Chapter 2 Review
Calculus
Quick Review
()
In Exercises 1 - 4, find f 2 .
1.
2.
3.
4.
()
f x =2x 3 -5x 2 +4
4x 2 -5
f x = 3
x +4
æ xö
f x =sin ç p ÷
è 2ø
ì3x -1, x <2
ï
f x =í 1
ï 2 , x ³2
î x -1
()
()
()
1.) f(2) = 0
2.) f(2) = 11/12
3.) f(2) = 0
4.) f(2) = 1/3
Quick Review
In Exercises 5 - 8, write the inequality in interval form:
5.
x <4
7.
x -2 <3
(-4, 4)
(-1, 5)
In Exercises 9 and 10, write the fraction in reduced form.
9.
x 2 -3x -18
x +3
(x - 6)(x + 3)
= x-6
=
x+3
Slide 2- 3
2x 2 - x
10.
2x 2 + x -1
=
x(2x -1)
(2x -1)(x +1)
x
=
x +1
Example Limits
(
Find the lim 4x 2 - 2x+6
x®5
)
() ()
=4 ( 25) -10 + 6
2
=4 5 -2 5 +6
=100 -10 + 6
=96
Slide 2- 4
Remember to always try to plug
in what you are approaching, if
you get a value then that is your
limit.
Example Limits
Find
1  sin x
x 0 cos x
lim
Solve graphically:
The graph of f  x  
1  sin x
suggests that the limit exists and is 1.
cos x
Confirm Analytically:
Find
1  sin x  1  sin 0 
1  sin x lim
x 0
lim


x  0 cos x
lim cos x
cos 0
x 0

1 0
1
1
Slide 2- 5
Example Limits
5
by graphing:
x®0 x
Find lim
5
Solve graphically: The graph of f  x   suggests that
x
the limit does not exist.
[-6,6] by [-10,10]
Confirm Analytically :
We can't use substitution in this example because when x is relaced by 0,
the denominator becomes 0 and the function is undefined.
This would suggest that we rely on the graph to see that the
limit does not exist.
Example One-Sided and
Two-Sided Limits
Find the following limits from the given graph.
a.
4
o
b.
c.
1 2 3
d.
e.
Slide 2- 7
lim f  x 
x  0
lim f  x 
x  2
lim f  x 
x2
lim f  x 
x2
lim f  x 
x  3
0
 Does Not Exist
4
 Does Not Exist
0
Quick Review Solutions
In Exercises 1 - 4, find f - 1 and graph f , f - 1 and y = x in the
same viewing window.
1.
f (x)= 2 x - 3
f
- 1
2.
x+ 3
(x)=
2
f - 1 (x )= ln x
[-12,12] by [-8,8]
Slide 2- 8
f (x )= e x
[-6,6] by [-4,4]
Quick Review Solutions
æ1ö
In Exercises 7 - 10, write a formula for a f - x and b f ç ÷.
èxø
() ( )
Simplify where possible.
7.
()
f x =cos x
( )
f - x =cos x,
ln x
9. f x =
x
æ 1ö
10. f x = ç x + ÷ sin x
è xø
()
()
Slide 2- 9
8.
æ1ö
1
f ç ÷ =cos
x
èxø
()
()
f x =e - x
( )
f - x =e ,
( ),
ln -x
x
æ 1 ö -1
f ç ÷ =e x
èxø
æ1ö
1
f - x =f ç ÷ =- x ln
x
x
èxø
æ 1ö
æ1ö æ 1ö 1
f - x = ç x + ÷ sin x, f ç ÷ = ç x + ÷ sin
è xø
èxø è xø x
( )
( )
Example Horizontal Asymptote
Use a graph and tables to find  a  lim f  x  and
x 
 c  Identify all horizontal asymptotes.
f  x 
f  x .
 b  xlim

x 1
x
f  x  1
 a  lim
x 
f  x  1
 b  xlim

 c  Identify all horizontal asymptotes.
[-6,6] by [-5,5]
y 1
Slide 2- 10
Example Sandwich
Theorem Revisited
The sandwich theorem also holds for limits as x  .
cos x
Find lim
graphically and using a table of values.
x 
x
The graph and table suggest that the function oscillates about the x -axis.
cos x
Thus y  0 is the horizontal asymptote and lim
0
x 
x
Slide 2- 11
Example Vertical Asymptote
Find the vertical asymptotes of the graph of f ( x) and describe the behavior
of f ( x) to the right and left of each vertical asymptote.
8
f  x 
4  x2
The values of the function approach   to the left of x   2.
The values of the function approach + to the right of x   2.
The values of the function approach + to the left of x  2.
The values of the function approach   to the right of x  2.
8
8
lim



