AP Calculus
1004 Continuity (2.3)
C CONVERSATION: Voice level 0. No talking!
HHELP: Raise your hand and wait to be called on.
AACTIVITY: Whole class instruction; students in seats.
M MOVEMENT: Remain in seat during instruction.
PPARTICIPATION: Look at teacher or materials being discussed. Raise hand to contribute; respond to questions, write or perform other actions as directed.NO SLEEPING OR PUTTING HEAD DOWN, TEXTING, DOING OTHER WORK.
S
Activity: Teacher-Directed Instruction
Content:SWBAT calculate limits of any functions and apply properties of continuity Language: SW complete the sentence โLocal linearity meansโฆโ
General Idea:
General Idea: ________________________________________
We already know the continuity of many functions:
Polynomial (Power), Rational, Radical, Exponential, Trigonometric, and Logarithmic functions
DEFN: A function is continuous on an interval if it is continuous at each point in the interval.
DEFN: A function is continuous at a point IFF
a)
b)
c)
Can you draw without picking up your pencil
Has a point f(a) exists
Has a limit lim๐ฅโ๐
๐ (๐ฅ )๐๐ฅ๐๐ ๐ก๐
Limit = value lim๐ฅโ๐
๐ (๐ฅ )= ๐ (๐)
Limits Review:
PART 1: LOCAL BEHAVIOR (1). General Idea: Behavior of a function very near the point where
(2). Laymanโs Description of Limit (Local Behavior)
L
amust write every time
(3). Notation:
(4). Mantra
x ax a
Continuity Theorems
Interior Point: A function is continuous at an interior point of its
domain if
ONE-SIDED CONTINUITY
Endpoint: A function is continuous at a left endpoint of i
limx c
y f x c
y f x a
f x f c
ts domain
if lim
or
continuous at a right endpoint if lim .
x a
x b
f x f a
b f x f b
Continuity on a CLOSED INTERVAL.
Theorem: A function is Continuous on a closed interval if it is continuous at every point in the open interval and continuous from one side at the end points.
Example :The graph over the closed interval [-2,4] is given.
From the right
From the left
Discontinuity
No valuef(a) DNE hole
No limit
Limit does not equal valueLimit โ value
Vertical asymptotea)
c)
b)
jump
Discontinuity: cont.
Method:
(a).
(b).
(c).
Removable or Essential Discontinuities
Test the value = Look for f(a) =
Test the limitlim๐ฅโ๐โ
๐ (๐ฅ )=ยฟยฟlim
๐ฅโ๐+ยฟ ๐ (๐ฅ )=ยฟ ยฟยฟยฟ
Holes and hiccups are removableJumps and Vertical Asymptotes are essential
Test f(a) = lim๐ฅโ๐
๐ (๐ฅ) lim๐ฅโ๐
๐ (๐ฅ)f(a) =
00h๐๐๐
Vertical Asymptote
Lim DNEJump
= cont.โ hiccup
Examples:
EX:2
( )4
xf x
x
Identify the x-values (if any) at which f(x)is not continuous. Identify the reason for the discontinuity and the type of discontinuity. Is the discontinuity removable or essential?
removable or essential?
00
=
xโ 4
lim๐ฅโ 4
๐ (๐ฅ )=00
Hole discontinuous because f(x) has no valueIt is removable
Examples: cont.
2
1( )
( 3)f x
x
Identify the x-values (if any) at which f(x)is not continuous. Identify the reason for the discontinuity and the type of discontinuity. Is the discontinuity removable or essential?
removable or essential?
xโ 3
lim๐ฅโ 3
1(๐ฅโ3 )2
=10
VA discontinuous because no value
It is essential
Examples: cont.
3 , 1( )
3 , 1
x xf x
x x
Identify the x-values (if any) at which f(x)is not continuous. Identify the reason for the discontinuity and the type of discontinuity. Is the discontinuity removable or essential?
Step 1: Value must look at 4 equationf(1) = 4
Step 2: Limitlim๐ฅโ1โ
3+๐ฅ=4
lim๐ฅโ1+ยฟ 3โ๐ฅ=2ยฟ
ยฟ
lim๐ฅโ 1
๐ (๐ฅ )=๐ท๐๐ธ 2โ 4
It is a jump discontinuity(essential) because limit does not exist
Graph:
Determine the continuity at each point. Give the reason and the type of discontinuity.
x = -3
x = -2
x = 0
x =1
x = 2
x = 3
Hole discont. No valueremovableVA discont. Because no value no limit essential
Hiccup discont. Because limit โ valueremovable
Continuous limit = value
VA discont. No limitessentialJump discont. Because limit DNEessential
Algebraic Method
2
3 2 2( )
3 4 2
x xf x
x x
a.
b.
c.
Value: f(2) = 8
Look at function with equal
Limit:
lim๐ฅโ 2โ
๐ (๐ฅ )=8
lim๐ฅโ2+ยฟ ๐ (๐ฅ )=8ยฟ
ยฟ
lim๐ฅโ 2
๐ (๐ฅ )=8
Limit = value: 8=8
Limit = Value
Algebraic Method 2
2
2
1- 1
( ) - 2 1 3
9 3
3
x x
f x x x
xx
x
At x=1a.
b.
c.
Value: f(1) = -1
Limit: lim๐ฅโ 1โ๐ (๐ฅ )=1โ๐ฅ 2=0
lim๐ฅโ 1+ยฟ ๐ (๐ฅ )=๐ฅ 2โ2=โ 1ยฟ
ยฟ
lim๐ฅโ 1
๐ (๐ฅ )=๐ท๐๐ธ
Jump discontinuity because limit DNEessential
At x=3a.
b.
c.
Value: x=3 f(3) =
Hole discontinuity because no valueremovable
Consequences of Continuity:
A. INTERMEDIATE VALUE THEOREM
** Existence Theorem
EX: Verify the I.V.T. for f(c) Then find c.
on 2( )f x x 1,2 ( ) 3f c
If f(c) is between f(a) and f(b) there exists a c between a and b
ca b
f(a)
f(c)
f(b)
f(1) =1f(2) = 4Since 3 is between 1 and 4. There exists a c between 1 and 2 such that f(c) =3 x2=3 x=ยฑ1.732
Consequences: cont.
EX: Show that the function has a ZERO on the interval [0,1].3( ) 2 1f x x x
I.V.T - Zero Locator Corollary
CALCULUS AND THE CALCULATOR:
The calculator looks for a SIGN CHANGE between Left Bound and Right Bound
f(0) = -1f(1) = 2Since 0 is between -1 and 2 there exists a c between 0 and 1 such that f(c) = c
Intermediate Value Theorem
Consequences: cont.
EX: ( 1)( 2)( 4) 0x x x
I.V.T - Sign on an Interval - Corollary(Number Line Analysis)
EX:1 3
1 2x
Consequences of Continuity:
B. EXTREME VALUE THEOREM
On every closed interval there exists an absolute maximum value and minimum value.
x
y
x
y
Updates:
8/22/10
