ontinuity, End Behavior, and Limits 1

You found domain and range using the graph of a function. (Lesson 1-2)

NewVocabulary continuous function limit discontinuous function infinite discontinuity jump discontinuity removable discontinuity nonremovable discontinuity

end behavior

24 I L e s s o n 1-3

Use limits to determine the continuity of a function, and apply the Intermediate Value Theorem to continuous functions.

Use limits to describe end behavior of functions.

Since the early 1980s, the current minimum wage has jumped up several times.

The graph of the minimum wage as a function of time shows these jumps as breaks in the graph, such as those atx = 1990, x = 1996, and x = 2008.

7 3 £ 6 £ 5 & £ 4 re

=5 3

° 0

i >

> >

1980 1984 1988 1992 1996 2000 2004

Year

1 Continuity T h e g r a p h o f a c o n t i n u o u s f u n c t i o n has n o breaks , holes , or gaps. Y o u can trace the g r a p h o f a c o n t i n u o u s f u n c t i o n

w i t h o u t l i f t i n g y o u r p e n c i l .

O n e c o n d i t i o n f o r a f u n c t i o n / ( x ) to be c o n t i n u o u s at x = c is t h a t the f u n c t i o n m u s t a p p r o a c h a u n i q u e f u n c t i o n v a l u e as x -va lues a p p r o a c h c f r o m the le f t a n d r i g h t sides. The concept o f a p p r o a c h i n g a v a l u e w i t h o u t necessari ly ever r e a c h i n g i t is ca l l ed a l imi t .

fix) is continuous for all x.

'Concept Limits

Words If the value of f(x) approaches a unique value L as x approaches c from each side, then the limit of f{x) as x approaches c is L.

Symbols Jirn^ f{x) = L, which is read The

limit of f(x) as x approaches c isL.

To u n d e r s t a n d w h a t i t means f o r a f u n c t i o n to be c o n t i n u o u s f r o m a n algebraic perspec t ive , i t he lps to e x a m i n e the g r a p h s o f d i s c o n t i n u o u s f u n c t i o n s , or f u n c t i o n s t h a t are n o t c o n t i n u o u s . F u n c t i o n s can h a v e m a n y d i f f e r e n t t y p e s o f d i s c o n t i n u i t y .

Concept Types of Discontinuity A function has an infinite discontinuity at x = c if the function value increases or decreases indefinitely as x approaches cfrom the left and right.

Example

A function has a jump discontinuity at x = c if the limits of the function as xapproaches cfrom the left and right exist but have two distinct values.

Example

V 1 »

0 ,1 x

y=f(x)

0 ,1 x 0 ,1 x

A function has a removable discontinuity if the function is continuous everywhere except for a hole at x = c.

Example

y *

y=f(x)

y *

V y *

oT x

StudyTip Limits Whether f(x) exists at x = c has no bearing on the existence of the limit of f{x) as x approaches c.

1 N o t i c e t h a t f o r g r a p h s o f f u n c t i o n s w i t h a r e m o v a b l e d i s c o n t i n u i t y , ^> the l i m i t o f f ( x ) at p o i n t c exists , b u t e i ther the v a l u e o f the f u n c t i o n

at c is u n d e f i n e d , or, as w i t h the g r a p h s h o w n , the v a l u e of/(c ) is n o t the same as the v a l u e of the l i m i t at p o i n t c.

y=f(x)

°^ c

I n f i n i t e a n d j u m p d i s c o n t i n u i t i e s are c lass i f ied as nonremovab le d iscont inui t ies . A n o n r e m o v a b l e d i s c o n t i n u i t y c a n n o t be e l i m i n a t e d b y r e d e f i n i n g the f u n c t i o n at t h a t p o i n t , since the f u n c t i o n approaches d i f f e r e n t v a l u e s f r o m the l e f t a n d r i g h t sides at t h a t p o i n t o r does n o t a p p r o a c h a s ing le v a l u e at a l l . Ins tead i t is i n c r e a s i n g or decreas ing i n d e f i n i t e l y .

These observat ions l e a d to the f o l l o w i n g test f o r the c o n t i n u i t y o f a f u n c t i o n .

ConceptSummary Continuity Test

A function f(x) is continuous at x = c if it satisfies the following conditions.

• f(x) is defined at c. That is, f[c) exists.

• f(x) approaches the same value from either side of c. That is, Hm f(x) exists.

• The value that f(x) approaches from each side of c is f(c). That is, jirn^ f(x) = f(c).

y

1 / \ J (; )

\ /

0 / X

\ / \fix) = 2 x 2 - 3 x - 1 1

1 I I I I I I

Figure 1.3.1

Exam i Identify a Point of Continuity Determine w h e t h e r fix) = 2x2 — 3x — 1 is cont inuous at x = 2. Just i fy u s i n g the cont inuity test.

C h e c k the three c o n d i t i o n s i n the c o n t i n u i t y test.

1. D o e s / ( 2 ) exist?

Because f{2) = 1 , the f u n c t i o n is d e f i n e d at x = 2.

2. D o e s l i m / ( x ) exist?

C o n s t r u c t a table t h a t s h o w s v a l u e s o f f(x) f o r x -va lues a p p r o a c h i n g 2 f r o m the l e f t a n d f r o m the r i g h t .

x approaches 2 — • — x approaches 2

1.9 1.99 1.999 2.0 2.001 2.01 2.1

0.52 0.95 0.995 1.005 1.05 1.52

T h e p a t t e r n of o u t p u t s suggests t h a t as the v a l u e o f x gets closer to 2 f r o m the l e f t a n d f r o m

the r i g h t , / ( x ) gets closer to 1 . So, w e es t imate t h a t l i m / ( x ) = 1 .

