Solved Exercises on Domain from Graphs and Equations

Master finding domain from graphs and equations: polynomial, rational, radical, and piecewise functions through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Domain from Graphs
Exercise 1
From the graph of function f(x), determine the domain. The graph shows a continuous curve from x = -4 to x = 6 with an open circle at (-2, 1) and a closed circle at (4, 3).
Definition:

Domain from graph: All x-values where the function is defined (horizontal extent)

Open circle: Point is not included in the domain

Closed circle: Point is included in the domain

Method for finding domain from graphs:
  1. Identify the horizontal extent of the graph (from leftmost to rightmost point)
  2. Look for open/closed circles that indicate inclusion/exclusion of endpoints
  3. Check for any breaks or gaps in the graph
  4. Express the result using interval notation
Graph Features
Curve from x = -4 to x = 6
Open Circle
x = -2 not included
Closed Circle
x = 4 included
Step 1: Analyze the horizontal extent

The graph extends from x = -4 to x = 6

This is the basic range of x-values

Step 2: Check for excluded points

There is an open circle at x = -2, which means x = -2 is NOT in the domain

This creates a gap in the domain at x = -2

Step 3: Check for included points

There is a closed circle at x = 4, which means x = 4 IS in the domain

Endpoints: x = -4 (included) and x = 6 (included)

Step 4: Determine the complete domain

From x = -4 (included) to x = -2 (excluded): [-4, -2)

From x = -2 (excluded) to x = 6 (included): (-2, 6]

Combined: [-4, -2) ∪ (-2, 6]

Step 5: Express in interval notation

Domain: [-4, -2) ∪ (-2, 6]

This means x can be any value from -4 to 6, except x = -2

Domain: [-4, -2) ∪ (-2, 6]
Final answer:

Domain: [-4, -2) ∪ (-2, 6]

In set notation: {x | -4 ≤ x < -2 or -2 < x ≤ 6}

Applied rules:

• Graph analysis: Domain = horizontal extent of the graph

• Circle interpretation: Open circle = excluded, Closed circle = included

• Interval notation: Parentheses exclude, brackets include endpoints

2 Domain from Rational Functions
Exercise 2
Find the domain of f(x) = (x + 2)/(x² - 9). Factor the denominator and explain why certain values are excluded.
Definition:

Rational function: A function expressed as a fraction of polynomials

Domain restriction: Values that make the denominator zero are excluded

Function
f(x) = (x + 2)/(x² - 9)
Factored
f(x) = (x + 2)/[(x - 3)(x + 3)]
Excluded Values
x ≠ ±3
Step 1: Factor the denominator

x² - 9 = (x - 3)(x + 3) [difference of squares]

So f(x) = (x + 2)/[(x - 3)(x + 3)]

Step 2: Find values that make denominator zero

Set denominator equal to zero: (x - 3)(x + 3) = 0

This occurs when x - 3 = 0 OR x + 3 = 0

So x = 3 OR x = -3

Step 3: State the domain restrictions

Division by zero is undefined

Therefore, x ≠ 3 and x ≠ -3

Step 4: Express domain in interval notation

Domain: (-∞, -3) ∪ (-3, 3) ∪ (3, ∞)

This means x can be any real number except x = -3 and x = 3

Step 5: Verify with examples

f(0) = (0 + 2)/(0² - 9) = 2/(-9) = -2/9 ✓

f(3) = (3 + 2)/(3² - 9) = 5/0 → undefined ✗

f(-3) = (-3 + 2)/((-3)² - 9) = -1/0 → undefined ✗

Domain: (-∞, -3) ∪ (-3, 3) ∪ (3, ∞)
Final answer:

Domain: (-∞, -3) ∪ (-3, 3) ∪ (3, ∞)

Or in set notation: {x ∈ ℝ | x ≠ ±3}

Explanation: x = 3 and x = -3 make the denominator zero, causing division by zero which is undefined.

