1a
MATH 31 UNIT 1 –TANGENTS AND LIMITS
Tangents to a Curve
Slope = m =
rise y y2  y1 f ( x2 )  f ( x1 )



run x x2  x1
x2  x1
If we use a variable point P(x,y) and the slope, m, is known the equation of the line comes from
the
y  y1
point–slope formula:
m
x  x1
For parallel lines m 1 = m 2;
1.
for perpendicular lines
m 1  m 2 = –1 or
m1 
What is the equation of a line perpendicular to y = 2x + 4 if this perpendicular line
is on (6, 2)?
1
1
m2
1b
2.
What is the equation of a line parallel to 4x + 2y – 7 = 0 and having an x–intercept at –3.
Equation of a Tangent Line to a Curve
For a curve the slope of the tangent to the curve is taken as the limiting value of the
slopes of a sequence of secant lines for which the point of tangency is fixed as one end of all the
secant lines.
Tangent – touches but does not cut a curve
Secant – Cuts a curve in 2 or more points
Chord – Line segment between secant line intercepts
2
1c
1.
What is the equation of the tangent to y = x 2 at the point P(2, 4)?
We could use either side but we will find the slope of PQ n as Q n approaches
P from the left.
(As n takes values 1, 2, 3,...the point Q n approaches P)
5
Y
P(2,4)
Q(1.5, 2.25)
Q(1,1)
X
Q(0,0)
0
P(2, 4)
Slope of PQ n = m 
Qn
y2  y1 f ( x2 )  f ( x1 )
=
x2  x1
x2  x1
-5
4  0 22  ANS 2
Grapher
- http:/
/ www.alentum.com/
agrapher/  2
Q 1 (0, Created
0) with a trial version of Advanced
Slope
of PQ

1=
20
2  ANS
Q 2 (1, 1)
Slope of PQ 2 =
4 1
3
2 1
Q 3 (1.5, 2.25)
Slope of PQ 3 =
4  2.25 1.75

 3.5
2  1.5
.5
Q 4 (1.9, 3.61)
Slope of PQ 4 =
4  3.61 .39

 3.9
2  1.9
.1
Q 5 (1.99, 3.9601)
Slope of PQ 5 =
4  3.9601 .0399

 3.99
2  1.99
.01
Q 6 (1.999, 3.996001)
Slope of PQ 6 = 
3
.003996001
 3.999
.001
1d
As Q n approaches P, the slope of the secant line PQ n is getting closer and closer to
being the same as the slope of the tangent at P. In this case the slope seems to be
approaching 4 so we can say that the slope of the tangent to y = x 2 at P(2, 4) is
m tangent = 4.
What is the equation of the tangent at P(2, 4) on y = x 2 ?
m
y  y1
x  x1
4
y4
x2
4x – 8 = y – 4
y = 4x – 4
* If we had used Q n values of 5, 4, 3, 2.5, 2.1, 2.01, 2.001 etc the slope of the tangent
would still be 4 but we would have approached P from the right (from above 4)
Read p(5 – 9) Do p9 #1, 2, 4, 7, 9
4
2a
Limits
A limit is a value that an expression is approaching. By lim f ( x)  L we mean that as x
x a
gets closer and closer to having a value of a, f(x) gets closer and closer to having a value
of L.
The limit is said to equal L even if x will never be allowed to equal a.
1. lim
x 5
x
x2
x
4
4.5
4.9
4.999
4.99999
6
5.5
5.1
5.001
5.00001
y
x
y
Simple limit can be found by substitution as in lim f ( x)  f (a) .
x a
2.
lim  x  7  
3.
lim  x 2  7  
4.
 5x  3 
lim 

x 
 x 1 
x 3
x 5
5
2b
5.
 x2  4 
lim 

x 2
 x2 
For more complicated limits we need to understand the properties of limits.
Given c is a constant and lim f ( x ) and lim g ( x) both exist.
x a
xa
1.
lim  f ( x)  g ( x)   lim f ( x)  lim g ( x)
2.
lim  f ( x)  g ( x)   lim f ( x)  lim g ( x)
3.
lim  f ( x)  g ( x)   lim f ( x)  lim g ( x) 
x a
 x a
  xa