3. D o e s l i m / ( x ) = / ( 2 ) ?

x—*2

Because l i m (2x2 — 3x — 1) is e s t i m a t e d to be 1 a n d / ( 2 ) = 1 , w e c o n c l u d e t h a t / ( x ) is

c o n t i n u o u s at x = 2. T h e g r a p h o f / ( x ) s h o w n i n F i g u r e 1.3.1 s u p p o r t s th is c o n c l u s i o n .

^ Guided Practice D e t e r m i n e w h e t h e r each f u n c t i o n is c o n t i n u o u s at x = 0. J u s t i f y u s i n g t h e c o n t i n u i t y test.

1A. f{x) = x3 1B. f{x) j i f x < 0

x i f x > 0

connectED.mcgraw-hill.com j 25

I f j u s t one o f the c o n d i t i o n s f o r c o n t i n u i t y is n o t sa t i s f ied , the f u n c t i o n is d i s c o n t i n u o u s at x = c. E x a m i n i n g a f u n c t i o n can h e l p y o u i d e n t i f y the t y p e o f d i s c o n t i n u i t y at t h a t p o i n t .

12 y 12

( -3 , 5) P

b

- 4 I 2 L X

3 , - -r > 3 , - -r > \

fix) = < f 3 x - 2 i f x > - 3 [ 2 - x if x < - 3

Figure 1.3.2

fix) x+3

(-3-1):

Figure 1.3.3

Identify a Point of Discontinuity D e t e r m i n e w h e t h e r each f u n c t i o n is c o n t i n u o u s at t h e g i v e n x - v a l u e ( s ) . J u s t i f y u s i n g t h e c o n t i n u i t y test. I f d i s c o n t i n u o u s , i d e n t i f y t h e t y p e o f d i s c o n t i n u i t y as infinite, jump, o r removable.

a. fix) 3x x >

; at x = —3 if x < - 3

1. B e c a u s e / ( - 3 ) = 5 , / ( - 3 ) exists.

2. I n v e s t i g a t e f u n c t i o n va lues close to/(—3) .

x approaches -- 3 • x approaches - 3

-3 .1 -3.01 -3.001 -3.0 -2.999 -2.99 -2 .9

| m y 5.01 5.001 -10.997 -10.97 10.7

T h e p a t t e r n o f o u t p u t s suggests t h a t / ( x ) approaches 5 as x approaches —3 f r o m the l e f t a n d —11 asf(x) approaches —3 f r o m the r i g h t . Because these va lues are n o t the same, \im^f(x) does n o t exist . There fore ,/(x ) is d i s c o n t i n u o u s at x = —3. Because/(x)

approaches t w o d i f f e r e n t va lues w h e n x = —3,/(x) has a j u m p d i s c o n t i n u i t y at x = —3. T h e g r a p h o f / ( x ) i n F i g u r e 1.3.2 s u p p o r t s th is c o n c l u s i o n .

b. / ( * ) = ^ ± ! . ; at x = - 3 a n d x = 3

1. B e c a u s e / ( - 3 ) 0 a n d / ( 3 ) = jr, b o t h o f w h i c h are u n d e f i n e d , / ( — 3 ) a n d / ( 3 ) d o n o t exist. 0 — J 0

There fore ,/(x ) is d i s c o n t i n u o u s at b o t h x = —3 a n d at x = 3

I n v e s t i g a t e f u n c t i o n va lues close to/(—3) .

x approaches —3 x approaches - 3

-3 .1 -3.01 -3.001 -3.0 -2.999 -2.99 -2 .9

-0.164 -0.166 -0.167 -0.167 -0.167 -0.169

T h e p a t t e r n o f o u t p u t s suggests t h a t / ( x ) approaches a l i m i t close to —0.167 as x

•0.167 o r - \6

approaches —3 f r o m each side, so l i m /(x) = x—•—3

Inves t iga te f u n c t i o n va lues close to/(3 ) .

x approaches 3 • xapproaches 3

X 2.9 2.99 2.999 3.0 3.001 3.01 3.1

fix) -10 -100 -1000 1000 100 10

T h e p a t t e r n o f o u t p u t s suggests t h a t f o r va lues o f x a p p r o a c h i n g 3 f r o m the l e f t , / ( x ) becomes i n c r e a s i n g l y m o r e n e g a t i v e . For va lues o f x a p p r o a c h i n g 3 f r o m the r i g h t , / ( x ) becomes i n c r e a s i n g l y m o r e p o s i t i v e . There fore , l i m / ( x ) does n o t exist .

3. Because l i m ^ / ( x ) exists, b u t / ( — 3 ) is u n d e f i n e d , / ( x ) has a r e m o v a b l e d i s c o n t i n u i t y at

x = —3. Because/(x) decreases w i t h o u t b o u n d as x approaches 3 f r o m the le f t a n d increases w i t h o u t b o u n d as x approaches 3 f r o m the r i g h t , / ( x ) has a n i n f i n i t e d i s c o n t i n u i t y at x = 3. T h e g r a p h o f / ( x ) i n F i g u r e 1.3.3 s u p p o r t s these conc lus ions .

w GuidedPractice

2A. /(x) -; at x = 0 2B. f{x) f 5 x +

l 2 "

x > 2

x < 2 ' at x

26 I L e s s o n 1-3 | Continuity, End Behavior, a n d Limits

I f a f u n c t i o n is c o n t i n u o u s , y o u can a p p r o x i m a t e the l o c a t i o n of i t s zeros b y u s i n g the I n t e r m e d i a t e Va lue T h e o r e m a n d its c o r o l l a r y T h e L o c a t i o n P r i n c i p l e .

