# f - jpiichsabcalculus

Section 1.4
Definition of Continuity
Continuity at a Point: A function f is continuous
at c if the following three conditions are met.
1. f (c) is defined.
2.
lim f  x  exists.
3.
lim f  x   f  c 
xc
xc
Continuity on an Open Interval: A function is
continuous on an open interval (a, b) if it is
continuous at each point in the interval. A function
that is continuous on the entire real line (−∞, ∞) is
continuous everywhere.
a
b
Discontinuity
Discontinuities fall into two categories:
removable and nonremovable.
A discontinuity at c is called removable if f can
be made continuous by appropriately defining
(or redefining) f (c). Otherwise it is considered
nonremovable.
Removable Discontinuities
To be a removable discontinuity there is a hole in
the graph.
lim f  x   f  c 
xc
Formally, a removable discontinuity is one at which
the limit of the function exists but does not equal the
value of the function at that point; this may be
because the function does not exist at that point.
Nonremovable Discontinuity
Jump Discontinuity
Infinite Discontinuity
The limit does not exist for all nonremovable
discontinuities.
Formally, it is a discontinuity for which the limits
from the left and right both exist but are not equal to
each other.
1. Jump Discontinuity is when the right-hand
limit and the left-hand limit are unequal.
2.
Infinite Discontinuity is when at least one of
the one-sided limits either does not exist or is
infinite.
Example 1
Discuss the continuity of each function.
2
1
x 1
a.
f  x 
b.
g  x 
x
x 1
c.
 x  1,
h x   2
 x  1,
x0
x0
a.
1
f  x 
x
The domain of f is all
non zero real numbers.
We can conclude that f
is continuous at every
value x in its domain.
However, at x = 0, f has
a nonremovable
discontinuity because
the limit does not exist
at x = 0.








b.
x2  1
g  x 
x 1
The domain of g is all
real numbers except 1.
The function is
continuous at every x
value in its domain.
It has a removable
discontinuity at x = 1
because there is a hole
at x = 1.
4
2
-4
-2
2
-2
-4
4
c.
 x  1,
h x   2
 x  1,
The domain for h is all
real numbers. The
function h is continuous
on (−∞, 0) and (0, ∞),
and because
lim h  x   h  0   1
x0
h is continuous on the
entire real line.
x0
x0








Example 2
Find the value of c which makes g(x) continuous
for all x.
cx  1, x  3
g  x   2
cx  1, x  3
*Type of question on the Free Response
Questions on the AP Exam.
lim g  x   g  3
cx  1, x  3
g  x   2
cx  1, x  3
x3
g  3  3c  1
lim g  x   3c  1
x3
lim+ g  x   9c  1
x3
lim g  x   lim+ g  x 
x3
x3
3c  1  9c  1
1
c
3
1
 3 x  1, x  3
g  x  
 1 x 2  1, x  3
 3




   
Remember to always find the
one-sided limits for the x – value
in question. You may also want
graphing it.









Definition of Continuity on a Closed
Interval
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 f is continuous from the right at a
and continuous from the left at b.
lim+ f  x   f  a 
x a
lim f  x   f  b 
xb
(b, f(b))
(a, f(a))
f(x)
The following types of functions are continuous at
every point in their domain.
1. Polynomial functions
2. Rational functions
4. Trigonometric functions
Intermediate Value Theorem
If f is continuous on the closed interval [a, b],
f (a) ≠ f (b), and k is any number between f (a)
and f (b), then there is at least one number c in
[a, b] such that f (c) = k.
This is an existence theorem and will not provide
a solution. It just tells us of the existence of a
solution.
find c.
It just guarantees the existence of at least one
number c in the closed interval [a, b].
If f is not continuous on [a, b], it may not exhibit
the intermediate value property.
k
Example 3
The function is continuous for −2 ≤ x ≤ 1. If
f (−2) = −5 and f (1) = 4, decide if the following
statements are true by the Intermediate Value
Theorem?
A.
There exists c, where −2 ≤ x ≤ 1, such that
f (c) ≥ f (x) for all x on the closed interval
−2 ≤ x ≤ 1.

True

y



x










B.
There exists c, −2 < c < 1, such that f (c) = 0.
True. Since f is continuous and f (−2) = −5 and
f (1) = 4, then by the IVT there must be a c
such that f (c) = 0

y




x










C.
There exists c, where −2 ≤ x ≤ 1, such that
f (c) = 3.
True. Since f is continuous, by the IVT there
must be a c such that f (c) = 3. 
y




x










Example 4
Use Intermediate Value Theorem to show that
there is a root of f (x) = x3 + 2x – 1 [0, 1] on the
given interval.
f (x) = x3 + 2x – 1 [0, 1]
Since f is a polynomial, we know that the function is
continuous over the real number line.
f (0) = −1 and f (1) = 1 + 2 – 1 = 2
Since f (0) < 0 and f (1) > 0, using the Intermediate
Value Theorem you can conclude that there must be
a c in the closed interval such that f (c) = 0.
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