ap calculus 1004 continuity (2.3). c conversation: voice level 0. no talking! h help: raise your...
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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 “Local linearity means…”
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 lim𝑥→𝑎
𝑓 (𝑥 )𝑒𝑥𝑖𝑠𝑡𝑠
Limit = value lim𝑥→𝑎
𝑓 (𝑥 )= 𝑓 (𝑎)
Limits Review:
PART 1: LOCAL BEHAVIOR (1). General Idea: Behavior of a function very near the point where
(2). Layman’s 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 ≠ 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 limitlim𝑥→𝑎−
𝑓 (𝑥 )=¿¿lim
𝑥→𝑎+¿ 𝑓 (𝑥 )=¿ ¿¿¿
Holes and hiccups are removableJumps and Vertical Asymptotes are essential
Test f(a) = lim𝑥→𝑎
𝑓 (𝑥) lim𝑥→𝑎
𝑓 (𝑥)f(a) =
00h𝑜𝑙𝑒
Vertical Asymptote
Lim DNEJump
= cont.≠ 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
=
x≠ 4
lim𝑥→ 4
𝑓 (𝑥 )=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?
x≠3
lim𝑥→ 3
1(𝑥−3 )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: Limitlim𝑥→1−
3+𝑥=4
lim𝑥→1+¿ 3−𝑥=2¿
¿
lim𝑥→ 1
𝑓 (𝑥 )=𝐷𝑁𝐸 2≠ 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 ≠ 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:
lim𝑥→ 2−
𝑓 (𝑥 )=8
lim𝑥→2+¿ 𝑓 (𝑥 )=8¿
¿
lim𝑥→ 2
𝑓 (𝑥 )=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: lim𝑥→ 1−𝑓 (𝑥 )=1−𝑥 2=0
lim𝑥→ 1+¿ 𝑓 (𝑥 )=𝑥 2−2=− 1¿
¿
lim𝑥→ 1
𝑓 (𝑥 )=𝐷𝑁𝐸
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=±1.732
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