KeyConcept Intermediate Value Theorem

If f(x) is a continuous function and a < b and there is a value n such that n is between f(a) and f(b), then there is a number c, such that a < c < b and f(c) = n.

fib) y

(b, f ( b ) ) ^ - ^

f(c) ic, n ) / \ I

0 a / \

/(a) 1 / o b X

M 3 , f(a))

Corollary: The Location Principle If f(x) is a continuous function and f(a) and f(b) have opposite signs, then there exists at least one value c, such that a < c < b and f(c) = 0. That is, there is a zero between a and 6.

StudyTip Approximating Zeros with No Sign Changes While a sign change on an interval does indicate the location of a real zero, the absence of a sign change does nor indicate that there are no real zeros on that interval. The best method of checking this is to graph the function.

Example 3 Approximate Zero! D e t e r m i n e b e t w e e n w h i c h c o n s e c u t i v e i n t e g e r s t h e r e a l zeros o f each f u n c t i o n are l o c a t e d o n the g i v e n i n t e r v a l .

a. fix) = x3 - 4x + 2; [ - 4 , 4]

1 1 - 4 - 3 - 2 - 1 0 1 2 3 4

- 4 6 - 1 3 2 5 2 -1 2 17 50

Because/(—3) is n e g a t i v e and/(—2) is p o s i t i v e , b y the L o c a t i o n P r i n c i p l e , / ( x ) has a zero b e t w e e n —3 a n d —2. T h e v a l u e o f f ( x ) also changes s i g n f o r 0 < x < 1 a n d 1 < x < 2. T h i s indica tes the existence o f rea l zeros i n each o f these i n t e r v a l s .

T h e g r a p h of f(x) s h o w n at the r i g h t s u p p o r t s the c o n c l u s i o n t h a t there are rea l zeros b e t w e e n - 3 a n d - 2 , 0 a n d 1 , a n d 1 a n d 2.

b. fix) = xz + x + 0.16; [ - 3 , 3 ]

1 I -3 - 2 - 1 0 1 2 3

I 6.16

2.16 0.16 0.16 2.16 6.16 12.16

T h e va lues o f / ( x ) d o n o t change s i g n f o r the x -va lues used . H o w e v e r , as the x -va lues a p p r o a c h —1 f r o m the l e f t , / ( x ) decreases, t h e n b e g i n s i n c r e a s i n g at x = 0. So, there m a y be real zeros b e t w e e n consecut ive integers —1 a n d 0. G r a p h the f u n c t i o n to v e r i f y .

T h e g r a p h o f f(x) crosses the x-axis t w i c e o n the i n t e r v a l [—1,0] , so there are rea l zeros b e t w e e n —1 a n d 0.

p GuidedPractice x 2 - 6 .

0 y u j / /

y - 4 0 v. \x

/ k x ) = x 3 - 4 x + 2

-8 -8

\ y 4 \ n I \ o

\ -2 V -2 / \ — -1 A / 0

\ l\X) = X - -f- X + U. ID

- 2 -1 1 2

I 1

3A. /(x) x + 4

•3,4] 3B. f{x) = 8 x 3 - 2 x 2 - 5x - 1 ; [ - 5 , 0]

connectED.mcgraw-hill.com 1 2 7 k

ReadingMath Limits The expression

lim fix) is read the limit

of f(x) as x approaches positive infinity. The expression lim fix)

X—>-00

is read the limit of f(x) as x approaches negative infinity.

2 End BehaVIOr T h e e n d b e h a v i o r o f a f u n c t i o n describes h o w a f u n c t i o n behaves at e i ther end o f the g r a p h . T h a t is , e n d b e h a v i o r is w h a t h a p p e n s t o the v a l u e o f f(x) as x increases or

decreases w i t h o u t b o u n d — b e c o m i n g greater a n d greater or m o r e a n d m o r e n e g a t i v e . To descr ibe the e n d b e h a v i o r of a g r a p h , y o u c a n use the concept of a l i m i t .

L e f t - E n d B e h a v i o r R i g h t - E n d B e h a v i o r

l i m fix)

O n e p o s s i b i l i t y f o r the e n d b e h a v i o r o f the g r a p h o f a f u n c t i o n is f o r the v a l u e o f f ( x ) to increase or decrease w i t h o u t b o u n d . T h i s e n d b e h a v i o r is d e s c r i b e d b y s a y i n g t h a t / ( x ) approaches p o s i t i v e o r n e g a t i v e i n f i n i t y .

l i m / ( x )

\lim fix) = co \

\ X —• —oo \

fix) - oo

X—• oo

0

fix) - -oo

1 \ \m fix) = -oo

0

fix) - -oo \ ^ y = f i x )

0

fix) - -oo • 1

Graphs that Approach Infinity Use t h e g r a p h o f fix) = — x 4 + 8x3 + 3x2 + 6x — 80 t o d e s c r i b e i t s e n d b e h a v i o r . S u p p o r t t h e c o n j e c t u r e n u m e r i c a l l y .