Applied rules:

• Rational function domain: Exclude zeros of the denominator

• Difference of squares: a² - b² = (a - b)(a + b)

• Division by zero: Never allowed in mathematics

3 Domain from Radical Functions
Exercise 3
Find the domain of f(x) = √(2x - 6) and g(x) = √(x² - 4). Explain the restrictions for each.
Definition:

Radical function: Function containing a square root

Domain restriction: Expression under square root must be ≥ 0

Function 1
f(x) = √(2x - 6)
Function 2
g(x) = √(x² - 4)
Domains
f: x ≥ 3, g: x ≤ -2 or x ≥ 2
Step 1: Find domain of f(x) = √(2x - 6)

Expression under square root must be ≥ 0

So: 2x - 6 ≥ 0

2x ≥ 6

x ≥ 3

Domain: [3, ∞)

Step 2: Find domain of g(x) = √(x² - 4)

Expression under square root must be ≥ 0

So: x² - 4 ≥ 0

Factor: (x - 2)(x + 2) ≥ 0

Step 3: Solve (x - 2)(x + 2) ≥ 0

Find critical points: x = 2 and x = -2

Test intervals: (-∞, -2), (-2, 2), (2, ∞)

For x < -2: (negative)(negative) = positive ≥ 0 ✓

For -2 < x < 2: (negative)(positive) = negative < 0 ✗

For x > 2: (positive)(positive) = positive ≥ 0 ✓

Domain: (-∞, -2] ∪ [2, ∞)

Step 4: Verify with examples

For f(x): f(3) = √(0) = 0 ✓, f(2) = √(-2) → undefined ✗

For g(x): g(3) = √(5) ✓, g(0) = √(-4) → undefined ✗

Step 5: Express domains clearly

f(x) domain: [3, ∞) - all x-values ≥ 3

g(x) domain: (-∞, -2] ∪ [2, ∞) - all x-values ≤ -2 OR x-values ≥ 2

f(x) domain: [3, ∞)
g(x) domain: (-∞, -2] ∪ [2, ∞)
Final answer:

f(x) = √(2x - 6): Domain is [3, ∞)

g(x) = √(x² - 4): Domain is (-∞, -2] ∪ [2, ∞)

Restrictions: For f(x), need 2x - 6 ≥ 0; for g(x), need x² - 4 ≥ 0.

Applied rules:

• Radical function domain: Expression under √ ≥ 0

• Quadratic inequality: Factor and test intervals

• Sign analysis: Determine where product is non-negative

Domain from Graphs and Equations: Complete Guide
\text{Domain} = \{x | f(x) \text{ is defined}\}
Domain Definition
Graph Method
Horizontal Extent
Left to Right
Rational Function
Exclude Zeros
Denominator ≠ 0
Radical Function
Expression ≥ 0
Under Root
Key definitions:

Domain: The set of all possible input values (x-values) for which a function is defined

Function: A relation where each input has exactly one output

Graph analysis: Examining the visual representation to determine valid x-values

Algebraic analysis: Using equations to identify restrictions

Domain finding methodology:
  1. From graphs: Examine horizontal extent, note open/closed circles
  2. From equations: Identify restrictions based on function type
  3. For rational functions: Exclude zeros of denominator
  4. For radical functions: Ensure expression under root ≥ 0
  5. For polynomial functions: Domain is typically all real numbers
  6. Express results: Use interval notation appropriately
Tip 1: Always check for division by zero in rational functions.
Tip 2: For radicals, ensure the expression under the root is non-negative.
Tip 3: On graphs, open circles indicate excluded points, closed circles indicate included points.
Tip 4: Factor denominators and expressions under roots to identify restrictions.
Common errors: Forgetting domain restrictions, confusing open and closed circles, misfactoring expressions.
Key insights: Domain restrictions depend on function type, always verify with examples.
Essential formulas and rules:

• Interval notation: [a, b] includes endpoints, (a, b) excludes endpoints

• Rational functions: Domain excludes zeros of denominator

• Radical functions: Expression under even root ≥ 0

• Polynomial functions: Domain is all real numbers

• Graph interpretation: Open circle = excluded, Closed circle = included

Solution: Exercises 4 to 5
4 Domain from Piecewise Functions
Exercise 4
Find the domain of f(x) = { x² if x < 1, 2x + 1 if 1 ≤ x < 4, 5 if x ≥ 4 }. Explain how each piece contributes to the overall domain.
Definition:

Piecewise function: A function defined by different expressions over different intervals

Domain union: The overall domain is the union of domains of all pieces

Piecewise Function
f(x) = {x² if x < 1, 2x + 1 if 1 ≤ x < 4, 5 if x ≥ 4}
Piece Domains
(-∞, 1), [1, 4), [4, ∞)
Overall Domain
(-∞, ∞)
Step 1: Analyze the first piece: f(x) = x² when x < 1

Domain for this piece: x < 1, or (-∞, 1)

This piece is defined for all x-values less than 1

Step 2: Analyze the second piece: f(x) = 2x + 1 when 1 ≤ x < 4

Domain for this piece: 1 ≤ x < 4, or [1, 4)

This piece is defined for x-values from 1 (inclusive) to 4 (exclusive)

