1
2
4
1  2
There must be some c value in (a,b) that yields the true area under the curve if
using rectangles.

b
a
f ( x)dx  f (c)(b  a)  height base
Average Value Theorem for Integrals:
The f(c) (height) that yields the true area
1 b
f (c ) 
f ( x)dx
b  a a
Example:
or
Find the value of c guaranteed by the MVT if f ( x) 
Integral

Interval

b
a
f ( x)dx
ba
9
on 1,3
x3
1 3 3 1  9 2  1  1 9 
AveValue  f (c) 
9x   x       2
3  1 1
2 2
1 2  2 2 
3
so… f (c)  2 
3
9
9
9

2

2
x
 9  x  3 so c  3
3
x
2
2
Integration by U-Substitution:
1) let u = interior function
2) find du
3) Substitute u & du into original integral
4) power up
5) Substitute back in original function
Example:

x2
dx   x 2 16  x3  dx
2
16  x 
3 2
Let u  16  x3
2
1
3 2
16

x
 3x dx



3
1
1
1
   u 2 du   1u 1  c 
c
3
3
3 16  x3 

2nd Fundamental Theorem of Calculus:
(dump and derive)
if F ( x) 

x
a
f (t )dt then F '( x)  f ( x)
du  3x 2 dx