Analyze Graphically

I n the g r a p h o f / ( x ) , i t appears t h a t l i m f(x) = —oo

a n d l i m fix) = —oo.

fix) = - x 4 + 8x 3 + 3x 2 + 6x - 80

y f -

A 30 H 30

L JU

- 8 - 4 L \ i

30 30

Support Numerically

C o n s t r u c t a table o f va lues to i n v e s t i g a t e f u n c t i o n va lues as |x| increases. T h a t i s , inves t iga te the v a l u e o f f{x) as the v a l u e o f x becomes greater a n d greater or m o r e a n d m o r e n e g a t i v e .

x approaches — oo x approaches oo

-10,000 -1000 -100 0 100 1000 10,000

- 1 - 1 0 1 6 - 1 . 1 0 1 2 —1 - 1 0 8 - 8 0 —1 - 1 0 8 - 1 . 1 0 1 2 - 1 - 1 0 1 6

T h e p a t t e r n o f o u t p u t s suggests t h a t as x approaches —oo,f(x) approaches — oo a n d as x approaches oo,/(x) approaches —oo. T h i s s u p p o r t s the conjecture.

• Guided Practice U s e t h e g r a p h o f each f u n c t i o n t o d e s c r i b e i t s e n d b e h a v i o r . S u p p o r t t h e c o n j e c t u r e n u m e r i c a l l y .

4A. I y 1 c

I / - 4 - 2 o 2 / 4 x

e -D

12 c 12 c I\M = A — 3/1 -f C

4B. / 1 .J

- f fix) = - x 3 , 3x2 x

4 + 4 2

0 A - v

Q —O — n \j

—4 -' A -

O X

• + -- 8 •

1 4 Ins tead of fix) b e i n g u n b o u n d e d , a p p r o a c h i n g oo or - o o as \x\, some f u n c t i o n s approach , b u t never reach, a f i x e d v a l u e .

28 | L e s s o n 1-3 | Cont inui ty , End Behavior , and Limits

Graphs that Approach a Specific Value

to d e s c r i b e i t s e n d Use the g r a p h o f fix) = ——-x — 1x -+- 8

b e h a v i o r . S u p p o r t t h e c o n j e c t u r e n u m e r i c a l l y .

Analyze Graphically

I n the g r a p h of fix), i t appears t h a t l i m f(x) = 0 a n d l i m f(x) = 0.

Support Numerically

-< x approaches —oo

0.8

0.4

-0.4

- 0 .8 J

fix) x'-2x +

01 4 8x

x approaches oo

X -10,000 -1000 -100 0 100 1000 10,000

fix) - 1 - 1 0 - 4 -0.001 -0.01 0 0.01 0.001 1 . i o - 4

T h e p a t t e r n o f o u t p u t s suggests t h a t as x approaches —oo,f(x) approaches 0 a n d as x approaches oo,fix) approaches 0. T h i s s u p p o r t s the conjecture .

• Guided Practice Use t h e g r a p h o f each f u n c t i o n to d e s c r i b e i t s e n d b e h a v i o r . S u p p o r t t h e c o n j e c t u r e n u m e r i c a l l y .

5A.

4 O

3

8x

5B.

fix) 3x-2 x+1

Q y U

f(x) =

1 2x2+ x+ 1

J X —I ) - 4 % \ L \ 8x

- 4 I I

P —o I

K n o w i n g the e n d b e h a v i o r o f a f u n c t i o n can h e l p y o u solve r e a l - w o r l d p r o b l e m s .

Real-World Example 6

Real-WorldLmk The form U(r) = - ^ for gravitational potential energy is most useful for calculating the velocity required to escape Earth's gravity, 25,000 miles per hour.

Source: The Mechanical Universe

Apply End Behavior

PHYSICS G r a v i t a t i o n a l p o t e n t i a l e n e r g y o f a n o b j e c t is g i v e n

b y Uir) = — G m ^ e f w h e r e G is N e w t o n ' s g r a v i t a t i o n a l c o n s t a n t , m i s t h e mass o f t h e o b j e c t , Me i s t h e mass o f E a r t h , a n d r i s t h e d i s t a n c e f r o m t h e o b j e c t to t h e center o f E a r t h as s h o w n . W h a t h a p p e n s t o t h e g r a v i t a t i o n a l p o t e n t i a l e n e r g y o f t h e o b j e c t as i t m o v e s f a r t h e r a n d f a r t h e r f r o m Earth?

W e are asked to descr ibe the e n d b e h a v i o r L7(r) f o r large va lues o f r. T h a t is , w e are asked to f i n d l i m Uir). Because G, m, a n d Me are

r—*oo constant va lues , the p r o d u c t GmMc is also a constant v a l u e . For

i n c r e a s i n g va lues o f r, t h e f r a c t i o n — G m ^ e w [ \ h 0, so l i m Uir) = 0. There fore , as a n object m o v e s f a r t h e r f r o m E a r t h ,

r—*oo '

i ts g r a v i t a t i o n a l p o t e n t i a l e n e r g y approaches 0.

GuidedPractice 6. PHYSICS D y n a m i c pressure is the pressure genera ted b y the v e l o c i t y o f the m o v i n g f l u i d a n d

pxP" is g i v e n b y qiv) = ——, w h e r e p is the d e n s i t y o f the f l u i d a n d v is the v e l o c i t y o f the f l u i d . W h a t w o u l d h a p p e n to the d y n a m i c pressure o f a fluid i f the v e l o c i t y w e r e to c o n t i n u o u s l y increase?

