Practice problems for this concept can be found at Limits and Continuity Practice Problems for AP Calculus.

### Continuity of a Function at a Number

A function *f* is said to be continuous at *a* number *a* if the following three conditions are satisfied:

*f(a)*exists

The function *f* is said to be discontinuous at *a* if one or more of these three conditions are not satisfied and *a* is called the point of discontinuity.

### Continuity of a Function over an Interval

A function is continuous over an interval if it is continuous at every point in the interval.

### Theorems on Continuity

- If the functions
*f*and*g*are continuous at*a*, then the functions*f*+*g*,*f*–*g*,*f*·*g*and*f*/*g*,*g(a)*≠ 0, are also continuous at*a*. - A polynomial function is continuous everywhere.
- A rational function is continuous everywhere, except at points where the denominator is zero.
- Intermediate Value Theorem: If a function f is continuous on a closed interval [
*a, b*] and*k*is a number with*f (a) ≤ k ≤ f (b)*, then there exists a number*c*in [*a, b*] such that*f (c )=k*.

#### Example 1

Find the points of discontinuity of the function

Since *f (x)* is a rational function, it is continuous everywhere, except at points where the denominator is 0. Factor the denominator and set it equal to 0: (*x* – 2)(*x* +1)=0. Thus *x* =2 or *x* = – 1. The function *f (x)* is undefined at *x* = –1 and at *x* = 2. Therefore, *f (x)* is discontinuous at these points. Verify your result with a calculator. (See Figure 5.3-1.)

#### Example 2

Determine the intervals on which the given function is continuous:

Check the three conditions of continuity at *x* = 2:

Condition 1: *f*(2) = 10.

Condition 2:

Condition 3: . Thus, *f (x)* is discontinuous at *x* = 2.

The function is continuous on (–∞, 2) and (2, ∞). Verify your result with a calculator. (See Figure 5.3-2.)

#### Example 3

For what value of *k* is the function continuous at *x* = 6?

For *f (x)* to be continuous at *x* = 6, it must satisfy the three conditions of continuity.

Condition 1: *f*(6) = 6^{2} – 2(6) = 24.

Condition 2: must also be 24 in order for the to equal 24. Thus, which implies 2(6) + *k* =24 and *k* =12. Therefore, if *k* = 12,

Condition 3: is also satisfied.

#### Example 4

Given *f (x)* as shown in Figure 5.3-3, *(a)* find *f* (3) and *f (x)*, and *(b)* determine if *f (x)* is continuous at *x* = 3? Explain your answer.

The graph of *f (x)* shows that *f* (3) = 5 and the *f (x)* = 1. Since *f* (3) ≠ *f (x), f (x)* is discontinuous at *x* =3.

#### Example 5

If *g (x)* = *x*^{2} – 2*x* – 15, using the Intermediate Value Theorem show that *g (x)* has a root in the interval [1, 7].

Begin by finding *g*(1) and *g*(7), and *g*(1)= – 16 and *g*(7) = 20. If *g (x)* has a root, then *g (x)* crosses the *x*-axis, i.e., *g (x)* = 0. Since – 16 ≤ 0 ≤ 20, by the Intermediate Value Theorem, there exists at least one number *c* in [1, 7] such that *g (c)* = 0. The number *c* is a root of *g (x)*.

#### Example 6

A function *f* is continuous on [0, 5], and some of the values of *f* are shown below.

If *f (x)* = – 2 has no solution on [0, 5], then *b* could be

(A) 3 (B) 1 (C) 0 (D) –2 (E) –5

If *b* = – 2, then *x* = 3 would be a solution for *f (x)* = – 2.

If *b* = 0, 1, or 3, *f (x)* = – 2 would have two solutions for *f (x)*= – 2.

Thus, *b* = – 5, choice (E). (See Figure 5.3-4.)

Practice problems for this concept can be found at Limits and Continuity Practice Problems for AP Calculus.

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