Domain — Combinations (Root in Denominator, Root in Numerator, etc.)
Domain — Combinations (Root in Denominator, Root in Numerator, etc.). Practice questions to deepen understanding of domain combinations. Online math practice with full solutions and step-by-step explanations.
Domain combinations practice — root in numerator (≥ 0), root in denominator (> 0), combinations of root and rational functions. Complex cases.
This practice covers: root in denominator (> 0, not ≥ 0!).
Question 1
10.00 pts
√÷ Square root in denominator:
What is the domain of \(f(x) = \frac{1}{\sqrt{x-3}}\)?
Explanation:
| √÷ Square root in the denominator The function: \(f(x) = \frac{1}{\sqrt{x-3}}\) ⚠️ Two restrictions! Restriction 1 — square root: \(x - 3 \geq 0\) \(x \geq 3\) Restriction 2 — denominator: \(\sqrt{x-3} \neq 0\) \(x - 3 \neq 0\) \(x \neq 3\) Intersection: \(x \geq 3\) and \(x \neq 3\) \(x > 3\) ✓ Domain: \((3, \infty)\) ⚠️ Open bracket! 3 itself is not allowed! Because if \(x = 3\): \(\frac{1}{\sqrt{0}} = \frac{1}{0}\) ✗ Division by zero! The general rule: Square root in the denominator: Expression under root must > 0 (not just ≥ 0!) because the denominator cannot vanish |
Question 2
10.00 pts
√→ Square root in numerator:
What is the domain of \(f(x) = \frac{\sqrt{x+2}}{x-5}\)?
Explanation:
| √→ Square root in the numerator The function: \(f(x) = \frac{\sqrt{x+2}}{x-5}\) Two restrictions: Restriction 1 — square root in numerator: \(x + 2 \geq 0\) \(x \geq -2\) ✓ Restriction 2 — denominator: \(x - 5 \neq 0\) \(x \neq 5\) ✓ Intersection: \(x \geq -2\) and \(x \neq 5\) Domain: \([-2, \infty) \setminus \{5\}\) Or in interval notation: \([-2, 5) \cup (5, \infty)\) ✓ Important distinction: Square root in numerator: the root may equal 0 ✓ (the numerator can be 0) Square root in denominator: the root may NOT equal 0 ✗ (the denominator cannot be 0) Check: • \(x = -3\): \(\sqrt{-1}\) ✗ • \(x = -2\): \(\frac{0}{-7}\) ✓ • \(x = 0\): \(\frac{\sqrt{2}}{-5}\) ✓ • \(x = 5\): \(\frac{\sqrt{7}}{0}\) ✗ • \(x = 10\): \(\frac{\sqrt{12}}{5}\) ✓ |
Question 3
10.00 pts
√√ Two square roots:
What is the domain of \(f(x) = \sqrt{x-1} + \sqrt{5-x}\)?
Explanation:
| √√ Two square roots The function: \(f(x) = \sqrt{x-1} + \sqrt{5-x}\) Two conditions: Condition 1 — first root: \(x - 1 \geq 0\) \(x \geq 1\) ✓ Condition 2 — second root: \(5 - x \geq 0\) \(5 \geq x\) \(x \leq 5\) ✓ Intersection: \(x \geq 1\) and \(x \leq 5\) \(1 \leq x \leq 5\) ✓ Domain: \([1, 5]\) Graphical explanation: Condition 1: \([1, \infty)\) → Condition 2: \((-\infty, 5]\) ← Intersection: \([1, 5]\) ✓ Only the interval that satisfies both conditions! Check: • \(x = 0\): \(\sqrt{-1}\) ✗ • \(x = 1\): \(\sqrt{0} + \sqrt{4} = 2\) ✓ • \(x = 3\): \(\sqrt{2} + \sqrt{2}\) ✓ • \(x = 5\): \(\sqrt{4} + \sqrt{0} = 2\) ✓ • \(x = 6\): \(\sqrt{-1}\) ✗ |
Question 4
10.00 pts
√÷ Square root and rational:
What is the domain of \(f(x) = \sqrt{x+3} + \frac{1}{x-2}\)?
Explanation:
| √÷ Square root + rational The function: \(f(x) = \sqrt{x+3} + \frac{1}{x-2}\) Two restrictions: Restriction 1 — square root: \(x + 3 \geq 0\) \(x \geq -3\) ✓ Restriction 2 — rational: \(x - 2 \neq 0\) \(x \neq 2\) ✓ Intersection: \(x \geq -3\) and \(x \neq 2\) Domain: \([-3, \infty) \setminus \{2\}\) Or: \([-3, 2) \cup (2, \infty)\) ✓ Explanation: "from -3 upward, but not 2" • Allowed: -3, -2, -1, 0, 1 • Not allowed: 2 • Allowed: 3, 4, 5, ... The rule: When there are several restrictions: 1. Write each condition separately 2. Find the intersection 3. Remove forbidden points |
Question 5
10.00 pts
√(÷) Square root of a rational:
What is the domain of \(f(x) = \sqrt{\frac{x+1}{x-4}}\)?