29

p-by-Step Solutions begin on page R29.

D e t e r m i n e w h e t h e r each f u n c t i o n i s c o n t i n u o u s at t h e g i v e n x - v a l u e ( s ) . J u s t i f y u s i n g t h e c o n t i n u i t y test. I f d i s c o n t i n u o u s , i d e n t i f y t h e t y p e o f d i s c o n t i n u i t y as infinite, jump, o r removable. (Examples 1 and 2)

1. f(x) = V x 2 - 4 ; at x = - 5

2. f(x) = \/x + 5 ; at x = 8

at x = — 6 a n d x = 6 3. /z(x) x + 6

x 2 - 25, 4. /z(x) = — ; at x = — 5 a n d x = 5 v ' x + 5

5. £ t o

6-

7. h(x)

x-1

2-x 2 + x

x - 4

; at x = 1

; at x = —2 a n d x = 2

; at x = 1 a n d x = 4 xz - 5x + 4

x(x - 6) 8. h(x) = ; at x = 0 and x = 6

x-3

9. /(x) = { 4 * - J * ^ - 6 ; a t x = - 6 7 w l - x + 2 i f x > - 6

10. /(x) x 2 - 1 i f x > - 2

x - 5 i f x < - 2 ; at x = - 2

® Jgf PHYSICS A w a l l separates t w o r o o m s w i t h d i f f e r e n t

t e m p e r a t u r e s . T h e heat t ransfer i n w a t t s b e t w e e n the t w o

r o o m s can be m o d e l e d by f(zv) = w h e r e w is the w a l l

th ickness i n meters . (Examples 1 and 2)

Cool Room • W a r m Room 60°F 80°F

\ Wall

a. D e t e r m i n e w h e t h e r the f u n c t i o n is c o n t i n u o u s at w = 0.4. Jus t i fy y o u r a n s w e r u s i n g the c o n t i n u i t y test.

b. Is the f u n c t i o n c o n t i n u o u s ? Jus t i fy y o u r a n s w e r u s i n g the c o n t i n u i t y test. I f d i s c o n t i n u o u s , i d e n t i f y the t y p e o f d i s c o n t i n u i t y as infinite, jump, o r removable.

C. G r a p h the f u n c t i o n to v e r i f y y o u r conc lus ion f r o m p a r t b .

12. CHEMISTRY A s o l u t i o n m u s t be d i l u t e d so i t can be u s e d i n a n e x p e r i m e n t . A d d i n g a 4 - m o l a r N a C l s o l u t i o n to a 10-molar s o l u t i o n w i l l decrease the c onc e nt r a t i on . T h e c o n c e n t r a t i o n C of the m i x t u r e c a n be m o d e l e d b y

C(x) = ^ + 4 x , w h e r e x is the n u m b e r o f l i ters of w 50 + x

4 - m o l a r s o l u t i o n a d d e d . (Examples 1 and 2)

a. D e t e r m i n e w h e t h e r the f u n c t i o n is c o n t i n u o u s at x = 10. Jus t i fy the a n s w e r u s i n g the c o n t i n u i t y test.

b. Is the f u n c t i o n c o n t i n u o u s ? Jus t i fy y o u r a n s w e r u s i n g the c o n t i n u i t y test. I f d i s c o n t i n u o u s , i d e n t i f y the t y p e o f d i s c o n t i n u i t y as infinite, jump, or removable a n d descr ibe w h a t affect, i f any, the d i s c o n t i n u i t y has o n the c o n c e n t r a t i o n o f the m i x t u r e .

c. G r a p h the f u n c t i o n to v e r i f y y o u r conc lus ion f r o m p a r t b .

D e t e r m i n e b e t w e e n w h i c h c o n s e c u t i v e i n t e g e r s the r e a l zeros o f each f u n c t i o n are l o c a t e d o n t h e g i v e n i n t e r v a l . (Example 3)

13. /(x) = x 3 - x 2 - 3; [-2,4]

14. g(x) = - x 3 + 6x + 2; [ - 4 , 4 ]

15. /(x) = 2 x 4 - 3 x 3 + x 2 - 3; [ - 3 , 3 ]

16. h(x) = - x 4 + 4 x 3 - 5x - 6; [ 3 , 5 ]

17. f(x) = 3 x 3 - 6 x 2 - 2x + 2; [ - 2 , 4 ]

18. g(x) = * 2 + 3 x - 3 ; [

x 2 + 1

19. h(x)

20. f{x) = V x 2 - 6 - 6 ; [3, 8]

21- gix) = V * 3 + 1 - 5; [0, 5]

U s e t h e g r a p h o f each f u n c t i o n to d e s c r i b e i t s e n d b e h a v i o r . S u p p o r t t h e c o n j e c t u r e n u m e r i c a l l y . (Examples 4 and 5)

22. p64 y p64 J

0 \ ? 4 6 X .

f x) 4 x 4 - 6 x 3 - X

64 64

23. y 23. 23.

/ »

23.

A 23. 23.

O X

23.

< 1 fix) = X 3 + 7x < 1 /

i i M i i i n

32J f

v + 2x+1 L

32J

fix) = A + 2x+1 L

> r X - 3 16

- 4 8 - 32 - 1 6 / 16x 1 R I 0

32 32

25. I 1 R

y 25. I D

25.