4.
f ( x)
 f ( x)  lim
x a
lim 


x a g ( x)
g ( x)

 lim
x a
5.
lim  cf ( x)   c lim f ( x)
6.
lim  f ( x)   lim f ( x) 
x a
 x a

x a
x a
x a
x a
x a

xa
n
xa
x a

n
for n a positive integer or fraction
Simple limit can be found by substitution as in lim f ( x)  f (a) .
x a
If f(a) does not exist (causes division by zero) a limit may still exist but it will be found
only after some trick of factoring and simplifying, or rationalizing, to make finding the
limit possible.
6
2c
Find the limits:
1.
 x2  7 
 x 2  7  lim
x 5
lim 


x 5
x  2

 x  2  lim
x 5
2.
 3x 2  8 x  7 
lim 

x 4
 2x  5 
3.
1
lim   
x  x
 
4.
5.
1
lim   
x 0 x
 
6.
7.
 x 2  49 
lim 

x 7
 x7 
8.
 x3  27 
lim 

x 3
 x 3 
9.
  5  h  2  52 
lim 

h 0 

h


7
 1
lim  1
x  
 x7




 1 
lim 

x 3 x  3


2d
10.
11.
 x 9 
lim 

x 9
 x 3
 x  125 
lim  3

x 125
 x 5 
Read p(11 – 18)
Do p(18 – 20): #1, 8, (2 – 6) a, c, e
8
3a
One Sided Limits
10
 x2  4 
For limits such as lim 
 we found that
x 2
x

2


Y
5
  x  2  x  2  
 x 4
lim 
 lim 
 x  2  2  2  4
  lim

x 2
x2
 x  2  x 2 
 x 2
2
This can also be seen from the graph of
y=
-10
0
-5
5
10
x 2  4 ( x  2)( x  2)

 x  2, x  2
x2
( x  2)
-5
We see that whether x approaches 2 from the left (x < 2)
or x approaches 2 from the right (x > 2); the limit is still 4.
-10
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 x 4
 x2  4 
and
lim 

4
lim


  4.
x 2
x  2
 x2 
 x2 
2
We can write this as
Since
If
 x2  4 
 x2  4 
lim 

4

lim



x 2
x  2
 x2 
 x2 
 x2  4 
we can say lim 
  4.
x 2
x

2


lim f ( x)  L  lim f ( x) then lim f ( x)  L
x a
but if
x a
x a
lim f ( x)  lim f ( x) then lim f ( x)  Does not exist
x a
x a
x a
9
3b
10
1.
2

 x if x  1 

Consider the function: f ( x)  

  x  1 if x  1 


c)
5
lim f ( x) 
a)
b)
Y
x 1
lim f ( x) 
X
x 1
-10
0
-5
5
10
lim f ( x) 
x 1
-5
Y
5
2.
Some functions only have limits from 1 side.
-10
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for f(x) = x  1
lim f ( x)  lim x  1 
x 1
x 1
X
-10
-5
0
5
10
lim f ( x)  lim x  1 
x 1
x 1
Thus lim x  1 
x 1
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A function may be continuous if there are no breaks in the graph ( y = x, y = x 2, y = sinx)
A discontinuous function is one that has breaks or holes in it. ( # 1 was discontinuous at x = 1,
y=
x2  4
1
is discontinuous at x = 4, y =
is discontinuous at x = 2.)
x4
x2
10
3c
A function is continuous at x = a if lim f ( x)  f (a) .
x a
For this to happen:
1. f(a) must be defined
2. lim f ( x) must exist. i.e. lim f ( x)  lim f ( x)
xa
x a
x a
3. lim f ( x)  f (a )
xa
10
Y
y = f(x)
Which of the 3 conditions fail at each of the following discontinuities?
5
1.
1
f(x) =
x 3
X
-10
-5
0
5
10
-5
-10
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11
10
2.
f(x) =
x2  4
x2
3d
Y
5
X
-10
0
-5
5
10
-5
-10
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.
Y
3.
10
 3 for
x  2 and x  2 
f(x) = 

