2.3 Continuity and Intermediate
Value Theorem
Continuity
• A function is continuous if you can draw the
graph without picking up your pencil.
Definition:
A function y = f(x) is continuous at an interior point
c in its domain if
lim f ( x)  f (c)
x c
(the limit has to equal the value of the function at c)
Continuity (at endpoints)
• A function y = f(x) is continuous at a left
endpoint a or a right endpoint b in its domain if
lim f ( x)  f (a ) or lim f ( x)  f (b)
xa
x b
If a function is not continuous anywhere on its graph,
we say there is a discontinuity at that point.
Continuity - Graphically
Continuous at a?
lim f ( x)  f (a )
Yes
xa
Continuous at b?
lim f ( x)  f (b)
a
c
b
x b
Continuous at c?
lim f ( x)  f (c)
x c
Yes
Yes
Discontinuities
• Is
1
f ( x) 
continuous for all real numbers?
x
1
lim  f (0)
x 0 x
This is an example of a
nonremovable
discontinuity at x = 0.
(vertical asymptote)
Discontinuities
• Is f(x) = [x] continuous for all real numbers?
lim x   1
x 0
lim x   0
x 0
Since the limit does not
exist, we can say the
function is discontinuous.
This is another example
of a nonremovable
discontinuity at x = 0.
(jump)
Discontinuities
1
• Is the function f ( x )  sin
continuous for
x
all real numbers?
1
lim sin  sin( )
x 0
x
1
lim sin  sin( )
x 0
x
Sin constantly oscillates, so we cannot say that
these two limits are the same. Thus, the limit
as x approaches 0 does not exist and there is a
discontinuity here.
Another nonremovable discontinuity.
(infinite oscillation)
Discontinuities
x2 1
• Is f ( x) 
continuous everywhere?
x 1
x2 1
(1, 2)
lim
x 1
x 1
 f (1)
because f(1) does not exist.
This is an example of a
removable
discontinuity at x = 1.
(hole)
Removing a Discontinuity
(hole)
2
x 1
• How can we make f ( x) 
continuous?
x 1
The only place in question is at x = 1.
x 2  1 ( x  1)( x  1)
f ( x) 

x 1
x 1
 x 1
lim ( x  1)  2
x 1
When you can cancel out like
factors in the top and bottom,
this means there is a hole at
where the denominator was
equal to 0.
To be continuous, the limit as x
approaches 1 has to be the value
of the function at 1. Therefore,
to be continuous, f(1) = 2.
Removing a Discontinuity
(hole)
2
x 1
• How can we make f ( x) 
continuous?
x 1
We can now turn the original function into a piecewise so
that f(1) not only exists, but is equal to the limit as x
approaches 1, which was 2.
 x2 1
, x 1

f ( x)   x  1

2,
x 1
Properties of Continuous Functions
• If two functions are continuous, then
▫
▫
▫
▫
The sum of the functions is also continuous.
The difference of the functions is also continuous
The product of the functions is also continuous.
The result of multiplying one of the functions by a
constant is also continuous.
▫ The quotient of the functions is also continuous
provided the denominator ≠ 0
Theorem Involving Composites
• If f(x) = │x │ and g(x) = cos x,
what is f(g(x))?
f(g(x)) = │cos x│
If f and g are both continuous functions,
the composites of two continuous
functions is also always continuous.
Big Theorem #1:
Intermediate Value Theorem
• A function f(x) that is continuous on the interval
[a, b] takes on every value between f(a) and f(b).
f(b)
f(a)
a
b
Intermediate Value Theorem
2
x 5
• Show using the IVT that f ( x) 
has a
x 1
root between x = 2 and x = 3.
f(x) is discontinuous only at x = –1, so on the interval [2, 3]
the function is continuous.
Since f is cont. on
[2, 3] and f(2) and
2
2
(2)  5
(
3
)
 5 f(3) have opposite
f (2) 
f (3) 
2 1
3  1 signs, there is a
value c in the
45
95
Interval where


3
4
f(c) = 0 by the
Intermediate Value
1

1

Theorem.
3