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Determine the range of each function: (i) f(x) = 2/x; (ii) f(x) = 1/(x² - x + 1); (iii) f(x) = (x² - x + 1)/(x² + x + 1).

1. (i) R - {0}; (ii) (0, 4/3]; (iii) [1/3, 3]
2. (i) R; (ii) (0, 3/4]; (iii) [1/3, 3]
3. (i) R - {0}; (ii) (0, 3/4]; (iii) (1/3, 3)
4. (i) (0, infinity); (ii) [4/3, infinity); (iii) [1/3, 3]

Correct answer: (i) R - {0}; (ii) (0, 4/3]; (iii) [1/3, 3]

Solution

(i) 2/x takes every real value except 0, so range = R - {0}. (ii) The denominator x² - x + 1 = (x - 1/2)² + 3/4 has minimum 3/4 (always positive), so 1/(x² - x + 1) has maximum 1/(3/4) = 4/3 and approaches 0; range = (0, 4/3]. (iii) Put y = (x² - x + 1)/(x² + x + 1). Cross-multiplying and requiring the resulting quadratic in x to have real solutions gives discriminant >= 0, yielding 1/3 <= y <= 3; range = [1/3, 3].

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