Standards for Second Calculus Unit: CONTINUITY
Continuity as a property of functions
• An intuitive understanding of continuity. (The function values can be made as close as desired by
taking sufficiently close values of the domain.)
• Understanding continuity in terms of limits.
• Geometric understanding of graphs of continuous functions (Intermediate Value Theorem and Extreme
Value Theorem.)
Textbook: Section 2.3, Section 4.1 (EVT)
A continuous function is a function whose output varies continuously with the inputs and do not jump
from one value to another without taking on the values in between.
You have learned in previous classes that if you can sketch a function’s graph in one continuing motion
without lifting your pencil, it is a continuous function.
Officially, we define continuity at individual points.
DEFINITION:
A function f  x  is continuous at x = c if three conditions are met:
1. lim f  x   real number1 *
x c
2.
f  c   real number2
[the limit value exists at x = c]*
[the function value exists at x = c]
3. real number1  real number2
[the limit value and the function value are the same]
* For domain endpoints, remember that if the one-sided limit exists, the overall limit exists.
Another way to say it: A function f  x  is continuous at x = c if and only if lim f  x   f  c  .
x c
If a function f  x  is not continuous at a point x = c, we say that function is discontinuous at c and that
x = c is a point of discontinuity of f  x  .
YOU TRY:
(A) Can a finite set of points, such as  2,6 , 1,7  ,  4,6 , be continuous? Why or why not?
(B) Identify the domain of the function.
(C) At what point(s) is f  x  discontinuous? Why? [list x-values]
(D) At what point(s) is f  x  continuous [list x-values in interval
notation]?
Using the graph, explain whether the following points are continuous using the definition of continuity.
(E) x = -6
(F) x = -2
(G) x = 1
(H) x = 3
(I) x = 6
x = -6 in Example 3 is a JUMP discontinuity. Notice the left-hand and right-hand limits are both real
numbers but are not equal.
x = 1 and x = 6 in Example 3 are called REMOVABLE discontinuities. By changing only the function
value (y-value) at that point, you could make the function continuous at that point.
x = 3 in Example 3 is an INFINITE discontinuity. Notice the left-hand and/or right-hand limits have
infinity / nonexistent answers. Some books may call this an ESSENTIAL discontinuity.
YOU TRY: Find the point(s) of discontinuity, if any, on each graph.
x2  5x  6
(J)
x2  4
 x2 1
, x 1

(K) f  x    x  1
4,
x 1

(L) y 
x
x
A general statement: Polynomial, rational, trigonometric, exponential, and logarithmic functions
are always continuous over their entire domain. This is also called everywhere continuous. If the
function has domain all real numbers, then the function is continuous for all real numbers. If the
function has domain x > 0, then the function is continuous for all x > 0. If the function has
domain x  2,1 , then the function is continuous for all real numbers except x = -2 and x = 1.
YOU TRY: Determine the value of the variable a so the function is everywhere continuous.
 2 x  a , x  0
(M) g  x    2
 3x  1, x  0
2 x  3, x  2
(N) f  x   
 ax  1, x  2
The first important theorem associated with continuous functions is the Intermediate Value Theorem:
Preliminary condition for theorem to be in effect:
The function f  x  must be continuous over a specific closed interval [a, b].
Theorem states:
f  x  will take on all (y-) values between f  a  and f  b  . Or, if a specific y-value is between
f  a  and f  b  , I am guaranteed an x-value  between a and b such that f    that y-value.
Important note for using results of theorems on assessments and the A. P. Exam:
In order to use a Theorem, you must make a statement showing the preliminary conditions have
been met. Even if these conditions seem obvious, you are required to “activate” the theorem
before using it.
YOU TRY: Is any real number exactly 2 more than its cube?
(O) Write an equation to model this question.
(P) Set the equation equal to zero. Do not try to solve.
(Q) Is the Intermediate Value Theorem in effect for this equation? How do you know? Write a
statement similar to the preliminary condition above to “activate” the Int. Value Theorem.
(R) Find two x-values a and b such that the y-value 0 is between f  a  and f  b  .
(S) By the Intermediate Value Theorem, you have proven the existence of the number. Now use your
calculator to find the x-value (accurate to 3 decimal places).
YOU TRY: Prove that the number
20 must exist and has a value between 4 and 5.
Note that if x  20 , then x 2  20 . Since f  x   x2 is a polynomial and is continuous for all real
numbers, the Intermediate Value Theorem is in effect for any interval in its domain. Note that
f  4  16 and f  5  25 . By the Intermediate Value Theorem, there must exist some x-value  in the
interval [4, 5] such that f   20 .
Section 4.1
The second important theorem associated with continuous functions is the Extreme Value Theorem:
Preliminary condition for theorem to be in effect:
The function f  x  must be continuous over a specific closed interval [a, b].
Theorem states:
f  x  will have an absolute maximum (y-) value and an absolute minimum (y-) value on the interval [a,
b].
 It is possible that those extreme values might be repeated within that interval, but they will never
be exceeded.
 If the preliminary conditions above are not met, there may or may not be an absolute maximum
and may or may not be an absolute minimum – we are not guaranteed their existence by the
EVT, though.
YOU TRY: (T) The function
1
has an absolute minimum but no absolute maximum. Explain
4  x2
why this fact does not violate the Extreme Value Theorem.
(U) Does the function y  3x 2  12 x  5 on the interval [1, 4] have an absolute maximum and absolute
minimum? Explain how you know. If the value(s) exist, use your calculator to find the points where
they occur.
YOU TRY Answers:
(A) NO, no limit values
(B) [0, 4]
(D) [0, 1) U (1, 2) U (2, 4]
(E) NO, limit does not exist
(C) x = 1 (no limit value); x = 2 (function value  limit value)
(H) NO, fails everything (I) NO, (function value  limit value)
(L) x = 0
(M) a = -1
(N) a = 3
(F) YES
(G) NO, no function value
(J) x = 2, -2
(K) x = 1
(O) x  x3  2
(P) 0  x3  x  2
(Q) the polynomial is everywhere continuous for all real numbers, which means
the IVT can be used. (R) f(-3)=-22 and f(0)=2
(S) x = -1.5214
(T) domain is (-2, 2), not a
closed interval
(U) YES, polynomials are everywhere continuous for all real numbers, which
means it is also continuous for [1, 4]. Abs Min at (2, -7), Abs Max at (4, 5)