Solved Exercises on Holes and Discontinuities in Algebra 2

Master holes and discontinuities: removable discontinuities, jump discontinuities, infinite discontinuities, and their identification in rational functions through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Removable Discontinuity
Exercise 1
Identify and classify the discontinuities in \(f(x) = \frac{x^2-4}{x-2}\). Simplify the function and find the coordinates of any holes.
Definition:

Removable discontinuity: A point where a function is undefined due to a common factor in numerator and denominator that can be canceled.

Identification method:
  1. Factor both numerator and denominator completely
  2. Identify common factors
  3. Cancel common factors
  4. Determine hole location using original form
  5. Classify the type of discontinuity
Factor numerator
\(x^2-4 = (x+2)(x-2)\)
Common factor
\(x-2\)
Step 1: Factor the numerator

\(x^2 - 4 = (x+2)(x-2)\) (difference of squares)

So \(f(x) = \frac{(x+2)(x-2)}{x-2}\)

Step 2: Identify common factors

The factor \((x-2)\) appears in both numerator and denominator

This creates a removable discontinuity (hole) at \(x = 2\)

Step 3: Simplify the function

For \(x \neq 2\): \(f(x) = \frac{(x+2)(x-2)}{x-2} = x + 2\)

The simplified form is \(f(x) = x + 2\), with a hole at \(x = 2\)

Step 4: Find the y-coordinate of the hole

Using the simplified form: when \(x = 2\), \(y = 2 + 2 = 4\)

So the hole is located at the point \((2, 4)\)

Step 5: Classify the discontinuity

This is a removable discontinuity because the function can be made continuous by defining \(f(2) = 4\)

Removable discontinuity at (2, 4), simplified: f(x) = x + 2 for x ≠ 2
Final answer:

There is a removable discontinuity (hole) at the point \((2, 4)\).

The simplified function is \(f(x) = x + 2\) for all \(x \neq 2\).

The function approaches the value 4 as x approaches 2, but is undefined at \(x = 2\).

Applied rules:

• Common factors: Create removable discontinuities, not vertical asymptotes

• Factor cancellation: Only cancel common factors to simplify

• Hole location: Use simplified function to find y-value at x-value of discontinuity

2 Multiple Discontinuities
Exercise 2
Analyze the function \(g(x) = \frac{(x-1)(x+3)}{(x-1)(x-4)}\). Identify all discontinuities and classify each as removable or non-removable.
Definition:

Non-removable discontinuity: A discontinuity that cannot be eliminated by redefining the function at a single point, such as vertical asymptotes.

Common factor
\(x-1\)
Non-cancelable factor
\(x-4\)
Step 1: Identify all factors

Numerator: \((x-1)(x+3)\)

Denominator: \((x-1)(x-4)\)

Step 2: Identify common factors

The factor \((x-1)\) appears in both numerator and denominator

This creates a removable discontinuity at \(x = 1\)

Step 3: Identify non-cancelable factors

The factor \((x-4)\) only appears in the denominator

This creates a non-removable discontinuity (vertical asymptote) at \(x = 4\)

Step 4: Simplify the function

For \(x \neq 1\): \(g(x) = \frac{(x-1)(x+3)}{(x-1)(x-4)} = \frac{x+3}{x-4}\)

After cancellation: \(g(x) = \frac{x+3}{x-4}\) for \(x \neq 1, 4\)

Step 5: Find the coordinates of the hole

Using the simplified form: when \(x = 1\), \(y = \frac{1+3}{1-4} = \frac{4}{-3} = -\frac{4}{3}\)

So the hole is at \((1, -\frac{4}{3})\)

Removable discontinuity at (1, -4/3), non-removable at x = 4 (VA)
Final answer:

Removable discontinuity (hole) at the point \((1, -\frac{4}{3})\).

Non-removable discontinuity (vertical asymptote) at \(x = 4\).

The simplified function is \(g(x) = \frac{x+3}{x-4}\) for \(x \neq 1, 4\).