and
lim

2
2

x 2 4  x
x 2 4  x
8
8
lim



and
lim

2
2
x2 4  x
x  2 4  x
So, the vertical asymptotes are x  2 and x  2
[-6,6] by [-6,6]
Example “Seeing” Limits as
x→±∞
We can investigate the graph of y  f  x  as x   by investigating the
1
graph of y  f   as x  0.
x
1
Use the graph of y  f   to find lim f  x  and lim f  x 
x 
x 
x
1
for f  x   x cos .
x
cos x
1
The graph of y  f   =
is shown.
x
 x
1
lim f  x   lim f     
x 
x 0
x
1
lim f  x   lim f     
x 
x 0
x
Quick Quiz Sections 2.1 and 2.2
You may use a graphing calculator to solve the following problems.
1.
A
 B
C
D
E
x2  x  6
Find lim
if it exists
x 3
x 3
1
1
2
5
does not exist
Slide 2- 14
Quick Quiz Sections 2.1
and 2.2
2.
A
 B
C
 D
E
3 x  1,

Find lim f  x  =  5
x2
 x  1 ,
5
3
13
3
7

does not exist
x2
x2
if it exists
Slide 2- 15
Quick Quiz Sections 2.1
and 2.2
3.
Which of the following lines is a horizontal asymptote for
3x3  x 2  x  7
f  x 
2 x3  4 x  5
3
A
y

