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Continuity of a Function for AP Calculus

By — McGraw-Hill Professional
Updated on Oct 24, 2011

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:

  1. 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

  1. If the functions f and g are continuous at a, then the functions f + g, fg, f · g and f/g, g(a) ≠ 0, are also continuous at a.
  2. A polynomial function is continuous everywhere.
  3. A rational function is continuous everywhere, except at points where the denominator is zero.
  4. 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.)

Theorems on Continuity

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.)

Theorems on Continuity

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) = 62 – 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.

Theorems on Continuity

Example 5

If g (x) = x2 – 2x – 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.)

Theorems on Continuity

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

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