Q

25.

O

25.

L - 8 -

U i X

fix) \x

6 -- 5 - X

- 8 (

.—

^ - 1 6

26. <y

< A \ \ X

o . . 2 c . f [ x ) = o x - O A T .

ox

27.

12x

fix) 16x2

2x3 + 5x + 2

28. v

k V k - I ] 0 8 16 24x

29.

f ( x ) = 1 ^ ± 4 x _ 5

4 x 3 - 9

6 x'-\

:12 - 4

30 | L e s s o n 1-3 | Continuity, End Behavior, and Limits

30. POPULATION T h e U.S. p o p u l a t i o n f r o m 1790 to 1990 can be m o d e l e d b y p{x) = 0 .0057x 3 + 0.4895x 2 + 0.3236* + 3.8431, w h e r e x is the n u m b e r o f decades af ter 1790. Use the e n d b e h a v i o r o f the g r a p h to descr ibe the p o p u l a t i o n t r e n d . S u p p o r t the conjecture n u m e r i c a l l y . Does t h i s t r e n d seem realist ic? E x p l a i n y o u r r e a s o n i n g . (Example 4)

U.S. Population from 1790 to 1990 250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n

250

225

- £ 2 0 0 J 175 % 150 g" 125 "•g 100 1 75 o ° - 50

25 n u 2 4 6 8 10 12 14 16 1

Decades Since 1790 8 20

31. CHEMISTRY A cata lyst is u s e d to increase the rate of a c h e m i c a l reac t ion . T h e reac t ion rate R, or the speed at

0 5x w h i c h the reac t ion is o c c u r r i n g , is g i v e n b y R{x) = ^ +

w h e r e x is the c o n c e n t r a t i o n o f the s o l u t i o n i n m i l l i g r a m s of so lute p e r l i t e r of s o l u t i o n . (Example 5)

a. G r a p h the f u n c t i o n u s i n g a g r a p h i n g calculator .

b. W h a t does the e n d b e h a v i o r of the g r a p h m e a n i n the c o n t e x t o f t h i s e x p e r i m e n t ? S u p p o r t the conjecture n u m e r i c a l l y .

32. ROLLER COASTERS The speed of a r o l l e r coaster after i t d r o p s f r o m a h e i g h t A to a h e i g h t B is g i v e n b y f(h/) — \/2g(hA — hB), w h e r e hA is the h e i g h t at p o i n t A, hBis the h e i g h t at p o i n t B, a n d g is the accelerat ion d u e to g r a v i t y . W h a t h a p p e n s to f{hA) as hB decreases to 0? (Example 6)

Use l o g i c a l r e a s o n i n g to d e t e r m i n e t h e e n d b e h a v i o r o r l i m i t o f the f u n c t i o n as x a p p r o a c h e s i n f i n i t y . E x p l a i n y o u r r e a s o n i n g . (Example 6)

33. q(x)

35. P(x)

37. cix)

39. h(x)

_ 24 x

x + 1

34. /(x) = ° f x z

5x^ x5 + 2x + 1

2 x 5 + 7 x 3 -

36. m(x)

38. k(x)

4 + x 2x + 6

4x" 3 x - 1 l l x

40. g(x) = x 4 - 9 x 2 +

41. PHYSICS T h e k i n e t i c e n e r g y o f an object i n m o t i o n can be

v1

expressed as E(m) = — , w h e r e p is the m o m e n t u m a n d m is the mass o f the object. I f s a n d is a d d e d t o a m o v i n g r a i l w a y car, w h a t w o u l d h a p p e n as m c o n t i n u e s to increase? (Example 6)

Use each g r a p h to d e t e r m i n e t h e x -va lue (s ) at w h i c h each f u n c t i o n is d i s c o n t i n u o u s . I d e n t i f y t h e t y p e o f d i s c o n t i n u i t y . T h e n use the g r a p h t o d e s c r i b e i t s e n d b e h a v i o r . J u s t i f y y o u r a n s w e r s .

42. y \

A J ~\ 7 /I ! ~\ 7 /I !

— r X

A r 1

Q —ori

1

® 1 j i <-

Q I

%X) = - f -

—7 s 1 A

— L r — L r 0 i 5 1 fx A

i

44. PHYSICS T h e w a v e l e n g t h X o f a p e r i o d i c w a v e is the dis tance b e t w e e n consecut ive c o r r e s p o n d i n g p o i n t s o n the w a v e , s u c h as t w o crests or t r o u g h s .

crest

one 1 wavelength 1

trough

T h e f r e q u e n c y / , o r n u m b e r o f w a v e crests t h a t pass a n y

g i v e n p o i n t d u r i n g a g i v e n p e r i o d o f t i m e , is g i v e n b y

/(A) = 4> w h e r e c is the speed o f l i g h t or 2.99 • 1 0 8 meters

p e r second.

a. G r a p h the f u n c t i o n u s i n g a g r a p h i n g calculator .

b. Use the g r a p h to descr ibe the e n d b e h a v i o r o f the f u n c t i o n . S u p p o r t y o u r conjecture n u m e r i c a l l y .

C. Is the f u n c t i o n c o n t i n u o u s ? I f n o t , i d e n t i f y a n d descr ibe a n y p o i n t s o f d i s c o n t i n u i t y .