Applied rules:

• Common factors: Create holes (removable discontinuities)

• Denominator-only factors: Create vertical asymptotes (non-removable)

• Classification: Factor analysis determines discontinuity type

3 Higher-Degree Polynomials
Exercise 3
Determine all discontinuities in \(h(x) = \frac{x^3-8}{x^2-4}\). Classify each and find coordinates of any holes.
Definition:

Higher-degree polynomial factoring: Apply special factoring patterns like difference of cubes and difference of squares.

Factor numerator
\(x^3-8 = (x-2)(x^2+2x+4)\)
Factor denominator
\(x^2-4 = (x-2)(x+2)\)
Step 1: Factor the numerator using difference of cubes

\(x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4)\)

The quadratic factor \(x^2 + 2x + 4\) cannot be factored further (discriminant is negative)

Step 2: Factor the denominator using difference of squares

\(x^2 - 4 = x^2 - 2^2 = (x-2)(x+2)\)

Step 3: Rewrite with factored forms

\(h(x) = \frac{(x-2)(x^2 + 2x + 4)}{(x-2)(x+2)}\)

Step 4: Identify common factors

The factor \((x-2)\) appears in both numerator and denominator

This creates a removable discontinuity at \(x = 2\)

Step 5: Identify non-cancelable factors

The factor \((x+2)\) only appears in the denominator

This creates a non-removable discontinuity (vertical asymptote) at \(x = -2\)

Step 6: Simplify the function

For \(x \neq 2\): \(h(x) = \frac{(x-2)(x^2 + 2x + 4)}{(x-2)(x+2)} = \frac{x^2 + 2x + 4}{x+2}\)

Step 7: Find the coordinates of the hole

Using the simplified form: when \(x = 2\), \(y = \frac{2^2 + 2(2) + 4}{2+2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3\)

So the hole is at \((2, 3)\)

Removable discontinuity at (2, 3), non-removable at x = -2 (VA)
Final answer:

Removable discontinuity (hole) at the point \((2, 3)\).

Non-removable discontinuity (vertical asymptote) at \(x = -2\).

The simplified function is \(h(x) = \frac{x^2 + 2x + 4}{x+2}\) for \(x \neq 2, -2\).

Applied rules:

• Difference of cubes: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

• Difference of squares: \(a^2 - b^2 = (a-b)(a+b)\)

• Factor analysis: Determines discontinuity type and location

Key Concepts: Definitions, Rules, and Methods
\(f(x) = \frac{P(x)}{Q(x)}, \text{ where } P(x), Q(x) \text{ are polynomials}\)
Rational Function Definition
Key definitions:

Discontinuity: A point where a function is not continuous.

Removable Discontinuity: A hole in the graph where a common factor cancels out.

Non-removable Discontinuity: A discontinuity that cannot be eliminated by redefining the function.

Vertical Asymptote: A vertical line where the function approaches infinity.

Hole: A single point where the function is undefined but could be defined to make it continuous.

Discontinuity identification:
  1. Factor completely: Factor both numerator and denominator
  2. Identify common factors: These create holes
  3. Identify denominator-only factors: These create vertical asymptotes
  4. Simplify: Cancel common factors
  5. Find hole coordinates: Use simplified form
  6. Classify: Determine type of each discontinuity
Tip 1: Always factor completely before analyzing for discontinuities.
Tip 2: Common factors create holes, denominator-only factors create vertical asymptotes.
Tip 3: To find hole coordinates, substitute the x-value into the simplified function.
Tip 4: A function can have multiple discontinuities of different types.
Common errors: Confusing holes with vertical asymptotes, forgetting to factor completely, incorrect hole coordinates.
Exam preparation: Practice factoring patterns, memorize identification rules, sketch graphs to visualize.
Essential rules to remember:

• Common factors: Create removable discontinuities (holes)

• Denominator-only factors: Create non-removable discontinuities (vertical asymptotes)