Explanation:
| √(÷) Square root of a rational expression The function: \(f(x) = \sqrt{\frac{x+1}{x-4}}\) ⚠️ Two restrictions: Restriction 1 — square root: \(\frac{x+1}{x-4} \geq 0\) Restriction 2 — denominator: \(x - 4 \neq 0\) Step 1: critical values Numerator = 0: \(x = -1\) Denominator = 0: \(x = 4\) (forbidden) Step 2: sign table
Step 3: Domain \((-\infty, -1] \cup (4, \infty)\) ✓ ⚠️ Note the brackets: • \(x = -1\): fraction = 0, \(\sqrt{0}=0\) allowed → closed bracket ] • \(x = 4\): division by 0, forbidden → open bracket ( Verification: • \(x = -2\): \(\frac{-1}{-6} = \frac{1}{6}\), \(\sqrt{1/6}\) ✓ • \(x = 0\): \(\frac{1}{-4}\) negative, \(\sqrt{-1/4}\) ✗ • \(x = 5\): \(\frac{6}{1} = 6\), \(\sqrt{6}\) ✓ |
Question 6
10.00 pts
🔢 More complex:
What is the domain of \(f(x) = \frac{1}{\sqrt{x^2-9}}\)?
Explanation:
| 🔢 Quadratic square root in denominator The function: \(f(x) = \frac{1}{\sqrt{x^2-9}}\) ⚠️ Two restrictions! Restriction 1 — square root: \(x^2 - 9 \geq 0\) Restriction 2 — denominator: \(\sqrt{x^2-9} \neq 0\) \(x^2 - 9 \neq 0\) Intersection: \(x^2 - 9 > 0\) (stricter!) ✓ Factor: \((x-3)(x+3) > 0\) Sign table:
Domain: \((-\infty, -3) \cup (3, \infty)\) ✓ ⚠️ Open brackets! Both -3 and 3 are forbidden! because both would cause division by 0 |
Question 7
10.00 pts
💡 Key rule:
What is the difference between \(\sqrt{x}\) in the numerator and \(\sqrt{x}\) in the denominator?
Explanation:
| 💡 The golden rule The essential difference:
Explanation: Square root in numerator: \(f(x) = \frac{\sqrt{x}}{5}\) At \(x=0\): \(\frac{0}{5} = 0\) ✓ The numerator can be 0! Square root in denominator: \(f(x) = \frac{5}{\sqrt{x}}\) At \(x=0\): \(\frac{5}{0}\) ✗ The denominator cannot be 0! Examples: \(\frac{\sqrt{x-3}}{2}\) Domain: \([3, \infty)\) ✓ \(\frac{2}{\sqrt{x-3}}\) Domain: \((3, \infty)\) ✓ Notice the bracket! Quick rule: Numerator: [ closed bracket Denominator: ( open bracket Remember this! |
Question 8
10.00 pts
³√÷ Three parts:
What is the domain of \(f(x) = \sqrt{x+1} + \frac{1}{x-2} + \sqrt{4-x}\)?
Explanation:
| ³√÷ Three expressions The function: \(f(x) = \sqrt{x+1} + \frac{1}{x-2} + \sqrt{4-x}\) Three conditions: Condition 1 — first root: \(x + 1 \geq 0\) \(x \geq -1\) ✓ Condition 2 — rational: \(x - 2 \neq 0\) \(x \neq 2\) ✓ Condition 3 — second root: \(4 - x \geq 0\) \(x \leq 4\) ✓ Intersection of all three: \(x \geq -1\) and \(x \neq 2\) and \(x \leq 4\) \(-1 \leq x \leq 4\), but \(x \neq 2\) Domain: \([-1, 2) \cup (2, 4]\) ✓ Graphical explanation: Condition 1: →→→ from -1 upward Condition 2: ✗ not 2 Condition 3: ←←← up to 4 Together: [-1, 2) ∪ (2, 4] |
Question 9
10.00 pts
√√ Nested square root:
What is the domain of \(f(x) = \sqrt{\sqrt{x-1}}\)?
Explanation:
| √√ Square root of a square root The function: \(f(x) = \sqrt{\sqrt{x-1}}\) Solution from the inside out: Step 1: inner root \(\sqrt{x-1}\) Condition: \(x - 1 \geq 0\) \(x \geq 1\) ✓ Step 2: outer root \(\sqrt{...}\) Condition: the expression inside ≥ 0 But \(\sqrt{x-1} \geq 0\) always! (whenever it is defined) So there is no additional condition ✓ Domain: \([1, \infty)\) Explanation: A square root always returns a value ≥ 0 so a square root of a square root: the outer root is always "happy"! Only the inner root determines the domain Check: • \(x = 0\): \(\sqrt{\sqrt{-1}}\) ✗ • \(x = 1\): \(\sqrt{\sqrt{0}} = 0\) ✓ • \(x = 5\): \(\sqrt{\sqrt{4}} = \sqrt{2}\) ✓ |
Question 10
10.00 pts
📚 Summary:
What is the correct way to handle multiple constraints?
Explanation:
📚 Summary — Combinations The full method: 1️⃣ Identify the restrictions: • Square root in numerator? ✓ • Square root in denominator? ✓ • Rational? ✓ • Logarithm? ✓ 2️⃣ Write a condition for each: • sqrt in numerator: ≥ 0 • sqrt in denominator: > 0 • rational: denominator ≠ 0 • logarithm: > 0 3️⃣ Find the intersection: All conditions must hold together! 4️⃣ Write the domain: In interval notation Memory table:
Quick example: \(f(x) = \frac{\sqrt{x+2}}{x-3}\) Condition 1: \(x \geq -2\) Condition 2: \(x \neq 3\) Domain: \([-2, 3) \cup (3, \infty)\) ✓ ⚠️ Remember: Always intersection, not union! All conditions together! |