1/27/2013
Math 103 – Rimmer
2.5 Continuity
A function is continuous at the point x = a if
lim f ( x ) = lim+ f ( x ) = f ( a )
x →a−
x→a
this says that the limit
must equal the function
this says that the limit exists
value at x = a
Types of discontinuities
removable
discontinuity
jump
discontinuity
Math 103 – Rimmer
2.5 Continuity
infinite
discontinuity
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1/27/2013
Math 103 – Rimmer
2.5 Continuity
One - Sided Continuity
A function f is continuous from the left at a if
A function f is continuous from the right at a if
lim f ( x ) = f ( a )
lim f ( x ) = f ( a )
x →a−
x →a+
f ( x)
exercise # 3
f ( x ) is discontinuous at:
x = −4
x = −2
x=2
x=4
f ( x ) is continuous from the left at: x = −2
f ( x ) is continuous from the right at: x = 2
x=4
Continuous on an interval
Math 103 – Rimmer
2.5 Continuity
A function f is continuous on an interval if its continuous at every number in the interval.
f ( x)
exercise # 3
State the intervals on which f is continuous.
( −∞, −4 ) ( −4, −2]
( −2, 2 )
this is used to say that the
function is continuous from
the left at x = −2.
[ 2, 4 )
[ 4, ∞ )
this is used to say that the
function is continuous from
the right at x = 2.
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1/27/2013
Math 103 – Rimmer
2.5 Continuity
Combining continuous functions
Let k be a constant. If f and g are continuous functions at x = a,
then the following functions are also continuous at x = a :
1. f + g
f
provided g ( a ) ≠ 0
g
4.
2. f − g
3. fg
5. kf
If g is continuous at x = a and f is continuous at g ( a ) , then
f ( g ( x ) ) is contiuous at x = a.
Math 103 – Rimmer
2.5 Continuity
The following functions are continuous at every number in their domain:
1. polynomial functions
5. exponential functions
2. rational functions
6. logarithmic functions
3. root functions
4. trigonometric functions
7. inverse trigonometric functions
Find where the following functions are continuous:
f ( x) =
x
x2 + 5x + 6
g ( x ) = 4x2 − 9
x2 + 5x + 6 = 0
( x + 2 )( x + 3) = 0
All real numbers
except x = −2, −3
4 x2 − 9 ≥ 0
9
3
x ≥
⇒ x≥
4
2
2
3 3 

 −∞, −  ∪  , ∞ 
2 2 

h ( x ) = tan ( 2 x )
Domain of tan ( x ) :
All real numbers except
π
2
+ nπ , n any integer
Domain of tan ( 2 x ) :
All real numbers except
π
4
+n
π
2
, n any integer
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Find the values of a and b that make f continuous everywhere.
Math 103 – Rimmer
2.5 Continuity
 1 + x2
if
x<2

f ( x ) =  x + a + b if 2 ≤ x < 4

x≥4
 x + 5a if
f ( x ) is continuous away from x = 2 and 4, because the individual parts
of the function are continuous there. We just need to check x = 2 and x = 4
lim f ( x ) = lim− (1 + x 2 ) = 5
x → 2−
x→2
lim f ( x ) = lim+ ( x + a + b ) = 2 + a + b
x→ 2+
x→2
⇒ 5 = 2+a +b
⇒3= a+b
lim f ( x ) = lim− ( x + a + b ) = 4 + a + b
x→ 4−
x→4
lim f ( x ) = lim+ = x + 5a = 4 + 5a
x→ 4+
x→4
⇒ 4 + a
+ b = 4 + 5a
3
⇒ 7 = 4 + 5a
49 = 4 + 5a
45 = 5a
a =9
a=9
b = −6
b = 3 − 9 = −6
The Intermediate Value Theorem
Math 103 – Rimmer
2.5 Continuity
Suppose f ( x ) is continuous on [ a, b ] .
Let f ( a ) < N < f ( b ) with f ( a ) ≠ f ( b ) .
Then there exists a number c in ( a, b ) such that f ( c ) = N .
There could be more than one such c.
The Intermediate Value Theorem can be used to locate roots of equations.
f ( a ) < 0 and f ( b ) > 0 ( or vice versa ) ,let N = 0
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1/27/2013
Math 103 – Rimmer
2.5 Continuity
f (b)
f (a)
c
a
a
b
b
f (b)
f (a)
f is continuous on [ a, b ]
f (a) > 0
f (a) < 0
or vice versa
f (b) < 0
f (b ) > 0
With these conditions, the Intermediate Value Theorem guarantees that there
is at least one value c in the interval ( a, b ) that is a root of the function f i.e. f ( c ) = 0.
Math 103 – Rimmer
2.5 Continuity
Use the Intermediate Value Theorem to show that there
is a root of equation in the given interval.
2 x3 + x 2 + 2 = 0,
( −2, −1)
3
2
3
2
f ( −2 ) = 2 ( −2 ) + ( −2 ) + 2 = −16 + 4 + 2 = −10
f ( −1) = 2 ( −1) + ( −1) + 2 =
−2 + 1 + 2 = 1
f ( −2 ) < 0
f ( −1) > 0
By the Intermediate Value Theorem, there exists c in ( −2, −1) with f ( c ) = 0.
2 sin x = 3 − 2 x,
( 0, π )
2 sin x − 3 + 2 x = 0
f ( 0 ) = 2 sin ( 0 ) − 3 + 2 ( 0 ) = 0 − 3 + 0 = −3
f ( π ) = 2 sin (π ) − 3 + 2 (π ) = 0 − 3 + 2π
2π > 3 so
f ( 0) < 0
f (π ) > 0
By the Intermediate Value Theorem, there exists c in ( 0, π ) with f ( c ) = 0.
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