Section 1.4

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 exists.

x cf x

3. lim

x c

f x f 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

x c

f x f 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

The limit does not exist for all nonremovable discontinuities.

Infinite Discontinuity

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.

1a. f x

x

2 1b.

1

x

g xx

2

1, 0c.

1, 0

x xh x

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.

1a. f x

x

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.

2 1

b.1

x

g xx

-4 -2 2 4

-4

-2

2

4

The domain for h is all real numbers. The function h is continuous on (−∞, 0) and (0, ∞), and because

h is continuous on the entire real line.

0

lim 0 1

x

h x h

2

1, 0c.

1, 0

x xh x

x x

Example 2

Find the value of c which makes g(x) continuous for all x.

*Type of question on the Free Response Questions on the AP Exam.

2

1, 3

1, 3

cx xg x

cx x

3

lim 3

x

g x g

3

lim

x

g x

3 3 1 g c

2

1, 3

1, 3

cx xg x

cx x

3 1c

3

lim

x

g x+

9 1c

3 3

lim lim

x x

g x g x+

3 1 9 1 c c1

3c

2

11, 3

31

1, 33

x xg x

x x

Remember to always find the one-sided limits for the x – value in question. You may also want to check your solution by 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

The function f is continuous from the right at a and continuous from the left at b.

lim and lim .

x a x b

f x f a f x f b

(a, f(a))

(b, f(b))

lim

x a

f x f a+

f(x)

lim

x b

f x f b

The following types of functions are continuous at every point in their domain.

1. Polynomial functions

2. Rational functions

3. Radical functions

4. Trigonometric functions

Intermediate Value TheoremIf 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.

The Intermediate Value Theorem does not help you 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

x

y

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

x

y

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.

x

y

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.