GRAPHING CALCULATOR G r a p h each f u n c t i o n a n d d e t e r m i n e w h e t h e r i t i s c o n t i n u o u s . I f d i s c o n t i n u o u s , i d e n t i f y a n d d e s c r i b e a n y p o i n t s o f d i s c o n t i n u i t y . T h e n descr ibe i t s e n d b e h a v i o r a n d locate a n y zeros .

45. / ( X ) :

46. g(x)

47. h(x)

AS. h(x)

49. h(x)

4x z + x + 6

x x J - 5x z - 18x + 72

' 2 - l l l x - 3 4x^

x •3x

4 x 2

18

29x 24 c l - 2x - 15

5x 2 - 26x + 120 x 2 + x - 12

50. VEHICLES T h e n u m b e r A o f a l t e r n a t i v e - f u e l e d vehic les i n use i n the U n i t e d States f r o m 1995 to 2004 c a n be a p p r o x i m a t e d b y / ( f ) = 2044f 2 - 3388f + 206,808, w h e r e t represents the year a n d t = 5 c o r r e s p o n d s to 1995.

a. G r a p h the f u n c t i o n .

b. A b o u t h o w m a n y a l t e r n a t i v e - f u e l e d vehic les w e r e there i n the U n i t e d States i n 1998?

c. A s t i m e goes by, w h a t w i l l the n u m b e r of a l t e r n a t i v e -f u e l e d vehic les a p p r o a c h , a c c o r d i n g to the m o d e l ? D o y o u t h i n k t h a t the m o d e l is v a l i d after 2004? E x p l a i n .

GRAPHING CALCULATOR G r a p h each f u n c t i o n , a n d d e s c r i b e i t s e n d b e h a v i o r . S u p p o r t t h e c o n j e c t u r e n u m e r i c a l l y , a n d p r o v i d e a n e f f e c t i v e v i e w i n g w i n d o w f o r each g r a p h .

- x 4 + 1 2 x 3 1 a J 1 51. /(x)

52. g(x)

53. /(x)

54. g(x)

4 x z - 4

2 0 x 4 + 2 x 3 - 5

16x^ x 2 + 15x

8x - 24x 3

14 + 2x3

55. BUSINESS G a b r i e l is s t a r t i n g a s m a l l business screen-p r i n t i n g a n d s e l l i n g T-shir ts . Each s h i r t costs $3 to p r o d u c e . H e i n i t i a l l y i n v e s t e d $4000 f o r a screen p r i n t e r a n d o t h e r business needs.

a. W r i t e a f u n c t i o n to represent the average cost p e r s h i r t as a f u n c t i o n o f the n u m b e r o f sh i r t s s o l d n.

b. Use a g r a p h i n g ca lcu la tor t o g r a p h the f u n c t i o n .

C. A s the n u m b e r of s h i r t s s o l d increases, w h a t v a l u e does the average cost approach?

56. ^ M U L T I P L E REPRESENTATIONS I n t h i s p r o b l e m , y o u w i l l

i n v e s t i g a t e l i m i t s . C o n s i d e r / ( x ) = a x + b , w h e r e a a n d c cxd + d

are n o n z e r o integers , a n d b a n d d are integers .

a. TABULAR L e t c = 1 , a n d choose three d i f f e r e n t sets of va lues f o r a, b, a n d d. W r i t e the f u n c t i o n w i t h each set o f va lues . C o p y a n d c o m p l e t e t h e table b e l o w .

c = 1

lim fix) X—yoo

lim f(x)

b. TABULAR Choose three d i f f e r e n t sets o f va lues f o r each v a r i a b l e : one set w i t h a > c, one set w i t h a < c, a n d one set w i t h a = c. W r i t e each f u n c t i o n , a n d create a table as y o u d i d i n p a r t a.

c. ANALYTICAL M a k e a conjecture a b o u t the l i m i t o f ax3 + b

/ ( * ) cx-

i n f i n i t y .

as x approaches p o s i t i v e a n d n e g a t i v e

57. GRAPHING CALCULATOR G r a p h several d i f f e r e n t f u n c t i o n s of the f o r m / ( x ) = xn + ax" ~1 + bxn - 2 , w h e r e n, a, a n d b are n o n n e g a t i v e in tegers .

a. M a k e a conjecture a b o u t the e n d b e h a v i o r o f the f u n c t i o n w h e n n is p o s i t i v e a n d e v e n . I n c l u d e a g r a p h t o s u p p o r t y o u r conjecture .

b. M a k e a conjecture a b o u t the e n d b e h a v i o r o f the f u n c t i o n w h e n n is p o s i t i v e a n d o d d . I n c l u d e a g r a p h to s u p p o r t y o u r conjecture.

H.O.T. Problems Use Higher-Order Thinking Skills

REASONING D e t e r m i n e w h e t h e r each f u n c t i o n has a n infinite,

jump, o r removable d i s c o n t i n u i t y at x = 0. E x p l a i n .

58. /(x) 59. f(x) = ±r

60. ERROR ANALYSIS K e e n a n a n d George are d e t e r m i n i n g w h e t h e r the r e l a t i o n g r a p h e d b e l o w is c o n t i n u o u s at p o i n t c. K e e n a n t h i n k s t h a t i t is the g r a p h of a f u n c t i o n fix) t h a t is d i s c o n t i n u o u s at p o i n t c because l i m / ( x ) =/(c )

f r o m o n l y one s ide o f c. G e o r g e t h i n k s t h a t the g r a p h is n o t a f u n c t i o n because w h e n x = c, the r e l a t i o n has t w o d i f f e r e n t y - v a l u e s . Is e i ther o f t h e m correct? E x p l a i n y o u r r e a s o n i n g .

y

(S) CHALLENGE D e t e r m i n e the va lues o f a a n d b so t h a t /

is c o n t i n u o u s .

x 2 + a i f x > 3

REASONING F i n d l i m fix) f o r each o f t h e f o l l o w i n g . E x p l a i n y o u r r e a s o n i n g .

