Continuity
2.4
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.
A function is continuous at a point if the limit is the same
as the value of the function.
This function has discontinuities
at x=1 and x=2.
2
1
1
2
3
4
It is still considered continuous
at x=0 and x=4, because the
one-sided limits match the value
of the function

Continuity
Continuity TEST (MEMORIZE!)
Show
g ( x)  x 2  1
is continuous at x = 1
1. g (1) exists. [ g (1)  2]
2. lim g ( x)exists. [lim g ( x)  2]
x 1
x 1
3. lim g ( x)  g (1)  2
x 1
 g ( x) is continuous at x  1
 x 1 x  2
Show the function f(x)  
continuous at x  2?
2x - 1 x  2
1. f (2) exists. [ f (2)  3]
2. lim f ( x)  (2)  1  3
x 2
lim f ( x)  2(2)  1  3
x 2
 lim f ( x) exists  3
x 2
3. lim f ( x)  f (2)  3
x2
 f ( x) is continuous at x  2
Key locations to test for continuity

Rational Expression


Piecewise Functions


Changes in interval
Absolute Value Functions


Values that make denominator = 0
Use piecewise definition and test changes in
interval
Step Functions

Test jumps at each step.
Types of Discontinuities

There are 3 types of discontinuities





Jump
Infinite (Asymptotic)
Point
The first two are considered non-removable.
Point discontinuity is often referred to as a
removable discontinuity since we can make the
function continuous if we alter our function at the
location of the hole.
Jump Discontinuity

Occurs when the graph breaks at a
particular point and starts somewhere else

Left hand limit does not equal Right hand limit
lim f ( x)  lim f ( x)
x c 
x c
Infinite (Asymptotic) Discontinuity

Occurs when curve has a vertical asymptote.

Continuity does not exist due to asymptote!
Point Discontinuity

Can occur when you have a rational expression
with common factors in the numerator and
denominator.

Can occur when the curve has a “hole” because
the function has a value that is off the curve at
that point.
This will result when
lim f ( x)  f (c)
x c
Identify the discontinuity on the graph and
state whether it is removable.
Find and identify the discontinuity.
x  3 x  2
f(x)   2
x2
x
1. f (2)  5
2. lim f ( x)  5
x2
lim f ( x)  4
x2
 lim f ( x)  DNE
x2
Jump discontinuity (non-removable)
Find and identify the discontinuity.
5
f ( x) 
x4
Infinite discontinuity (non-removable)
(VA at x = 4)
Find and identify the discontinuity.
x  8 x  15
f ( x)  2
x  6x  5
2

x  5x  3
f ( x) 
x  5x  1
2 Types of Discontinuity
Point discontinuity (removable) at x=5. [Hole at (5,.5)]
Infinite discontinuity (non-removable) at x=1. (VA)
Find and identify the discontinuity
5
f(x)   2
x
x2
x2
1. f (2)  5
2. lim f ( x)  4
x2
lim f ( x)  4
x2
3. lim f ( x)  f (2)
x 2
 lim f ( x)  4
x2
Point discontinuity (removable)
For what value(s) of a is the function
continuous at 4?
ax  5 x  4
f ( x)   2
x  x x  4
Test for continuity at 4
1. Does f(4) exist? f(4)  42  4  12
2. lim f ( x)  lim f ( x)  12
x4
x 4
So set ax + 5 = 12,
a(4)+5=12
This value automatically satisfies 3rd condition!
7
a
4
Determine the constant c,
so the function is continuous everywhere.
 cx  1, x  3
f ( x)   2
cx  1, x  3
2. lim f ( x)  lim f ( x)
x 3
x 3
c(3)  1  c(3) 2  1
3c  1  9c  1
2  6c
1
c
3
Test for continuity at 3!
Continuity : (Check )
1
1. f (3)  (3)  1  2
3
2. lim f ( x)  lim f ( x)
x 3
x 3
1
1 2
(3)  1  (3)  1
3
3
22
This ensures the 3rd condition is satisfied!
Intermediate Value
Theorem
2.4 cont.
Simple examples:

If between 7am and 2pm the temperature
went from 55 to 70.



At some time it reached 62.
Time is continuous
If between his 14th and 15th birthday, a
boy went from 150 to 165 lbs.


At some point he weighed 155lbs.
It may have occurred more than once.
Show that a “c” exists such that f(c)=2 for
f ( x)  x 2  2 x  3 in the interval [0, 2]
f(x) is continuous on the interval
f(0)= -3
f(2)= 5
By the IVT, there must be y-value of 2 since f is a
polynomial (continuous) and the y-values are in the
range from [-3,5].
Determine if f(x) has any real roots on
the stated interval: f ( x)  x  x  1 on [1,2]
2
f(x) is continuous on the interval
f (1)  1  2  -.414
f (2)  4  3  2.268
By the IVT, there must be a real root between x=1 and
x=2 where f(c)=0 because f is continuous and 0 is
between f(1) and f(2).
What you will write:
By the IVT, there exists a value (x=c) between [1,2]
where f(c)=0.