Function Continuity

Grade 12 · algebra · 98 practice problems · read aloud

🔊 Listen to this explanation

Function Continuity: The Unbroken Path

A function is continuous at a point if you can draw its graph at that point without lifting your pencil. It's a fundamental concept in calculus because it's a prerequisite for differentiability and is essential for understanding limits and the behavior of functions.

🔍 The 3-Part Continuity Test

A function f(x) is continuous at x = c if and only if:

  1. f(c) is defined: The point exists on the graph.
  2. limx→c f(x) exists: The limit from the left and right are equal.
  3. limx→c f(x) = f(c): The limit value equals the function's value.

If any condition fails, the function is discontinuous at x = c.

📚 Worked Examples

Example 1: Polynomial

Is f(x) = x² + 3 continuous at x = 1?

  1. f(1) is defined: f(1) = (1)² + 3 = 4 ✅
  2. Limit exists: limx→1 (x² + 3) = 4 ✅
  3. Limit equals value: 4 = 4 ✅

Conclusion: f(x) is continuous at x = 1.

Example 2: Piecewise Function

Is g(x) continuous at x = 2?
g(x) = { x + 1 for x < 2; 5 for x = 2; x² - 1 for x > 2 }

  1. g(2) is defined: g(2) = 5 ✅
  2. Limit exists?
    Left-hand limit: limx→2⁻ = 2 + 1 = 3
    Right-hand limit: limx→2⁺ = (2)² - 1 = 3
    Limits are equal, so overall limit = 3 ✅
  3. Limit equals value? 3 ≠ 5 ❌

Conclusion: Discontinuous at x = 2 (a "removable" discontinuity).

🚨 Common Mistakes

  • Assuming all functions are continuous: Always check the three conditions, especially with piecewise functions.
  • Only checking f(c): A defined point doesn't guarantee continuity. The limit must also exist and match.
  • Confusing "hole" vs. "jump": A hole (removable discontinuity) fails condition 3. A jump fails condition 2.

💡 Tips & Tricks

  • Pencil Test: Visualize tracing the graph. If you must jump or lift your pencil, it's discontinuous.
  • Polynomials are friends: All polynomials are continuous everywhere. Rational functions are continuous except where the denominator is zero.
  • Check the "seams": For piecewise functions, always check the x-values where the formula changes.

🎯 Practice Suggestions

To master continuity:

  1. Start with simple polynomials and rational functions.
  2. Practice with piecewise functions, identifying the type of discontinuity.
  3. Use graphing software (like Desmos) to plot functions and visually confirm your analytical conclusions.
  4. Create your own piecewise functions with specific discontinuities (hole, jump, infinite).

Practice problems

6 of the 98, worked through step by step — try them before opening the answer.

1 ∫₀¹ (3x² + 2x) dx = ?

Hint: To evaluate a definite integral, find the antiderivative of the function and then apply the Fundamental Theorem of Calculus.

Show the answer

Answer: 2

  1. Find the antiderivative of 3x² + 2x. The antiderivative of 3x² is (3/3)x³ = x³. The antiderivative of 2x is (2/2)x² = x². So the antiderivative F(x) = x³ + x².
  2. Apply the Fundamental Theorem of Calculus: F(1) - F(0). F(1) = (1)³ + (1)² = 1 + 1 = 2. F(0) = (0)³ + (0)² = 0 + 0 = 0.
  3. Calculate the definite integral: F(1) - F(0) = 2 - 0 = 2.

The answer is 2.

2 Is the function continuous at x=2? (check graph for break)

  1. A) no
  2. B) yes

Hint: Examine the graph carefully at x=2 to see if there is a gap, jump, or hole in the function

Show the answer

Answer: A) no

  1. Look at the graph at x=2
  2. Observe that there is a jump discontinuity at x=2 where the function value suddenly changes
  3. The left-hand limit and right-hand limit at x=2 are not equal
  4. Since the limits don't match, the function is not continuous at x=2

The answer is No.