1 for x  2


y = f(x)
5
X
-10
-5
0
5
10
-5
-10
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12
10
Y
y = f(x)
3e
eg
Given the function y = f(x)
1.
f(2) =
2.
f(–2) =
3.
f(3) =
4.
5.
6.
7.
8.
5
X
-10
-5
0
5
10
-5
lim f ( x) 
x 2
lim f ( x) 
x 2
-10
lim f ( x) 
x 2
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lim f ( x) 
x  2
lim f ( x) 
x  2
9.
lim f ( x) 
10.
There are discontinuities at x = –2, 2 and 4
x2
Read p(21 – 26)
Do p 27 # (1, 2, 4)( a, e, f, h), 3, 7, 10(a,b), 12
13
4a
Using Limits to Find Tangents
For a function such as y = x2 we can find the limiting value


of the slopes of a sequence of secants mPQn and use it to
Y
approximate the slope of the tangent at P.
i.e. As Q n changes from Q 1 to Q 2 to Q 3 etc. and ‘n’
increases as a natural number as Q n gets closer and closer to P

P(x,y)
Q4
Q1
Q3

Q2
the slope of the secant PQ mPQn will get closer and
closer to the slope of the tangent at P.
m
tan gent at P
  lim  m
sec ant PQn
Qn P

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This idea can be used to find the slope of the tangent to the curve at:
1. A fixed point
eg. (3,f(3))
2. A generic point
eg. (a,f(a))
3. A variable point
eg. (x,f(x))
The process is essentially the same but it can be done using 2 different formulas:
14
4b
eg.
  f ( x)  f (a)  
y
 lim 

x 0 x
x a
 x  a 

Formula 1
m  lim
Formula 2
m  lim
 f ( a  h)  f ( a ) 
y
 f ( x  h)  f ( x ) 
 lim 
 or m  lim


x 0 x
h 0
h

0
h
 a  h  a 
Use Formula 1 and 2 to find the slope of the tangent to f(x) = x2 at the point A(3,9)
Formula 1
  f ( x)  f (a)  
y
Q2
 lim 

x 0 x
x a
x

a

 

Y
m  lim
B(x,f(x))
slope of tangent at A(3,9) =
f(x) - f(3)
A(3,9)
x-3
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15
4c
Formula 2
 f ( x  h)  f ( x ) 
y
 lim 

x  0 x
h 0
  x  h  x 
m  lim
Y
Q2
slope of tangent at A(3,9) =
B(3+h,f(3+h))
f(3+h) - f(3)
A(3,9)
(3+h) - 3 = h
X
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eg.
Use Formula 1 to find the slope of the tangent to f(x) = 3x2 + 5 at A(2,17)
slope of tangent at A(2,17) = lim
x 0
y
 f ( x)  f (2) 
 lim 

x

2
x
x3

16
4d
eg
Slope of tangent to y = x3  5x + 5 at x = 2
10
9
8
7
6
5
4
3
2
1
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
Read p (30 – 35) Do p35 # 1, 2, 6, 8
Y
1 2 3 4 5 6 7
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17
4e
Slopes of Tangents at any Point
To avoid separate calculation when we want the slope of the tangent to a curve at more than 1
point we can obtain an expression for the slope of the tangent at a generic point (a, f(a)) and then
find specific slopes by substituting specific values of ‘a’ into this slope expression.
eg
Use Formula 1 and the point (a, f(a)) to obtain an expression for the slope of the tangent
to f(x) = 2x2 – 4x + 1, then find the slopes of the tangents to this function at x = 0, 1 and
4.
Formula 1
  f ( x)  f (a)  
y
 lim 

x 0 x
x a
 x  a 

mtan gent  lim
at x = 0
at x = 1
at x = 4
18
4f
Where on this curve (f(x) = 2x2 – 4x + 1) would the tangent have a slope of 24?
eg
Use Formula 2 and the point (a, f(a)) to obtain an expression for the slope of the tangent
3
to y 
at the point (a, f(a)); then find the equations of the tangents to this function
x2
at x = –5 and x = 0.
Formula 2
  f ( x  h)  f ( x )  
y
 lim 

