2.3 Continuity

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Continuity
2.3
Quick Algebra Review
How can you tell whether a graph has a vertical
asymptote or a hole?
x2
f ( x)  2
x  3x  10
f ( x) 
 x  2
( x  5)( x  2)
There is a vertical aysmptote when x  5.
There is a hole when x  2. (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.
Continuous
Not continuous

A function is continuous at a point if the limit is the same
as the value of the function.
lim f ( x) must exist and be equal to f (c)
xc
Is the function continuous at each value
of x? Explain your answer.
x 1
x2
x3
2
1
1
2
3
4
x0
x4
The function has one-sided continuity at x=0.

Is the function continuous at x = -3?
 x  19  4
, x  3

f ( x)   x  3
1
, x  3
 8
lim f ( x) must equal f ( 3)
x 3
Types of Discontinuity
(removable discontinuity)
Jump Discontinuity is typically caused by a
piece-wise function whose pieces don’t meet.
The limit does not exist because the left-hand
limit does not equal the right-hand limit.
Example:
 x  3,
f ( x)  
 x  1,
x0
x0
Find the real number, c that makes the function everywhere
continuous.
 x  2,
f ( x)   2
 x  c,
x3
x3
y  x2  8
(3,1)
y  x2
Point Discontinuity occurs
when a function has a hole.
The limit exists but it does not
equal the function value.
Example:
x 2  11x  28
f ( x) 
x4
x 2  11x  28  lim ( x  4)( x  7)  lim( x  7)  3
lim
x4
x 4
x 4
x4
x4
There is a hole at (-4,3).
Point Discontinuity is also called Removable Discontinuity.
To remove the discontinuity, just “fill in the hole”.
Make the last problem everywhere continuous.
x 2  11x  28
f ( x) 
x4
There is a hole at (-4,3).
 x 2  11x  28
, x  4

f ( x)  
x4

x  4
3,
Infinite/Essential Discontinuity occurs
when there is a vertical asymptote.
The limit does not exist
because it is either  or  .
Example:
3x
f ( x) 
x2
Example:
Give all x-values for which the function is discontinuous,
and classify each instance of discontinuity.
9 x 2  3x  2
(3x  1)(3 x  2)
f ( x)  2

3x  13x  4
(3x  1)( x  4)
There is an infinite discontinuity
(vertical asymptote) at:
There is a removable discontinuity
(hole) at: (
)
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  cos x

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

Use the Intermediate Value Theorem to explain
2
why the function f ( x)  x  3x  6 must have a
root (x-intercept) on the closed interval [1,2].
f (1) 
f (2) 
If f (1)  2 and f (2)  4, then by the
Intermediate Value Thm, the function
must take on all y values between -2 and 4,
so a y value of 0 must exist and it will
happen when x is between 1 and 2.
You can use your calculator to find a better
approximation of the root. (1.372)
Example 5:
Is any real number exactly one less than its cube?
(Note that this doesn’t ask what the number is, only if it exists.)
f 1  1
x  x3  1
0  x3  x  1
Guess and check.
f  2  5
Since f is a continuous function, by the
intermediate value theorem it must
take on every value between -1 and 5.
Therefore there must be at least one
solution between 1 and 2.
To find a better approximation, use your
calculator to find the zeros.
f  x   x3  x 1
1.32472

pg 80 #11-28,35-41 (no calc)
Wkst: Continuity
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