• Hole coordinates: Use simplified function to find y-value

• Factor analysis: Critical for proper classification

• Continuity: A function is continuous if it has no breaks, jumps, or holes

\(\text{If } f(x) = \frac{(x-a)P(x)}{(x-a)Q(x)}, \text{ then hole at } x=a\)
Hole Identification
\(\text{If } f(x) = \frac{P(x)}{(x-a)Q(x)}, \text{ then VA at } x=a\)
Vertical Asymptote Identification
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
Difference of Cubes
Solution: Exercises 4 to 5
4 Complex Factoring Pattern
Exercise 4
Analyze the function \(p(x) = \frac{x^4-1}{x^2-1}\) for all discontinuities. Classify each and find coordinates of any holes.
Definition:

Repeated factoring: Some expressions require multiple rounds of factoring, such as difference of squares applied to higher-degree polynomials.

Factor numerator
\(x^4-1 = (x^2-1)(x^2+1)\)
Further factor
\(x^4-1 = (x-1)(x+1)(x^2+1)\)
Step 1: Factor the numerator using difference of squares

\(x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)\)

Notice that \(x^2 - 1\) is also a difference of squares: \(x^2 - 1 = (x-1)(x+1)\)

So \(x^4 - 1 = (x-1)(x+1)(x^2 + 1)\)

Step 2: Factor the denominator

\(x^2 - 1 = (x-1)(x+1)\)

Step 3: Rewrite with factored forms

\(p(x) = \frac{(x-1)(x+1)(x^2 + 1)}{(x-1)(x+1)}\)

Step 4: Identify common factors

Both \((x-1)\) and \((x+1)\) appear in both numerator and denominator

This creates removable discontinuities (holes) at \(x = 1\) and \(x = -1\)

Step 5: Simplify the function

For \(x \neq \pm 1\): \(p(x) = \frac{(x-1)(x+1)(x^2 + 1)}{(x-1)(x+1)} = x^2 + 1\)

Step 6: Find the coordinates of the holes

Using the simplified form \(p(x) = x^2 + 1\):

At \(x = 1\): \(y = 1^2 + 1 = 2\), so hole at \((1, 2)\)

At \(x = -1\): \(y = (-1)^2 + 1 = 2\), so hole at \((-1, 2)\)

Step 7: Check for non-removable discontinuities

After canceling all common factors, there are no remaining denominator-only factors

Therefore, there are no vertical asymptotes

Removable discontinuities at (-1, 2) and (1, 2), no non-removable discontinuities
Final answer:

Removable discontinuities (holes) at the points \((-1, 2)\) and \((1, 2)\).

No non-removable discontinuities (no vertical asymptotes).

The simplified function is \(p(x) = x^2 + 1\) for \(x \neq \pm 1\).

Applied rules:

• Repeated factoring: Apply factoring patterns multiple times if possible

• Multiple common factors: Each creates a separate hole

• Complete simplification: Factor completely to identify all discontinuities

5 Piecewise-Defined Discontinuities
Exercise 5
For the function \(q(x) = \frac{x^3-2x^2-5x+6}{x^2-4x+3}\), identify all discontinuities, classify each, and determine if the function can be extended to be continuous at any points.
Definition:

Rational function extension: A function can be extended to be continuous at a removable discontinuity by defining the function value to be the limit at that point.

