Download AP Calculus

1004
Continuity
(2.3)
AP Calculus
Activity: Teacher-Directed Instruction
CONVERSATION: Voice level 0. No talking!
C
HELP: Raise your hand and wait to be called on.
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ACTIVITY: Whole class instruction; students in seats.
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M
P
S
MOVEMENT: Remain in seat during instruction.
PARTICIPATION: 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.
Content:
SWBAT calculate limits of any functions and apply
properties of continuity
Language:
SW complete the sentence “Local linearity means…”
General Idea:
General Idea: ________________________________________
Can you draw without picking up your pencil
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)
Has a point
f(a) exists
b)
Has a limit
lim 𝑓 𝑥 𝑒𝑥𝑖𝑠𝑡𝑠
c)
Limit = value
𝑥→𝑎
lim 𝑓 𝑥 = 𝑓(𝑎)
𝑥→𝑎
Limits Review:
PART 1: LOCAL BEHAVIOR
(1). General Idea: Behavior of a function very near the point where
xa
(2). Layman’s Description of Limit (Local Behavior)
L
a
must write every time
(3). Notation: lim 𝒇(𝒙)
𝒙→𝒂
 x  a
Continuity Theorems
Interior Point: A function y  f  x  is continuous at an interior point c of its
domain if
lim
f x  f c 
xc  
ONE-SIDED CONTINUITY
Endpoint: A function y  f  x  is continuous at a left endpoint a of its domain
if lim f  x   f  a 
xa
or
continuous at a right endpoint b if
lim f  x   f  b  .
x 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
a)
b)
No value
f(a) DNE
hole
No limit
lim 𝑓 𝑥 𝐷𝑁𝐸
𝑥→𝑎
c)
Vertical
asymptote
Limit does not equal
value
Limit ≠ value
jump
Discontinuity: cont.
Method:
0
ℎ𝑜𝑙𝑒
0
1
0
(a). Test the value =
Vertical
Asymptote
Lim DNE
Jump
= cont.
≠ hiccup
(b). Test the limit
Look for f(a) =
lim 𝑓 𝑥 =
lim+ 𝑓 𝑥 =
𝑥→𝑎−
𝑥→𝑎
lim 𝑓(𝑥)
(c). Test f(a) = 𝑥→𝑎
Removable or
f(a) =
Essential Discontinuities
Holes and hiccups are removable
Jumps and Vertical Asymptotes are essential
lim 𝑓(𝑥)
𝑥→𝑎
Examples:
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?
EX:
0
x 2
=
f ( x) 
0
x4
x≠ 4
0
lim 𝑓 𝑥 =
𝑥→4
0
Hole discontinuous because f(x) has no value
It is removable
removable or essential?
Examples: cont.
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?
1
f ( x) 
( x  3) 2
x≠3
lim
𝑥→3
1
𝑥−3
=
2
1
0
VA discontinuous because no value
It is essential
removable or essential?
Examples: cont.
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?
3  x, x  1
f ( x)  
3  x, x  1
Step 1: Value must look at 4 equation
f(1) = 4
Step 2: Limit
lim 3 + 𝑥 = 4
𝑥→1−
lim 3 − 𝑥 = 2
𝑥→1+
lim 𝑓 𝑥 = 𝐷𝑁𝐸 2 ≠ 4
𝑥→1
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
Hole discont.
No value
removable
x = -2
VA discont. Because
no value no limit
essential
x=0
Hiccup discont.
Because limit ≠
value
removable
x =1
x=2
x=3
Continuous limit
= value
VA discont. No
limit
essential
Jump discont.
Because limit
DNE
essential
Algebraic Method
 3x  2 x  2
f ( x)   2
3x  4 x  2
Look at function with equal
a. Value: f(2) = 8
lim 𝑓 𝑥 = 8
b. Limit:
𝑥→2−
lim 𝑓 𝑥 = 8
𝑥→2+
lim 𝑓 𝑥 = 8
𝑥→2
c. Limit = value: 8=8
Limit = Value ∴ 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑠
Algebraic Method

 1- x 2
x 1
 2
f ( x)   x - 2 1  x  3
 x2  9

x3
 x  3
At x=1
a. Value: f(1) = -1
lim 𝑓
b. Limit: 𝑥→1−
𝑥 = 1 − 𝑥2 = 0
At x=3
0
a. Value: x=3 f(3) =
0
b.
lim 𝑓 𝑥 = 𝑥2 − 2 = −1
𝑥→1+
lim 𝑓 𝑥 = 𝐷𝑁𝐸
𝑥→1
c.
Jump discontinuity because limit DNE
essential
Hole discontinuity
c. because no value
removable
Consequences of Continuity:
A. INTERMEDIATE VALUE THEOREM
f(b)
If f(c) is between f(a) and
f(b) there exists a c
between a and b
f(c)
f(a)
** Existence Theorem
a
EX: Verify the I.V.T. for f(c)
f ( x)  x 2
on
c
b
Then find c.
1, 2
f (c )  3
f(1) =1
f(2) = 4
Since 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.
I.V.T - Zero Locator Corollary
Intermediate Value Theorem
EX: Show that the function has a ZERO on the interval [0,1].
f ( x)  x 3  2 x  1
f(0) = -1
f(1) = 2
Since 0 is between -1 and 2 there exists a c between
0 and 1 such that f(c) = c
CALCULUS AND THE CALCULATOR:
The calculator looks for a SIGN CHANGE between Left Bound and
Right Bound
Consequences: cont.
I.V.T - Sign on an Interval - Corollary
(Number Line Analysis)
EX:
EX:
( x  1)( x  2)( x  4)  0
1
3

x 1
2
Consequences of Continuity:
B. EXTREME VALUE THEOREM
On every closed interval there exists an absolute
maximum value and minimum value.
y
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Updates:
8/22/10