Step 3: Analyze the third piece: f(x) = 5 when x ≥ 4

Domain for this piece: x ≥ 4, or [4, ∞)

This piece is defined for all x-values greater than or equal to 4

Step 4: Find the overall domain

The domain is the union of all individual piece domains

Overall domain: (-∞, 1) ∪ [1, 4) ∪ [4, ∞) = (-∞, ∞)

This covers all real numbers

Step 5: Verify with examples

For x = 0 (x < 1): f(0) = 0² = 0 ✓

For x = 2 (1 ≤ x < 4): f(2) = 2(2) + 1 = 5 ✓

For x = 4 (x ≥ 4): f(4) = 5 ✓

For x = 10 (x ≥ 4): f(10) = 5 ✓

Step 6: Conclusion

Every real number falls into exactly one of the three intervals

Therefore, the function is defined for all real numbers

Domain: (-∞, ∞)
Final answer:

Domain: (-∞, ∞) or all real numbers

Explanation: The three pieces together cover all possible x-values, with no gaps or exclusions.

Applied rules:

• Piecewise domain: Union of domains of all pieces

• Interval union: Combine all intervals where function is defined

• Complete coverage: If pieces cover all real numbers, domain is (-∞, ∞)

5 Complex Function Domain Analysis
Exercise 5
Find the domain of f(x) = √(x - 2)/(x² - 5x + 6). Factor the denominator and combine restrictions.
Definition:

Complex function: A function with multiple types of restrictions

Combined restrictions: All individual restrictions must be satisfied simultaneously

Function
f(x) = √(x - 2)/(x² - 5x + 6)
Factored
f(x) = √(x - 2)/[(x - 2)(x - 3)]
Domain
x > 2 and x ≠ 3
Step 1: Identify restrictions

Two types of restrictions:

1. Radical: x - 2 ≥ 0 (expression under √ ≥ 0)

2. Rational: x² - 5x + 6 ≠ 0 (denominator ≠ 0)

Step 2: Solve radical restriction

x - 2 ≥ 0

x ≥ 2

Step 3: Factor and solve rational restriction

x² - 5x + 6 = (x - 2)(x - 3)

So: (x - 2)(x - 3) ≠ 0

Therefore: x ≠ 2 AND x ≠ 3

Step 4: Combine restrictions

We need: x ≥ 2 AND x ≠ 2 AND x ≠ 3

Since x ≥ 2 AND x ≠ 2, we have x > 2

Also, x ≠ 3

So: x > 2 AND x ≠ 3

Step 5: Express in interval notation

From x > 2: (2, ∞)

Exclude x = 3: (2, 3) ∪ (3, ∞)

Step 6: Verify with examples

f(2.5) = √(0.5)/[(0.5)(-0.5)] = √(0.5)/(-0.25) → negative denominator ✓

f(4) = √(2)/[(2)(1)] = √(2)/2 ✓

f(2) = √(0)/0 → division by zero ✗

f(3) = √(1)/0 → division by zero ✗

Domain: (2, 3) ∪ (3, ∞)
Final answer:

Domain: (2, 3) ∪ (3, ∞)

Or in set notation: {x | x > 2 and x ≠ 3}

Explanation: Must satisfy both x ≥ 2 (radical) and x ≠ 2,3 (rational), resulting in x > 2 excluding x = 3.

Applied rules:

• Combined restrictions: All restrictions must be satisfied simultaneously

• Quadratic factoring: x² - 5x + 6 = (x - 2)(x - 3)

• Simultaneous conditions: Use logical AND to combine restrictions

Advanced Domain Analysis Techniques
\text{Domain}(f) = \{x \in \mathbb{R} | f(x) \text{ is well-defined}\}
Domain Definition
Key definitions:

Domain: The set of all possible input values (independent variable) for which the function produces a real output

Well-defined: The function expression evaluates to a real number

Restriction: A condition that limits the possible input values

Interval notation: A way to represent sets of real numbers using brackets and parentheses

Comprehensive domain analysis approach:
  1. Identify function type: Polynomial, rational, radical, logarithmic, etc.
  2. Look for restrictions: Division by zero, negative under radicals, etc.
  3. Solve restriction conditions: Set up and solve inequalities
  4. Combine multiple restrictions: Use logical AND for simultaneous conditions
  5. Express result: Use interval notation or set notation
  6. Verify: Test boundary values and examples
Tip 1: Always factor polynomials to easily identify zeros and restrictions.
Tip 2: For functions with multiple restrictions, solve each separately, then find intersection.
Tip 3: When graphing, look for vertical asymptotes and holes as indicators of domain restrictions.
Tip 4: Remember that polynomial functions have domain of all real numbers unless piecewise defined.