x 0 x
h 0
h


mtan gent  lim
19
eg Where on the curve y 
3
3
which has a slope at x = a of mtan gent 
is the slope
x2
(a  2) 2
3
?
64
eg.
What is the equation of the tangent line to y =
x at the point (a, a )using Formula 1.
What are the equations of the specific tangent lines at (4, 4 ) or (4, 2) , (9,3) and (49,
7)?
20
x have a slope of 4?
eg
Where does the tangent line to y =
eg.
Using the secant slope expressions what are the slopes of the secants from 4 to 9; 4 to 49?
Do p35: # 7(iv, v. vi), 9, 10
21
5a
Velocity, Rates of Change
On a graph of displacement, s, as a function of time, t, or s = f(t),
the slope of the secant line, m =
Ds
m
, would have units of
and
Dt
s
would represent the average velocity, vavg , over the time interval, D t .
Y
Q2
vavg =
st - st1
Ds
= 2
Dt
t 2 - t1
Q(t+dt,s+ds)
delta s
P(t,s)
If we take the points P(t,s), and Q (t + D t , s + D s ).
then
vavg =
st 2 - st1
t 2 - t1
=
delta t
X
(s + D s ) - s D s
=
Dt
(t + D t ) - t
As t approaches 0, Q approaches P and the secant line PQ becomes a tangent line at P.
The slope of the tangent at P(t,s) would be the limiting value of the slopes of the
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sequence of secants; i.e. the limiting value of
the average velocity over the time interval,
t, as t 0.
i.e. the slope of the tangent at P(t,s) will be the instantaneous velocity at P.
vavg =
Ds
Dt
but
æD s ö
vins t an t an eous = v = lim çç ÷
÷
D t ® 0 èD t ÷
ø
22
5b
eg.
If g = 10 and the distance a body falls is given by s = 0.5(10)t2 = 5t2.
a) What is the average velocity in the third second?
( third second is from t = 2 to t = 3)
b) What is the instantaneous velocity at t = 3 s?
Formula 1
  f (t )  f (3)  
s
 lim 

t 0 t
t 3
 t  3 

m  lim
23
5c
Formula 2
 f (3  h)  f (3) 
s
 lim 

t 0 t
h 0
 3  h  3 
m  lim
Read p (36 – 43) Skip pages (45 – 57);
Do p 43 # 1, 3, 7, 9
Do p 58
# 1, 2, 6, 8, 10, 3(f,g,h), 4(c,d,h)
Additional Questions
1. Given y = x3  2x + 1, find:
a. the slope of the tangent at any point.
ans: m = 3a2  2
b. the point(s) where the slope will be 10.
ans: (2, 5) and (2, 3)
24
5d
c. the equation of the tangent at the point where x = 1.
ans: x  y + 3 = 0
d. use the secant slope expression to obtain the slope of the secant from x = –2 to x = 2
ans: m = 2
2. Rationalize the numerator in
ans:
x  5  x2  3
. (simplify)
x 1
( x  2)
( x  5  x 2  3)
3. Evaluate the following limits.
3 6

5
x
b. lim
x 10 x  10
lim x  6  2
a.
x 6
ans: doesn't exist
ans:
4. Find the slope of the tangent at x = 3 for y =
ans:
x
.
x4
4
49
25
3
50
5e
5. Find the slope at any point for y =
ans: m 
6.
3x  1 .
3
2 3x  1
10
9
8
7
6
5
4
3
2
1
 4  x if x  0

Graph f ( x)   x 2  1 if 0  x  2 and find:
8  x if x  2

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
Y
X
1 2 3 4 5 6 7 8 9 10
a. lim f ( x )
ans: doesn't exist
lim with
f ( an
x)unregistered
b. Created
ans: 5version of Advanced Grapher - http:/ / www.serpik.com/ agrapher/
c.
ans: 6
d. lim f ( x )
x 0
e
lim f ( x)
x  2
f(0) =
x2
x2
f(2) = ans: 2, does not exist
f.
26
ans: doesn't exist
Any Discontinuities?
ans: x = 0, 2
6a
Infinite Limits
When we have functions that are undefined for a certain value of x, this expression has a
behaviour bout that number. To evaluate this behaviour, we look at the limits.
Vertical Asymptote

vertical line that graph gets closer to as "y values" gets very big
If
1.
2.
lim f (x)   or if lim f (x)   then x = a is a vertical asymptote.
x a 
lim
x 0
lim
x 0
x a
1
1
1
 , lim   lim  does not exist ,
x

