

InterviewSolution
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Find the domain of each of the following real valued functions of real variable :i. f(x) = 1/xii. f(x) = \(\frac{1}{x-7}\)iii. f(x) = \(\frac{3x-2}{x+1}\)iv. f(x) = \(\frac{2x+1}{x^2+9}\) v. f(x) = \(\frac{x^2+2x+1}{x^2-8x+12}\) |
Answer» i. f(x) = 1/x Clearly, f(x) is defined for all real values of x, Except for the case when x = 0. When x = 0, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {0} ii. f(x) = \(\frac{1}{x-7}\) Clearly, f(x) is defined for all real values of x, Except for the case when x – 7 = 0 or x = 7. When x = 7, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {7} iii. f(x) = \(\frac{3x-2}{x+1}\) Clearly, f(x) is defined for all real values of x, Except for the case when x + 1 = 0 or x = –1. When x = –1, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {–1} iv. f(x) = \(\frac{2x+1}{x^2+9}\) Clearly, f(x) is defined for all real values of x, Except for the case when x2 – 9 = 0. x2 – 9 = 0 ⇒ x2 – 32 = 0 ⇒ (x + 3)(x – 3) = 0 ⇒ x + 3 = 0 or x – 3 = 0 ⇒ x = ±3 When x = ±3, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {–3, 3} v. f(x) = \(\frac{x^2+2x+1}{x^2-8x+12}\) Clearly, f(x) is defined for all real values of x, except for the case when x2 – 8x + 12 = 0. x2 – 8x + 12 = 0 ⇒ x2 – 2x – 6x + 12 = 0 ⇒ x(x – 2) – 6(x – 2) = 0 ⇒ (x – 2)(x – 6) = 0 ⇒ x – 2 = 0 or x – 6 = 0 ⇒ x = 2 or 6 When x = 2 or 6, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {2, 6} |
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