MATH 12002 - CALCULUS I 1.5: Continuity Professor Donald L. White - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §1.5: Continuity

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 12

Definition of Continuity

Intuitively, a function is continuous at a point if the graph can be drawn through the point without lifting your pencil; that is, there are no holes or jumps in the graph at that point. This is not precise enough to accurately define continuity, however. We will use the following definition:

Definition

A function y = f (x) is continuous at x = a if f (a) is defined and lim

x→a f (x) = f (a).

A consequence of this definition is that when x is near a, a “small” change in x results in a “small” change in y, and therefore no break or jump in the graph is possible.

D.L. White (Kent State University) 2 / 12

Definition of Continuity

We will use the following checklist to determine if a given function y = f (x) is continuous at a given point x = a:

1 Is f (a) defined? 2 Does lim

x→a f (x) exist?

3 Does lim

x→a f (x) = f (a)?

If the answer to each of these is YES, then f is continuous at x = a. If the answer to any one of these is NO, then f is not continuous at x = a.

D.L. White (Kent State University) 3 / 12

Examples

Example

Determine whether the function f (x) = x2 − 4 x − 2 is continuous at x = 2 and justify your answer.

Solution

Notice that when x = 2, the denominator of the function is 0. Therefore f (2) is not defined, and so f is NOT continuous at x = 2. Note: Even though f (x) = (x−2)(x+2)

x−2

, we cannot “cancel” the x − 2 factors in this situation. The function f (x) is undefined at x = 2 (and has a hole in its graph at x = 2), but the function g(x) = x + 2 has g(2) = 4. Cancelling changes the function; f (x) = g(x).

D.L. White (Kent State University) 4 / 12

Examples

Example

Determine whether the function f (x) =        x2 − 4 x − 2 if x = 2 3 if x = 2 is continuous at x = 2 and justify your answer.

D.L. White (Kent State University) 5 / 12

Examples

Solution

1 By the definition of the function, f (2) = 3, so f is defined at x = 2. 2 We have

lim

x→2 f (x)

= lim

x→2

x2 − 4 x − 2 , since x = 2, = lim

x→2

(x − 2)(x + 2) x − 2 = lim

x→2(x + 2), since x = 2,

= 2 + 2 = 4, and so lim

x→2 f (x) exists.

3 Since f (2) = 3 = 4 = lim

x→2 f (x), we have lim x→2 f (x) = f (2), and so f is

NOT continuous at x = 2.

D.L. White (Kent State University) 6 / 12

Examples

Example

Determine if f (x) = x2 + 3x + 1 if x < 1 2x4 + 6x2 − 3 if x 1 is continuous at x = 1 and justify your answer.

D.L. White (Kent State University) 7 / 12

Examples

Solution

1 We have f (1) = 2(14) + 6(12) − 3 = 2 + 6 − 3 = 5, and so f (1) is

defined.

2 To determine if lim

x→1 f (x) exists, we compute the one-sided limits.

lim

x→1− f (x) = lim x→1−(x2 + 3x + 1) = 12 + 3 · 1 + 1 = 5

and lim

x→1+ f (x) = lim x→1+(2x4 + 6x2 − 3) = 2 · 14 + 6 · 12 − 3 = 5.

Since lim

x→1− f (x) = lim x→1+ f (x), lim x→1 f (x) does exist.

3 Since lim

x→1− f (x) = lim x→1+ f (x) = 5, we have lim x→1 f (x) = 5 = f (1), and

so f is continuous at x = 1.

D.L. White (Kent State University) 8 / 12

Examples

Example

Determine if g(x) =    x2 + 3x + 1 if x < 1 7 if x = 1 2x4 + 6x2 − 3 if x > 1 is continuous at x = 1 and justify your answer.

D.L. White (Kent State University) 9 / 12

Examples

Solution

1 By the definition of the function, we have g(1) = 7, and so g(1) is

defined.

2 To determine if lim

x→1 g(x) exists, we compute the one-sided limits.

lim

x→1− g(x) = lim x→1−(x2 + 3x + 1) = 12 + 3 · 1 + 1 = 5

and lim

x→1+ g(x) = lim x→1+(2x4 + 6x2 − 3) = 2 · 14 + 6 · 12 − 3 = 5.

Since lim

x→1− g(x) = lim x→1+ g(x), lim x→1 g(x) does exist.

3 Since lim

x→1− g(x) = lim x→1+ g(x) = 5, we have lim x→1 g(x) = 5 = 7 = g(1),

and so g is NOT continuous at x = 1.

D.L. White (Kent State University) 10 / 12

Polynomials and Rational Functions

Continuity for polynomials and rational functions is very easy to

• determine. From §1.4, we have

Direct Substitution Property

If f is a polynomial or rational function and a is in the domain of f , then lim

x→a f (x) = f (a).

Notice that this condition is precisely the definition of continuity at the point x = a, and so we have

Theorem

If f is a polynomial or rational function and a is in the domain of f , then f is continuous at x = a. This means that if f is a polynomial, then f is continuous at x = a for every real number a, and if f is a rational function, then f is continuous at x = a if and only if the denominator of f is not 0.

D.L. White (Kent State University) 11 / 12

Polynomials and Rational Functions

Example

Determine the values of x at which the function f (x) = x2 + x − 6 x2 + 2x − 3 is not continuous.

Solution

Since f is a rational function, it is discontinuous precisely at the values of x for which the denominator is 0. The denominator is x2 + 2x − 3 = (x + 3)(x − 1), and so is 0 when x = −3 or x = 1. Therefore, f is discontinuous at x = −3 and at x = 1 and continuous at every other real number.

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