0
x 0 x
x
x
1
x
10
1
 doesn ' t exist, lim
x 0
x
 ,
lim
x 0
1
x
x = 0 is a vertical asymptote
 does not exist , x = 0 is a vert. asy.
Y
8
3.
lim
x 2
Y
1
1
1
10


,
lim


,
lim
 ,
2
2

x

2
x  2 (x  2)
(x6 2)
(x  2) 2
Y
16 asymptote
x = 2 is a vertical
8
12
4
6
8
2
4
X
-10
-8
-6
-4
-2
0
2
4
6
8
4
10 2
-2
-4
X
-10
-8
-6
-4
-2
0
2
-2
4-12
6
-10
-88
-610 -4
-2
0
2
4
6
8
10
12
-4
-6
-4
-8
-8
Determining Vertical Asymptotes
-6
-10

determine when the denominator equals -8zero
Created with a trial version of Advanced Grapher - http:/ / www.alentum.com/ agrapher/
-10
-12
 find limit of a function from left and right, if either is  or –  then a V.A exists at this
Created with a trial version of Advanced Grapher - http:/
/ www.alentum.com/
agrapher/
Created
with a trial version
of Advanced Grapher - http:/ / www.alentum.com
point
27
-16
6b
1. y 
2

x  2x  8
2
2, Sketch
f(x) =
x 1

x2  9
intercepts at
Assignment
pg. 204 # 1 a, c, f, g, i, 2
pg. 206 # 1, 2 b, d, f, h
pg. 212 # 1 a, b, 2 a, b, e, k, 3 a, b, d, e, f, g (draw graph near asymptote on # 3 d, f)
28
7a
Horizontal Asymptotes

the horizontal line that a graph gets closer and closer to but usually doesn’t touch
as "x" gets very big (+ or –)
lim f (x)  a
If
x 
or if
lim f (x)  a
x 
, then y = a is a horizontal asymptote.
* A function could have two different horizontal asymptotes (one for x   and one for x 
)
1
0
x  x r
Note that for any positive rational number, r, lim
eg lim
1
x 
x
1
4
0
1. Find the following limits. (* divide each term by the highest degree variable)
2x

x  x  3
a. lim
b.
2x

x  x  3
c. lim
d.
2
x3

x  x  3
lim
29
3x 3  3x

x  7x 3  2
lim
7b
2. Find the horizontal asymptotes.
a. y 
2x
x 5
b. y 
x2
x3  2
3. Find the vertical and horizontal asymptotes, x & y-intercepts, and sketch the graph.
a. y 
2x  6
x5
 x-int:

possible vertical asymptote:

horizontal asymptote:
b. y 
y-int:
x 2  3x
x 2  7x  12
Assignment: pg.223 # 2 a, c, e, g, i, 3, 4 b, e, 5 a, d
30
8a
Limits of Trig Expressions
lim sin x  0
x 0
lim cos x  1
x 0
  3 2 1   15 1  17
lim  5cos x  sin x    5  
    

  2 
2  4 2 4
x
6


2
lim
x 0
sin x
 ?  (1)
x
Proof: (not necessary):
C
B
sin

O
A
D
OD = OC = 1
cos
AsectorOBA  AOCA  AsectorOCD
1
1
1
(OA) 2 ()  (OA)(AC)  (OD) 2 ()
2
2
2
1
1
1
2 

cos 2 ()  cos  sin   12 ()
 multiply by

2
2
2
 cos  

sin 
1
cos  


cos 
sin 
1 
sin 
sin 

lim cos  

 l  lim
 1, so lim
1

 0

0

0

cos  



31
8b
sin(3x)
x 0
x
eg 1. lim
sin 5x
x 0
8x
2. lim
3. lim
2sin x cos x

x
4. lim
cosh  1

h
x 0
h 0
32
8c
cosh  1

h 0
h
4*. lim
sin 3x

x 0 sin 4x
5. lim
1  cos 2x

x 0
x2
6. lim
7.
sin(3x)