x
 
2
 B y  0
C y 
 D
E
2
3
7
5
3
y
2
y
Slide 2- 16
Quick Review Solutions
1.
3x 2 -2x +1
Find lim
x®-1
x 3 +4
2.
Let f x =int x. Find each limit.
(a )
( c)
3.
2
()
()
lim f ( x )
lim- f x
x®-1
x®-1
-2
+
x®-1
no limit
ì x 2 - 4x +5,
Let f x = í
î4- x,
x ³2
Find each limit.
Slide 2- 17
()
lim f ( x )
lim- f x
x®2
x®2
-1
x <2
()
(a )
( c)
( b) lim f ( x)
(d) f (-1) -1
1
no limit
( b) lim f ( x )
( d ) f ( 2) 2
x®2
+
2
Quick Review Solutions
In Exercises 4 - 6, find the remaining functions in the list of functions:
f , g, f o g, g o f .
2x- 1
1
4. f (x )=
, g (x )= + 1
x+ 5
x
x+ 2
3x + 4
, x¹ 0
, x¹ 5
( f o g )(x) =
(g o f )(x) =
6x + 1
2x- 1
5.
f (x )= x 2 ,
(g o f )(x)= sin x 2 , domain of g = [0, ¥ )
g (x )= sin x, x ³ 0
( f o g )(x)= sin 2 x, x ³ 0
Slide 2- 18
Quick Review Solutions
6.
g (x)=
x - 1,
f (x)=
1
+ 1,
2
x
1
(g o f )(x)= ,
x
x> 0
x> 0
( f o g )(x)=
7.
Use factoring to solve
2 x2 + 9 x - 5= 0
8.
Use graphing to solve
x 3 + 2 x - 1= 0
Slide 2- 19
x
,
x- 1
x=
x> 1
1
, - 5
2
x » 0.453
Quick Review Solutions
ì5- x,
x £3
In Exercises 9 and 10, let f x = í 2
î-x +6x -8, x >3
()
9.
()
Solve the equation f x =4
x =1
()
10. Find a value of c for which the equation f x =c
has no solution.
Slide 2- 20
Any c in [1,2)
Example Continuity at a Point
Find the points at which the given function is continuous and the points at
which it is discontinuous.
o
Points at which f is continuous
At x  0
At x  6
lim f  x   f  0 
x 0
lim f  x   f  6 
x  6
At 0 < c < 6 but not 2  c  3
lim f  x   f  c 
x c
Points at which f is discontinuous
At x  2
lim f  x  does not exist
x2
At c  0, 2  c  3, c  6
these points are not in the domain of
f
Slide 2- 21
Continuity at a Point x = 0
Continuous
Jump
Slide
2- Dis.
22
Removable Discontinuity
Infinite Dis.
Removable
Oscillating functions
are not continuous,
infinite discontinuity
Example Continuity at a Point
Find and identify the points of discontinuity of y 
3
 x  1
2
There is an infinite discontinuity at x 1.
[-5,5] by [-5,10]
Continuous Functions
The given function is a continuous function because it is
continuous at every point of its domain. It does have a
point of discontinuity at x   2 because it is not defined there.
y
[-5,5] by [-5,10]
Slide 2- 24
2
 x  2
2
Quick Review Solutions
In Exercises 1 and 2, find the increments Dx and Dy from point A
to point B.
1.
(
)
( )
A -5,2 , B 3,5
Dx =8,
2.
Dy =3
( )
( )
A 1,3 , B a,b
Dx =a -1,
Dy =b-3
In Exercises 3 and 4, find the slope of the line determined by the
points.
3.
(-2,3) , (5,-1)
-
4
7
4.
(-3,-1) , (3,3)
2
3
Quick Review Solutions
In Exercises 5 - 9, write an equation for the specified line.
3
3
5. through (- 2,3) with slope =
y= x+ 6
2
2
6.
through (1,6) and (4, - 1)
7.
through (1, 4) and parallel to y = -
Slide 2- 26
y= -
3
x+ 2
4
7
25
x+
3
3
y= -
3
19
x+
4
4
Quick Review Solutions
8.
through (1, 4) and perpendicular to y = -
3
x+ 2
4
4
8
y= x+
3
3
9.
through (- 1,3) and parallel to 2 x + 3 y = 5
2
7
x+
3
3
10. For what value of b will the slope of the line through (2,3)
y= -
and (4, b) be
Slide 2- 27
5
?
3
b=
19
3
Average Rates of Change
The average rate of change of a quantity over a period
of time is the amount of change divided by the time it
takes.
In general, the average rate of change of a function over an interval is the amount
of change divided by the length of the interval.
Also, the average rate of change can be thought of as the slope of a secant line to
a curve.
Slide 2- 28
Example Average Rates of
Change
Find the average rate of change of f  x   2 x 2  3x  7
over the interval  -2,4
( ) ()
f -2 - f 4
-2 - 4
2
2
æ
ö æ
ö
ç 2 -2 - 3 -2 + 7÷- ç 2 4 -3 4 + 7÷
è
ø è
ø
=
-2 - 4
21- 27
=
-6
-6
=
-6
=1
( )
Slide 2- 29
( )
()
()
Example Tangent to a Curve
Given y  x 2  2 at x   1 find:
the slope of the curve and an equation of the tangent line.
Then draw a graph of the curve and tangent line in the
same viewing window.
a 
Write an expression for the slope of the secant line and find the
limiting value of the slope as Q approaches P along the curve.
When x   1, y  x 2  2  3 so =P  1,3
 1  h 
2


2


1
 2


y  1  h   y  1


lim
 lim
h 0
h 0
h
h
h  h  2
3  2h  h 2  3
h 2  2h
lim
 lim
 lim
 lim  h  2    2
h 0
h

0
h

0
h 0
h
h
h
2
Example Tangent to a
Curve
 b  The tangent line has slope
 2 and passes through  1,3 .
The equation of the tangent line is
y  3   2  x   1 
y   2  x  1  3
curve
y   2x  2  3
y  x2  2
y   2x  1
tangent
y   2x  1
Slope of a Curve
Slide 2- 32
Slope of a Curve at a Point
()
( ( )) is the number
The slope of the curve y = f x at the point P x, f x
m=lim
h®0
(
)
()
f x+h - f x
h
provided the limit exists.
The tangent line to the curve at P is the line through P with this slope.
Slide 2- 33
Normal to a Curve
The normal line to a curve at a point is the line perpendicular to the tangent at
the point.
The slope of the normal line is the negative reciprocal of the slope of the
tangent line.
Slide 2- 34
Example Normal to a Curve
Given y  x 2  2 at x   1 write the equation of the normal line.
Draw a graph of the curve, the tangent line and the normal line in the
same viewing window.
From an earlier example, the slope of the tangent line was found
to be  2 so the slope of the normal is
y  3
1
x   1 