3 Given the function f(x) = (x² - 4)/(x - 2) for x ≠ 2 and f(2) = 4. Is f continuous at x = 2?

  1. A) yes
  2. B) no

Hint: Check if the limit as x approaches 2 equals the function value at x = 2

Show the answer

Answer: A) yes

  1. Simplify the function for x ≠ 2: f(x) = (x² - 4)/(x - 2) = (x - 2)(x + 2)/(x - 2) = x + 2
  2. Find the limit as x approaches 2: lim(x→2) f(x) = lim(x→2) (x + 2) = 2 + 2 = 4
  3. Check the function value at x = 2: f(2) = 4
  4. Compare: lim(x→2) f(x) = f(2) = 4 Since the limit equals the function value, f is continuous at x = 2.

4 Given the piecewise function: f(x) = { x^2 - 9 for x < 3; 2x + 3 for x ≥ 3 }. Is f(x) continuous at x = 3?

  1. A) yes
  2. B) no

Hint: Check if the function value, left-hand limit, and right-hand limit all exist and are equal at the point in question.

Show the answer

Answer: B) no

  1. Check if f(3) exists. f(3) = 2(3) + 3 = 6 + 3 = 9.
  2. Find the left-hand limit as x approaches 3. lim (x→3⁻) f(x) = lim (x→3⁻) (x^2 - 9) = (3)^2 - 9 = 9 - 9 = 0.
  3. Find the right-hand limit as x approaches 3. lim (x→3⁺) f(x) = lim (x→3⁺) (2x + 3) = 2(3) + 3 = 6 + 3 = 9.
  4. Compare the values. f(3) = 9, left-hand limit = 0, right-hand limit = 9. Since the left-hand limit (0) is not equal to the function value (9), the function is not continuous at x = 3.

The answer is No.

5 Given the graph of Sophia's function f(x), determine if f(x) is continuous at x=7, x=11, and x=15. Classify any discontinuities.

Hint: Check if the function value equals the limit from both sides at each point. Look for breaks, jumps, or holes in the graph.

Show the answer

Answer: Continuous at x=7, jump discontinuity at x=11, removable discontinuity at x=15

  1. At x=7, the graph shows f(7) exists, the limit from the left equals f(7), and the limit from the right equals f(7). Therefore, f(x) is continuous at x=7.
  2. At x=11, the graph shows f(11) exists, but the limit from the left is 8 and the limit from the right is 3. Since the left and right limits are not equal, there is a jump discontinuity at x=11.
  3. At x=15, the graph shows the limit from both sides equals 9, but f(15) is defined as 5. Since the limit exists but does not equal the function value, there is a removable discontinuity at x=15.

The answer is: Continuous at x=7, jump discontinuity at x=11, removable discontinuity at x=15.

6 Given the graph of Aroha's function f(x), identify all x-values where f(x) is discontinuous and classify each discontinuity type.

Hint: Look for points where the graph has breaks, holes, jumps, or vertical asymptotes. A function is discontinuous at points where there is no single, connected y-value.

Show the answer

Answer: x = -3 (jump), x = 1 (removable), x = 5 (infinite)

  1. Examine the graph at x = -3. The left-hand limit approaches 7, while the right-hand limit approaches 3. Since the left and right limits exist but are not equal, this is a jump discontinuity.
  2. Examine the graph at x = 1. There is a hole at the point (1, 4), but the function is defined at f(1) = 2. Since the limit exists (approaches 4) but does not equal the function value (2), this is a removable discontinuity.
  3. Examine the graph at x = 5. The function approaches positive infinity from both sides, creating a vertical asymptote. Since the limit does not exist (approaches infinity), this is an infinite discontinuity.
  4. The function is continuous at all other x-values shown on the graph. The discontinuities are: x = -3 (jump), x = 1 (removable), x = 5 (infinite).
Practise this topic — 10 free problems, no signup →