Common misconceptions: Assuming all functions have unlimited domains, forgetting to check multiple restrictions, misapplying factoring techniques.
Memory aids: "Division by zero is forbidden," "Negative under even root is not real," "Check all function components."
Essential formulas and relationships:

• Interval notation: [a,b] closed, (a,b) open, [a,b) half-open

• Rational functions: Domain excludes zeros of denominator

• Radical functions: Even index: expression ≥ 0; Odd index: all reals

• Logarithmic functions: Argument > 0

• Combined restrictions: Intersection of individual restriction sets

Visualization: Domain Analysis Examples
Exercise 6: Domain Comparison
Compare the domains of:
f(x) = x² (polynomial)
g(x) = 1/(x - 2) (rational)
h(x) = √(x + 3) (radical)

Analysis: Different function types have different domain restrictions.

  • Polynomial: Domain all reals (-∞, ∞)
  • Rational: Domain excludes x = 2
  • Radical: Domain x ≥ -3

Questions & Answers

Question: How do I find the domain when a function has both a radical and a denominator? Do I consider both restrictions?

Answer: Yes, you must consider ALL restrictions simultaneously! When a function has multiple types of restrictions, you find each one separately, then combine them using logical AND (intersection).

For example, for f(x) = √(x - 1)/(x - 3):

  • Radical restriction: x - 1 ≥ 0, so x ≥ 1
  • Rational restriction: x - 3 ≠ 0, so x ≠ 3

Both must be true: x ≥ 1 AND x ≠ 3

This gives us: [1, 3) ∪ (3, ∞)

Always solve each restriction separately, then find values that satisfy ALL conditions.

Question: How do I interpret open and closed circles on graphs when finding domain?

Answer: The type of circle tells you whether that specific x-value is included in the domain:

Open circle (○): The point is NOT included in the domain

  • Means the function is undefined at that x-value
  • Use parentheses in interval notation
  • Example: If there's an open circle at x = 2, then x = 2 is not in the domain

Closed circle (●): The point IS included in the domain

  • Means the function is defined at that x-value
  • Use brackets in interval notation
  • Example: If there's a closed circle at x = 2, then x = 2 is in the domain

Think of open circles as "holes" in the graph and closed circles as "solid" points.

Question: How do I find domain restrictions for rational functions? Should I always factor the denominator?

Answer: Yes, factoring the denominator is usually the best approach! Here's the systematic method:

1. Factor the denominator completely

2. Set each factor equal to zero

3. Solve for x to find excluded values

4. Express the domain excluding these values

Example: For f(x) = (x + 1)/(x² - 5x + 6)

  • Factor: x² - 5x + 6 = (x - 2)(x - 3)
  • Set factors to zero: x - 2 = 0 → x = 2, x - 3 = 0 → x = 3
  • Domain: All real numbers except x = 2 and x = 3
  • Notation: (-∞, 2) ∪ (2, 3) ∪ (3, ∞)

Factoring reveals all values that would cause division by zero!

Question: How do I handle radical functions when the expression under the root is quadratic, like √(x² - 4)?

Answer: For radical functions with quadratic expressions, you need to solve the inequality where the expression ≥ 0:

For f(x) = √(x² - 4):

1. Set up the inequality: x² - 4 ≥ 0

2. Factor: (x - 2)(x + 2) ≥ 0

3. Find critical points: x = 2 and x = -2

4. Test intervals created by critical points:

  • For x < -2: (negative)(negative) = positive ≥ 0 ✓
  • For -2 < x < 2: (negative)(positive) = negative < 0 ✗
  • For x > 2: (positive)(positive) = positive ≥ 0 ✓

5. Include critical points since inequality is ≥ (not >)

Domain: (-∞, -2] ∪ [2, ∞)

The key is factoring and sign analysis of the intervals!

Question: How do I find the domain of piecewise functions? Do I combine the domains of each piece?

Answer: Exactly! The domain of a piecewise function is the UNION of the domains of all its pieces. Here's how to approach it:

1. Find the domain for each individual piece

2. Take the union of all these domains

3. Be careful with interval notation at boundary points

Example: f(x) = {x² if x < 0, x + 1 if x ≥ 0}

  • First piece domain: (-∞, 0)
  • Second piece domain: [0, ∞)
  • Overall domain: (-∞, 0) ∪ [0, ∞) = (-∞, ∞)

The function is defined wherever ANY piece applies, so use union (not intersection).

If pieces don't cover all real numbers, the domain will have gaps.