62. l i m fix) = —oo a n d / i s a n e v e n f u n c t i o n .

63. l i m f{x £ — • 0 0

a n d / i s an o d d f u n c t i o n .

64. l i m fix) = oo a n d the g r a p h o f / i s s y m m e t r i c w i t h respect

to the o r i g i n .

65. l i m f{x) = oo a n d the g r a p h o f / i s s y m m e t r i c w i t h respect

to the y-ax is .

66. WRITING IN MATH P r o v i d e a n e x a m p l e o f a f u n c t i o n w i t h a r e m o v a b l e d i s c o n t i n u i t y . E x p l a i n h o w th is d i s c o n t i n u i t y can be e l i m i n a t e d . H o w does e l i m i n a t i n g the d i s c o n t i n u i t y affect the f u n c t i o n ?

32 | L e s s o n 1-3 | Continuity, E n d Behavior , and Limits

Spiral Review

GRAPHING CALCULATOR G r a p h each f u n c t i o n . A n a l y z e t h e g r a p h to d e t e r m i n e w h e t h e r each f u n c t i o n is even, odd, o r neither. C o n f i r m a l g e b r a i c a l l y . I f o d d o r e v e n , d e s c r i b e the s y m m e t r y o f t h e g r a p h o f t h e f u n c t i o n . (Lesson 1 -2)

67. h(x) = V - r 2 - 16 68. fix) = 2 * ± I 69. g(x) = x5 - 5x3 + x

State t h e d o m a i n o f each f u n c t i o n . (Lesson 1-1)

70- fix) = / * + 6 71. S(x) = 2 * + 3 72. &) = V 2 ^ x z + 3x + 2 x — 2x - 10

73. POSTAL SERVICE T h e U.S. Postal Service uses f i v e - d i g i t Z I P codes t o r o u t e letters a n d packages to t h e i r des t ina t ions . (Lesson 0-7)

a. H o w m a n y Z I P codes are poss ib le i f the n u m b e r s 0 t h r o u g h 9 are u s e d f o r each of the f i v e d ig i t s?

b. Suppose t h a t w h e n the f i r s t d i g i t is 0, the second, t h i r d , a n d f o u r t h d i g i t s c a n n o t be 0. H o w m a n y f i v e - d i g i t Z I P codes are poss ib le i f the f i r s t d i g i t is 0?

C. I n 1983, the U.S. Postal Service i n t r o d u c e d the Z I P + 4, w h i c h a d d e d f o u r m o r e d i g i t s to the e x i s t i n g f i v e - d i g i t Z I P codes. U s i n g the n u m b e r s 0 t h r o u g h 9, h o w m a n y a d d i t i o n a l Z I P codes w e r e possible?

G i v e n A = \ ^ L 3 - 3

s o l v e each e q u a t i o n f o r X . (Lesson 0-6) 8 - 5 4 4 9 - 3

74. 3X-B = A 75. 2B + X = 4A 76. A - 5 X = B

^ o l v e each s y s t e m o f e q u a t i o n s . (Lesson 0-5)

77. 4x - 6y + 4z = 12 78. x + 2y + z = 10 79. 2x - y + 3z = - 2

6x - 9y + 6z = 18 2x - y + 3z = - 5 x + 4 y - 2z = 16

5x - 8y + lOz = 20 2x - 3y - 5z = 27 5x + y - z = 14

Skills Review for Standardized Tests SAT/ACT A t L i n c o l n C o u n t y H i g h School , 36 s t u d e n t s are t a k i n g e i ther c a l c u l u s or p h y s i c s or b o t h , a n d 10 s tudents are t a k i n g b o t h ca lculus a n d phys ics . I f there are 31 s t u d e n t s i n the ca lculus class, h o w m a n y s t u d e n t s are there i n the p h y s i c s class?

A 5 C 11 E 21

B 8 D 15

81. W h i c h o f the f o l l o w i n g s tatements c o u l d be u s e d to descr ibe the e n d b e h a v i o r of/(x)?

F l i m fix) = — oo a n d X—• — OCT

l i m fix) = — oo X—>oo

G l i m fix) = — oo a n d X—>—oo

l i m fix) = oo

H l i m fix) — oo a n d l i m fix) = — c X—*-ooJ x—too

J l i m fix) = oo a n d l i m /(x) = oo

82. REVIEW A m y ' s l o c k e r code i n c l u d e s three n u m b e r s b e t w e e n 1 a n d 45, i n c l u s i v e . N o n e o f the n u m b e r s can repeat . H o w m a n y poss ible l o c k e r p e r m u t a t i o n s are there?

83. REVIEW Suppose a f i g u r e consists of three concentr ic circles w i t h r a d i i o f 1 f o o t , 2 feet, a n d 3 feet. F i n d the p r o b a b i l i t y t h a t a p o i n t chosen at r a n d o m lies i n the o u t e r m o s t r e g i o n ( b e t w e e n the second a n d t h i r d circles).

C A I

B D