Find the domain of the function and write your answer in set notation. \( \begin{array}{lll}\text { i } f(t)=\frac{1}{t^{2}-81} & \text { ii } f(a)=\sqrt[4]{a^{2}-a-6} & \text { iii } f(b)=\sqrt[3]{b^{2}-b-6} \\ \text { iv } f(x)=\ln \left(x^{3}+x^{2}-2 x\right) & \text { v } f(z)=\sqrt{z}+\sqrt{10-z} & \text { vi } f(x)=\ln \left(4-x^{2}\right)\end{array} \)

We will find the domain of each function by determining the restrictions on the variable. 1. f(t) = 1/(t² – 81) The function is undefined when the denominator is zero. Solve:   t² – 81 = 0   (t – 9)(t + 9) = 0   t = 9 or t = –9 In set notation, the domain is:   { t ∈ ℝ : t ≠ 9 and t ≠ –9 } 2. f(a) = √[4](a² – a – 6) Since we have an even indexed root, the radicand must be nonnegative. Set:   a² – a – 6 ≥ 0 Factor the quadratic:   (a – 3)(a + 2) ≥ 0 Find the critical points: a = –2 and a = 3. Use a sign chart: - For a < –2, both factors are negative so their product is positive. - For –2 < a < 3, one factor is negative and the other is positive so the product is negative. - For a > 3, both factors are positive so the product is positive. Include the points where the expression equals 0. The domain is:   { a ∈ ℝ : a ≤ –2 or a ≥ 3 }   In interval notation: (–∞, –2] ∪ [3, ∞) 3. f(b) = ∛(b² – b – 6) The cube root function is defined for all real numbers, so there are no restrictions. Domain: ℝ 4. f(x) = ln(x³ + x² – 2x) The natural logarithm requires its argument to be positive, so:   x³ + x² – 2x > 0 Factor the expression:   x(x² + x – 2) > 0   x(x + 2)(x – 1) > 0 The critical points are x = –2, 0, and 1. Test intervals: - For x < –2, choose x = –3: (–3)(–3+2=–1)(–3–1=–4) → negative × negative × negative = negative. - For –2 < x < 0, choose x = –1: (–1)(1)(–2) → negative × positive × negative = positive. - For 0 < x < 1, choose x = 0.5: (0.5)(2.5)(–0.5) → positive × positive × negative = negative. - For x > 1, choose x = 2: (2)(4)(1) → all positive = positive. Also, the expression is zero at x = –2, 0, and 1; these cannot be included since ln(0) is undefined. Therefore, the domain is:   { x ∈ ℝ : –2 < x < 0 or x > 1 }   In interval notation: (–2, 0) ∪ (1, ∞) 5. f(z) = √z + √(10 – z) Both square roots require their radicands to be nonnegative: - For √z:  z ≥ 0 - For √(10 – z): 10 – z ≥ 0 → z ≤ 10 Hence, the domain is:   { z ∈ ℝ : 0 ≤ z ≤ 10 }   In interval notation: [0, 10] 6. f(x) = ln(4 – x²) The natural logarithm requires:   4 – x² > 0   x² < 4 Solving, we get:   –2 < x < 2 Thus, the domain in set notation is:   { x ∈ ℝ : –2 < x < 2 } Summary of Answers: i. { t ∈ ℝ : t ≠ 9 and t ≠ –9 } ii. { a ∈ ℝ : a ≤ –2 or a ≥ 3 } iii. ℝ iv. { x ∈ ℝ : –2 < x < 0 or x > 1 } v. [0, 10] vi. { x ∈ ℝ : –2 < x < 2 }