5
x
x
lim
6
33
8d
sin( 5x)

x 0
2x
8. lim
Assignment Read p 296 – 306
pg. 299 # 1
pg. 306 * Do not use calculator. # 1, 2, 4, 8, 12, 15, 18, 19, 28
34
9a
Exponential Functions
When looking at the limits of exponential functions, we need to evaluate the graphs of our basic
exponential functions.
Look at the graph of y = 2x
5
Y
X
0
5
We notice that this graph as an asymptote at y = 0, so using this idea we know that
lim 2 x = 0. If we are approaching -∞ along the x, we are approaching the asymptote of 0.
x
What is happening to the graph as x is getting bigger?
lim 2 x   .
x
Do all exponential graphs look this way? What if the base is between 0 and 1?
1
Look at the graph of y   
2
x
35
5
Y
X
0
5
This has different properties because it is reflected about the y – axis. Using this graph, we
x
x
1
1
can see that lim     and lim    0 . These are the basic properties for the limits of
x 2
x 2
 
 
exponential functions. Pay attention to the base of the exponential function. A negative
exponent will reciprocate the base (reflect about the y – axis)!
Ex. 1. lim(1.1) x 
2. lim(0.3) x 
x
x
1
3. lim (3.2) x 
4. lim(3) x 
x
x
5. lim(2)
cos x
x0
e)
7. lim(

x 1
2
x 1

6. lim (5)
x2

Assignment: Pg. 361 # 2, 3, 5 (all)
36
1
x2

10a
Log Function
y = logbx, b > 0, b  1
by = x  inverse of y = bx (reflected over line y = x)
y = 10x is the inverse of y = logx
y = ex is the inverse of y = lnx = logex
2. eln
Simplify: 1. ln e7
2
e 
4 x
3. 2ln 5e + ln 2e3  ln 50e = x
4.
5. ln 3 + ln 8 = ln 2 + log x
6. eln e  eln 1
37
 32
10b
When solving limits and graphing it is easier if you remember the basic
graphs of y = ax and its inverse y = loga x for the 2 cases:
Y
Y
a>1
0<a<1
y = log(x)
X
-10
-5
0
5
10
X
-10
0
-5
5
10
y = log(x)
Created with a trial version of Advanced Grapher - http:/ / www.alentum.com/ agrapher/ Created with a trial version of Advanced Grapher - http:/ / www.alentum.com/ agrapher/
1. y = log2x
domain:
range:
vertical asymptote:
10
9
8
7
6
5
4
3
2
1
(x = 2y)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
Y
X
1 2 3 4 5 6 7 8 9 10
Created with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ agrapher/
38
10c
2. Find the domain for log2(x2  9) = y
3. Draw the graph of y = log3(x  2)
4. Find the following limits.
y
a.
b.
lim log 0.5 (x  2) 
1
*  
2
lim ln(cos x  1) 
* ey 
x  2
x 
39
10d
c. lim ln x 
* ey = x, x > 0,
x 0
d.
lim e tan x =
x

2
e. lim e
x 5
x 5
f.
g.
=
lim ln x  3 =
x 3
lim e
x  2 
1 
cos x 
2 

5. Compare the graphs of y = log(x2) and y = 2 log(x).
y = 2 log(x). has a vertical stretch by 2 which does not affect the range y  R but the domain
is x > 0
y = log(x2)has the same range y  R, but the domain doubles ( x < 0 or x > 0) as the graph
opens both ways since x2 is also positive for x < 0.
40
10e
Assignment
pg. 353 # 3 b, g, j,
4 c, f
pg. 354 # 1 b, d,
2 a, b,
3 a, b
pg. 375 # 2 a, c, d,
3 a, b, e,
4 a, c,
5 a,
41
6 a,
7 b, c
9 a, c