2
1
y   x  1  3
2
1
1 6
y x 
2
2 2
1
7
y x
2
2
Slide 2- 35
1
.
2
tangent
y   2x  1
curve
y  x2  2
normal line
1
7
y x
2
2
Quick Quiz Sections 2.3 and 2.4
You may use a graphing calculator to solve the following problems.
1. Which of the following values is the average rate of change
of f  x   x  1 over the interval 0,3 ?
A
 B
1
C

3
1
3
1
3
D
E
3
Slide 2- 36
Quick Quiz Sections 2.3 and
2.4
2.
Which of the following statements is false for the function
3
0 x4
 4 x,

x4
f  x   2,
 x  7,
4 x6

6 x8 ?
1,
f  x  exists
 A  lim
x4
 B  f  4  exists
f  x  exists
 C  lim
x 6
f  x  exists
 D  xlim
8
 E  f is continuous at x  4

Slide 2- 37
Quick Quiz Sections 2.3 and
2.4
3.
Which of the following is an equation for the tangent line
to f  x   9  x 2 at x  2?
1
9
A
y

x

 
4
2
 B  y   4 x  13
C
D
E
y   4x  3
y  4x  3
y  4 x  13
Slide 2- 38
Chapter Test Solutions
Chapter Test Solutions
In Exercise 5, use the graph of the function with domain - 1£ x £ 3.
5. Determine
g (x )
(a ) xlim
®3
-
1
(b) g (3).
(c) whether g (x) is continuous at x = 3.
(d ) the points of discontinuity of g (x).
1.5
No
at x = 3 and points
not in the domain
(e) Writing to Learn whether any points of discontinuity
are removable. If so, describe the extended function. If not,
explain why not.
Removable at x = 3 by assigning the value 1 to g (3).
Slide 2- 40
Chapter Test Solutions
In Exercise 6, (a ) find the vertical asymptotes of the graph of
y = f (x ), and (b) describe the behavior of f (x ) to the right
and left of any vertical asymptote.
x- 1
6. f (x)= 2
vertical asymptotes: x = 0, x = - 2
x (x + 2)
At x = 0
Left-hand limit = limx® 0
x- 1
2
=- ¥
x ( x + 2)
Right-hand limit = lim+
x® 0
x- 1
2
=- ¥
x ( x + 2)
At x = - 2
Left-hand limit = limx® 2
x- 1
2
x ( x + 2)
=¥
Right-hand limit = lim+
x® 2
x- 1
2
x ( x + 2)
=- ¥
Chapter Test
7.
7. a 
Left-hand limit = lim f  x   lim 1 1
x 1
Given
ì1,
ï
- x,
ï
ï
f x = í1,
ï-x,
ï
ï
î1,
()
(a )
( b)
( c)
At x   1:
Right-hand limit = lim f  x   lim   x  1
x £-1
x 1
Left-hand limit = lim f  x   lim   x   0
x =0
x 0
x 0
x 0
At x 1
x ³1
Left-hand limit = lim f  x   lim   x    1
Find the right-hand and left-hand limits of f at
x 1
x 1
Right-hand limit = lim f  x   lim 1 1
x =-1, 0 and 1.
Is f continuous at x =-1? 0? 1? Explain.
x 0
Right-hand limit = lim f  x   lim   x   0
0 < x <1
If so, what is it? If not, why not?
x 1
At x  0 :
-1< x <0
Does f have a limit as x approaches -1? 0? 1?
x 1
x 1
7. b 
x 1
At x  1: Yes, the limit is 1.
At x  0 : Yes, the limit is 0.
At x  1: No, the limit doesn't exist because the two
one-sided limits are different.
c
At x  1: Continuous because f  1  the limit.
At x  0 : Discontinuous because f  0   the limit.
At x 1: Discontinuous because the limit does not exist.
Slide 2- 42
Chapter Test Solutions
()
In Exercise 8, find a a power function end behavior model and
( b) any horizontal asymptotes.
2x +1
a)
8. f ( x ) =
(
x -2x +1
2
9.
2
x
( b)
()
y =0
Find the average rate of change of f x =1+sin x
é pù
over the interval ê0, ú.
ë 2û
Slide 2- 43
2
p
Chapter Test Solutions
()
( ( ))
Find ( a ) the slope of the curve y = f ( x ) at P, ( b) an equation
of the tangent at P and ( c) an equation of the normal at P.
(a ) m=-1
( b) y =- x -1
( c) y = x - 3
10. Let f x = x 2 -3x and P = 1, f 1 .
Slide 2- 44