Topic 5 - Continuity

advertisement
1.4 Continuity
Calculus
Section 1
Understanding
Continuity
2
Most of the techniques of calculus
require that functions be continuous.
A function is continuous if you can draw it in
one motion without picking up your pencil.
Definition of Continuity
A function is continuous at a
point c if the following
three conditions are met.
1.
2.
3.
f(c) is defined
lim f ( x ) exists
xc
lim f ( x )  f ( c )
x c
Definition of Continuity
A function is continuous on
an open interval (a, b) if it is continuous
on each point in the interval. A function
that is continuous on the entire real line
is everywhere continuous.
f(x) is continuous on (-3,2)
One Sided Limits and Continuity
A function f is continuous on the closed interval [a,b] if
it is continuous on the open interval (a, b) and
lim f ( x )  f ( a ) and lim f ( x )  f (b )
xa
x b
The function is continuous from the right at a and
continuous from the left at b.
f(x) is continuous on [-3,2]
Example 1: Continuity of a
Function
Discuss the continuity of each function.
a.
b.
c.
d.
x2 1
f ( x) 
x 1
1
g ( x) 
x
 x  1, x  2
h( x )   2
 x  1, x  2
 x  1, x  0
i( x)   2
 x  1, x  0
7
Example 1a Solution
There is a removable
discontinuity at x = 1
because f(1) is not
defined (Definition of
Continuity Condition
1).
x2 1
f ( x) 
x 1
8
Example 1b Solution
g ( x) 
1
x
The function has a nonremovable
discontinuity at x = 0 because
lim g ( x)
x 0
does not exist
(Definition of Continuity Condition
2).
9
Example 1c Solution
The limit from the right of x = 2 does not
equal the limit from the left. Therefore,
the limit as x approaches 2 does not
exist.
Function has a discontinuity at x = 2
because lim g ( x) does not exist
x 0
(Definition of Continuity Condition 2).
 x  1, x  2
h( x )   2
 x  1, x  2
10
Example 1d Solution
The function is
continuous on the
entire real line.
 x  1, x  0
i ( x)   2
 x  1, x  0
11
Try These
Discuss the continuity of each function.
x2  4
a. f ( x) 
x2
1
b. g ( x) 
x3
 x  2, x  5
c. h( x)  
3 x  8, x  5
 x  4, x  0
d . i ( x)   2
 x  1, x  0
12
Removable Discontinuities:
(You can fill the hole.)
Nonremovable Discontinuities:
jump
infinite
oscillating
“Discussing Continuity”


Continuous or discontinuous?
If discontinuous



Removable or nonremovable discontinuity?
At what x-value is the discontinuity?
What condition is not met? (See slide 4 for
conditions)



Condition 1: f(c) is defined - means the function exists when
x =c
Condition 2: limit as x approaches c of f(x) existis – means
that the “two sides of the functions meet at c,” no “jumps, or
asymptotes
Condition 3: Limit and value are equal – means the is no
“hole” with a “dot” filled in elsewhere
14
Continuity by Function Type



Polynomials are everywhere continuous
Sine and Cosine are everywhere continuous
Rational functions and other trig functions are continuous except at xvalues where their denominators equal zero.




“Removable” discontinuity if factoring and canceling “removes” the zero in
the denominator
“Non-removable” otherwise. (Recall that vertical asymptotes occur where
numerator is nonzero and the denominator is zero.)
Root functions are continuous, except at x-values that would result in
a negative value under an even root
For piecewise functions, find the f(x) values at the x-value separating
the regions of the function.


If the f(x) values are equal, the function is continuous.
Otherwise, there is a (non-removable) discontinuity at this point.
15
Properties of Continuity
Continuous functions can be added, subtracted,
multiplied, divided and multiplied by a
constant, and the new function remains
continuous.
Also: Composites of continuous functions are continuous.
Examples:
y  sin  x 2 
y = 3x + cos(x)
Section 2
Finding the Intervals on
Which a Function is
Continuous
17
Examle 2: Describe the interval(s) on
which f is continuous.
f ( x)  x x  3
Solution: The function is continuous on the interval [-3,  )
18
Try This: Find the interval on which f is
continuous.
x2
2 x,
f ( x)   2
 x  4 x  1, x  2
Solution: f(x) has a non – removable discontinuity at x = 2.
The intervals on which f is continuous are (-  , 2] and [2,  )
19
Section 3
The Intermediate Value
Theorem
20
Suppose this pool holds 22,000 gal. of water. Then
there is some point in time, when it was being filled
that it held exactly 15,000 gal. It could not have
skipped over that number.
Intermediate Value Theorem
If a function is continuous between a and b, then it takes on
every value between f  a  and f  b  .
f b
Because the function is
continuous, it must take on
every y value between f(a) and
f(b)
f a
a
b
Here is a formal statement of the
Theorem:
23
Example 3: Intermediate Value Theorem
Without graphing, show that f(x) = 2x4 -3x2 + 5x +2
has at least one zero between -2 and -1.
Solution: This function is continuous on the interval
[-2, -1] (and everywhere else for that matter), and
f(-2) =12
f(-1) = -4
Therefore somewhere between
x = -2, and x = -1, f(x) passes through 0.
24
Try This
For the function f(x) = x2 -6x + 8 ,
a) use the Intermediate Value Theorem to
show that the function has a zero on the
interval [0, 3]
b) find the value of "c" guaranteed by the
theorem such that f(c) = 0.
25
Solution
Verify that the Intermediate Value theorem applies on [0, 3] and find the value of "c"
guaranteed by the theorem.
f(c) = 0
f(x) = x2 -6x + 8
f(0) = 8
f(3) = -1
0 is between -1 and 8.
f(c) = 0
c2 – 6c + 8 = 0
c = 2 or 4
Not 4, why?
26
Download
Related flashcards
Cybernetics

25 Cards

Create flashcards