Continuity Practice Problems with Answers and Solutions

These continuity practice problems cover the three-part definition, piecewise functions, finding a constant that makes a function continuous, classifying discontinuities, domains of continuity and the Intermediate Value Theorem. Try each one first. The answers come right after the list, then the full solutions.

A function f is continuous at x = a when all three hold:

For more on the idea, read the continuity guide.

Problems

Is it continuous?

1. Is f(x) = \dfrac{x^2 - 1}{x - 1} continuous at x = 1? If not, how could you fix it?
2. Is f(x) = \begin{cases} x^2 + 1 & x < 2 \\ 3x - 1 & x \ge 2 \end{cases} continuous at x = 2?
3. Is f(x) = \begin{cases} 2x + 1 & x \le 1 \\ x^2 + 4 & x > 1 \end{cases} continuous at x = 1?
4. Is f(x) = \begin{cases} \dfrac{\sin x}{x} & x \ne 0 \\ 1 & x = 0 \end{cases} continuous at x = 0?

Make it continuous

5. Find k so that f(x) = \begin{cases} kx + 3 & x < 2 \\ x^2 - k & x \ge 2 \end{cases} is continuous everywhere.
6. Find a and b so that f(x) = \begin{cases} 3x + a & x < 1 \\ x^2 + bx & 1 \le x \le 2 \\ 10 & x > 2 \end{cases} is continuous everywhere.

Types of discontinuity and domains

7. Find and classify the discontinuities of f(x) = \dfrac{x - 2}{x^2 - 5x + 6}.
8. Where is f(x) = \dfrac{\ln(x - 4)}{x - 6} continuous?
9. Where is f(x) = \sqrt{9 - x^2} continuous?

Intermediate Value Theorem

10. Show that x^3 + x - 1 = 0 has a solution between 0 and 1.
11. Show that \cos x = x has a solution in \left(0, \frac{\pi}{2}\right).
12. f(x) = \frac{1}{x} has f(-1) = -1 and f(1) = 1. Does the IVT say f has a zero in [-1, 1]?

Answers

Step-by-step solutions

1. At x = 1 the bottom is 0, so f(1) is undefined and f is not continuous there. The limit does exist:

This is a removable discontinuity, a hole at (1, 2). Defining f(1) = 2 fills the hole and makes f continuous.

2. Check all three conditions:

• Left limit: 2^2 + 1 = 5.
• Right limit: 3(2) - 1 = 5.
• Value: f(2) = 3(2) - 1 = 5.

All three agree, so f is continuous at x = 2.

3. The left limit is 2(1) + 1 = 3 and the right limit is 1^2 + 4 = 5. The one-sided limits differ, so the limit does not exist. This is a jump discontinuity.

4. The value is f(0) = 1, and the special limit gives \lim_{x\to 0}\frac{\sin x}{x} = 1. They match, so f is continuous at 0.

5. Each piece is a polynomial, so only x = 2 needs checking. Set the left limit equal to the right limit:

6. There are two break points, so you get two equations.

At x = 2: the middle piece gives 4 + 2b and the right piece gives 10, so b = 3.

At x = 1: the left piece gives 3 + a and the middle piece gives 1 + b = 4, so a = 1.

7. Factor the bottom: x^2 - 5x + 6 = (x - 2)(x - 3), so f is undefined at x = 2 and x = 3.

At x = 2 the limit is \frac{1}{2 - 3} = -1, so that is a removable discontinuity. At x = 3 the simplified form still has a zero bottom, so f\to\pm\infty: an infinite discontinuity with a vertical asymptote.

8. The log needs x - 4 > 0, so x > 4. The bottom needs x \ne 6. Combining these, f is continuous on (4, 6)\cup(6, \infty).

9. The root needs 9 - x^2 \ge 0, so -3 \le x \le 3. The function is continuous on the closed interval [-3, 3], using one-sided continuity at the endpoints.

10. Let f(x) = x^3 + x - 1. It is a polynomial, so it is continuous on [0, 1]. Since f(0) = -1 and f(1) = 1, the value 0 lies between them. By the Intermediate Value Theorem there is a c in (0, 1) with f(c) = 0.

11. Rewrite as g(x) = \cos x - x = 0. The function g is continuous everywhere. Check the endpoints:

The sign changes, so by the IVT there is a c in \left(0, \frac{\pi}{2}\right) with \cos c = c.

12. No. The IVT needs f to be continuous on the whole closed interval, and \frac{1}{x} is not defined at 0. The theorem says nothing here, and in fact \frac{1}{x} is never 0.

Common mistakes on these problems

• Checking only the limit. In problem 1 the limit exists, but the function is still discontinuous because f(1) is missing.
• Using the wrong piece at the break point. In problem 2, f(2) comes from the piece with x \ge 2. Read the inequality signs carefully.
• Checking only one break point. Problem 6 has two break points, and each one gives its own equation.
• Applying the IVT without continuity. Problem 12 shows that a sign change alone proves nothing.

Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new functions and instant feedback, try the continuity lesson and the Intermediate Value Theorem lesson. Continuity is also the first hypothesis of the Mean Value Theorem, which you can practice next.