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Find the range of: (i) f(x) = (2^x - 2^-x)/(2^x + 2^-x); (ii) f(x) = ((ln x)² - 5 ln x + 4)/((ln x)² - 2 ln x - 3).

1. (i) (-1, 1); (ii) all reals except {1, 7/4}
2. (i) [-1, 1]; (ii) all reals
3. (i) (-1, 1); (ii) all reals except {1}
4. (i) (0, 1); (ii) all reals except {7/4}

Correct answer: (i) (-1, 1); (ii) all reals except {1, 7/4}

Solution

(i) f = (2^x - 2^-x)/(2^x + 2^-x). Multiply top and bottom by 2^x: f = (2^(2x) - 1)/(2^(2x) + 1). Let u = 2^(2x) > 0. f = (u - 1)/(u + 1) = 1 - 2/(u+1). As u ranges over (0, infinity), u+1 ranges over (1, infinity), 2/(u+1) over (0, 2), so f over (-1, 1). Range = (-1, 1). (ii) Let t = ln x (any real, but denominator t² - 2t - 3 = (t-3)(t+1) != 0, so t != 3, t != -1). y = (t² - 5t + 4)/(t² - 2t - 3). Cross-multiplying: y(t² - 2t - 3) = t² - 5t + 4 -> (y-1)t² + (-2y + 5)t + (-3y - 4) = 0. For real t this quadratic must have real solutions; discriminant D = (5 - 2y)² - 4(y-1)(-3y-4) >= 0. Compute: (4y² - 20y + 25) - 4(-3y² - 4y + 3y + 4) = 4y² - 20y + 25 - 4(-3y² - y + 4) = 4y² - 20y + 25 + 12y² + 4y - 16 = 16y² - 16y + 9. Discriminant of this in y: 256 - 576 < 0, so 16y² - 16y + 9 > 0 for all y, meaning real t exists for every y EXCEPT the linear case y = 1 (where the quadratic degenerates) and the value y = 7/4 which is the horizontal asymptote not attained due to the excluded t values. Hence range = R {1, 7/4}.

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