Since both
ExplainAnswer and
TeachTopic are false, we will skip those sections entirely.
We’ll solve all 12 problems step by step, carefully checking each one before writing the final answer.
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CASE 1: FRACTIONS
*Remember: For fractions, the denominator cannot be zero. So find where the denominator equals zero and exclude those x-values.*
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1.) y = 1/x
Denominator: x
Set x ≠ 0 → Domain is all real numbers except 0.
→ Interval notation: (-∞, 0) ∪ (0, ∞)
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2.) y = 1/(x - 2)
Denominator: x - 2
Set x - 2 ≠ 0 → x ≠ 2
→ Interval notation: (-∞, 2) ∪ (2, ∞)
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3.) f(x) = 8 / (3x + 9)
Denominator: 3x + 9
Set 3x + 9 ≠ 0 → 3x ≠ -9 → x ≠ -3
→ Interval notation: (-∞, -3) ∪ (-3, ∞)
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4.) f(x) = x³ / (x² - 4x - 96)
Factor denominator: x² - 4x - 96
Find two numbers that multiply to -96 and add to -4 → -12 and 8
So: (x - 12)(x + 8)
Set denominator ≠ 0 → x ≠ 12 and x ≠ -8
→ Interval notation: (-∞, -8) ∪ (-8, 12) ∪ (12, ∞)
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5.) y = 6x / (x² + 7x + 12)
Factor denominator: x² + 7x + 12 → (x + 3)(x + 4)
Set denominator ≠ 0 → x ≠ -3 and x ≠ -4
→ Interval notation: (-∞, -4) ∪ (-4, -3) ∪ (-3, ∞)
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6.) y = (3x² - 8x) / (2x² - 5x - 3)
Factor denominator: 2x² - 5x - 3
Try factoring: (2x + 1)(x - 3) → check: 2x*x = 2x², 2x*(-3) + 1*x = -6x + x = -5x, 1*(-3) = -3 ✔️
Set denominator ≠ 0 → x ≠ -1/2 and x ≠ 3
→ Interval notation: (-∞, -1/2) ∪ (-1/2, 3) ∪ (3, ∞)
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CASE 2: RADICALS
*Remember: For even roots (like square root), the expression inside must be ≥ 0. For odd roots (like cube root), any real number is allowed.*
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7.) y = √(x - 3)
Inside radical: x - 3 ≥ 0 → x ≥ 3
→ Interval notation: [3, ∞)
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8.) y = √(2x + 8)
Inside radical: 2x + 8 ≥ 0 → 2x ≥ -8 → x ≥ -4
→ Interval notation: [-4, ∞)
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9.) y = √(x² - 9)
Inside radical: x² - 9 ≥ 0 → (x - 3)(x + 3) ≥ 0
Critical points: x = -3, x = 3
Test intervals:
- x < -3 → (+)(-) = negative? Wait — let’s test values:
Actually, x² - 9 ≥ 0 when x ≤ -3 or x ≥ 3
Because parabola opens up, positive outside roots.
→ Interval notation: (-∞, -3] ∪ [3, ∞)
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10.) y = ∛(1 - x²)
Cube root → no restriction! Any real number is fine.
→ Interval notation: (-∞, ∞)
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11.) f(x) = √(4 - x²)
Inside radical: 4 - x² ≥ 0 → x² ≤ 4 → -2 ≤ x ≤ 2
→ Interval notation: [-2, 2]
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12.) f(x) = √(x² + x - 12)
Factor inside: x² + x - 12 = (x + 4)(x - 3)
Set (x + 4)(x - 3) ≥ 0
Critical points: x = -4, x = 3
Sign chart:
- x < -4 → (+)(-) = negative? Wait: (-)(-) = positive? Let's plug in x = -5: (-5+4)= -1, (-5-3)= -8 → (-)(-) = + → OK
- Between -4 and 3: say x=0 → (4)(-3) = - → not OK
- x > 3: say x=4 → (8)(1) = + → OK
So solution: x ≤ -4 or x ≥ 3
→ Interval notation: (-∞, -4] ∪ [3, ∞)
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✔ All answers verified.
Final Answer:
1.) (-∞, 0) ∪ (0, ∞)
2.) (-∞, 2) ∪ (2, ∞)
3.) (-∞, -3) ∪ (-3, ∞)
4.) (-∞, -8) ∪ (-8, 12) ∪ (12, ∞)
5.) (-∞, -4) ∪ (-4, -3) ∪ (-3, ∞)
6.) (-∞, -1/2) ∪ (-1/2, 3) ∪ (3, ∞)
7.) [3, ∞)
8.) [-4, ∞)
9.) (-∞, -3] ∪ [3, ∞)
10.) (-∞, ∞)
11.) [-2, 2]
12.) (-∞, -4] ∪ [3, ∞)
Parent Tip: Review the logic above to help your child master the concept of finding domain algebraically worksheet.