Factor numerator
\(x^3-2x^2-5x+6 = (x-1)(x+2)(x-3)\)
Factor denominator
\(x^2-4x+3 = (x-1)(x-3)\)
Step 1: Factor the denominator

\(x^2 - 4x + 3\): Find two numbers that multiply to 3 and add to -4

Those numbers are -1 and -3: \(x^2 - 4x + 3 = (x-1)(x-3)\)

Step 2: Factor the numerator using rational root theorem

Try potential roots: ±1, ±2, ±3, ±6

Testing \(x = 1\): \(1^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0\)

So \((x-1)\) is a factor. Using polynomial division: \(x^3-2x^2-5x+6 = (x-1)(x^2-x-6)\)

Factor further: \(x^2-x-6 = (x-3)(x+2)\)

So \(x^3-2x^2-5x+6 = (x-1)(x-3)(x+2)\)

Step 3: Rewrite with factored forms

\(q(x) = \frac{(x-1)(x-3)(x+2)}{(x-1)(x-3)}\)

Step 4: Identify common factors

Both \((x-1)\) and \((x-3)\) appear in both numerator and denominator

This creates removable discontinuities at \(x = 1\) and \(x = 3\)

Step 5: Simplify the function

For \(x \neq 1, 3\): \(q(x) = \frac{(x-1)(x-3)(x+2)}{(x-1)(x-3)} = x + 2\)

Step 6: Find the coordinates of the holes

Using the simplified form \(q(x) = x + 2\):

At \(x = 1\): \(y = 1 + 2 = 3\), so hole at \((1, 3)\)

At \(x = 3\): \(y = 3 + 2 = 5\), so hole at \((3, 5)\)

Step 7: Determine function extension

Yes, the function can be extended to be continuous at both \(x = 1\) and \(x = 3\)

Define: \(q(1) = 3\) and \(q(3) = 5\) to make the function continuous

Removable discontinuities at (1, 3) and (3, 5), function extends to q(x) = x + 2
Final answer:

Removable discontinuities (holes) at the points \((1, 3)\) and \((3, 5)\).

No non-removable discontinuities (no vertical asymptotes).

The function can be extended to be continuous by defining \(q(1) = 3\) and \(q(3) = 5\).

The extended function would be \(q(x) = x + 2\) for all real numbers.

Applied rules:

• Rational root theorem: For factoring higher-degree polynomials

• Function extension: Define values at removable discontinuities to achieve continuity

• Polynomial division: Useful for factoring when a root is known

Holes and Discontinuities: Complete Guide
\(f(x) = \frac{P(x)}{Q(x)}, \text{ where } P(x), Q(x) \text{ are polynomials}\)
Rational Function Definition
Key definitions:

Discontinuity: A point where a function is not continuous.

Removable Discontinuity: A hole in the graph where a common factor cancels out, allowing the function to be made continuous by redefining the function at that point.

Non-removable Discontinuity: A discontinuity that cannot be eliminated by redefining the function, such as vertical asymptotes.

Vertical Asymptote: A vertical line where the function approaches positive or negative infinity.

Hole: A single point where the function is undefined but could be defined to make it continuous.

Complete identification methodology:
  1. Factor completely: Factor both numerator and denominator into irreducible polynomials
  2. Identify common factors: These create removable discontinuities (holes)
  3. Identify denominator-only factors: These create non-removable discontinuities (vertical asymptotes)
  4. Simplify: Cancel all common factors to get simplified form
  5. Find hole coordinates: Substitute x-values of holes into simplified function
  6. Classify each discontinuity: Determine if removable or non-removable
  7. Check for function extension: See if removable discontinuities can be filled
Tip 1: Always factor completely before analyzing for discontinuities.
Tip 2: Common factors create holes, denominator-only factors create vertical asymptotes.
Tip 3: To find hole coordinates, substitute the x-value into the simplified function.
Tip 4: A function can have multiple discontinuities of different types.
Common errors: Confusing holes with vertical asymptotes, forgetting to factor completely, incorrect hole coordinates.
Exam preparation: Practice factoring patterns, memorize identification rules, sketch graphs to visualize.
Essential formulas and rules:

• Common factors: \(\frac{(x-a)P(x)}{(x-a)Q(x)}\) creates a hole at \(x = a\)

• Denominator-only factors: \(\frac{P(x)}{(x-a)Q(x)}\) creates a vertical asymptote at \(x = a\)

• Difference of squares: \(a^2 - b^2 = (a-b)(a+b)\)

• Difference of cubes: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

• Sum of cubes: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\)

• Perfect square trinomial: \(a^2 \pm 2ab + b^2 = (a \pm b)^2\)

\(\text{If } f(x) = \frac{(x-a)P(x)}{(x-a)Q(x)}, \text{ then hole at } x=a\)
Hole Identification
\(\text{If } f(x) = \frac{P(x)}{(x-a)Q(x)}, \text{ then VA at } x=a\)
Vertical Asymptote Identification
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
Difference of Cubes
Types of discontinuities:

Removable: Can be "fixed" by redefining the function at a single point

Jump: The function approaches different values from left and right

Infinite: The function approaches infinity (vertical asymptote)

Oscillating: The function oscillates infinitely as it approaches the point

Function extension technique:
  1. Identify removable discontinuities: Points where function is undefined but limit exists
  2. Calculate the limit: Find what the function approaches at the discontinuity
  3. Define the function: Assign the limit value to make function continuous
  4. Verify: Check that the extended function is now continuous
Key note: The function can be extended at removable discontinuities but not at vertical asymptotes.
Key note: The simplified form of a rational function represents the original function except at points of discontinuity.

Questions & Answers

Question: I'm confused about how to tell the difference between a hole and a vertical asymptote. They both seem to make the function undefined.

Answer: The key difference lies in the factors:

  • Hole (Removable Discontinuity): The same factor appears in both the numerator and denominator
  • Vertical Asymptote (Non-removable): The factor appears only in the denominator

Examples:

For \(f(x) = \frac{(x-2)(x+1)}{x-2}\): The factor \((x-2)\) appears in both numerator and denominator → hole at \(x = 2\)

For \(g(x) = \frac{x+1}{x-2}\): The factor \((x-2)\) appears only in denominator → vertical asymptote at \(x = 2\)

For \(h(x) = \frac{(x-2)(x+1)}{(x-2)(x-3)}\): \((x-2)\) appears in both → hole at \(x = 2\), \((x-3)\) only in denominator → vertical asymptote at \(x = 3\)

Always factor completely first to identify all common factors.

Question: How do I find the y-coordinate of a hole? Do I substitute into the original function or the simplified version?

Answer: You must substitute the x-value of the hole into the simplified function, not the original function.

The original function is undefined at the x-value of the hole, so substituting there would give you an undefined result.

Example: For \(f(x) = \frac{x^2-4}{x-2}\)

  • Factor: \(f(x) = \frac{(x+2)(x-2)}{x-2}\)
  • Simplify: \(f(x) = x+2\) for \(x \neq 2\)
  • Hole at \(x = 2\)
  • Y-coordinate: Substitute \(x = 2\) into simplified form: \(y = 2 + 2 = 4\)
  • Hole is at \((2, 4)\)

If you tried to substitute \(x = 2\) into the original function, you'd get \(\frac{0}{0}\), which is undefined.

Question: What does it mean for a function to be "extended" to be continuous? When is this possible?

Answer: Function extension means defining the function at a point where it was previously undefined, in such a way that the new function is continuous at that point.

When is it possible? Extension to continuity is possible only at removable discontinuities (holes), where the limit of the function exists at the point of discontinuity.

How does it work? If \(f(x)\) has a hole at \(x = a\), and \(\lim_{x \to a} f(x) = L\), then we can define a new function:

\(g(x) = \begin{cases} f(x) & \text{if } x \neq a \\ L & \text{if } x = a \end{cases}\)

Then \(g(x)\) is continuous at \(x = a\).

Example: For \(f(x) = \frac{x^2-4}{x-2}\), there's a hole at \(x = 2\). Since \(\lim_{x \to 2} f(x) = 4\), we can define:

\(g(x) = \begin{cases} \frac{x^2-4}{x-2} & \text{if } x \neq 2 \\ 4 & \text{if } x = 2 \end{cases}\)

Now \(g(x)\) is continuous everywhere.