A-Level Functions: worked solution

Question

Define f:x↦[ln(x-1)]²+2 for x≥a, and g:x↦4+3x-x² for x≤3/2. (i) Given f⁻¹ exists, state the least a. (ii) Find g⁻¹(x) and its domain. (iii) Using that a, decide whether g⁻¹f⁻¹ exists.

(i) An inverse exists exactly when f is one-one, so find where f stops being one-one. f(x) = [ln(x − 1)]² + 2 needs x − 1 > 0, so x > 1 at the very least. Put u = ln(x − 1). As x increases from just above 1, u increases from −∞, passing through u = 0 when x − 1 = 1, i.e. x = 2, and on to +∞. Squaring, u² decreases while u < 0 and increases while u > 0, turning at u = 0. So f decreases on 1 < x ≤ 2 and increases on x ≥ 2: the turning point is at x = 2, and any domain containing points on both sides of it repeats values. Restricting to x ≥ a makes f one-one exactly when a ≥ 2, so the least value is a = 2. (ii) Complete the square so that x appears only once: g(x) = 4 + 3x − x² = −(x² − 3x) + 4 = −[(x − 3/2)² − 9/4] + 4 = 25/4 − (x − 3/2)² Set y = g(x) and make x the subject: (x − 3/2)² = 25/4 − y x − 3/2 = ±√(25/4 − y) The domain is x ≤ 3/2, so x − 3/2 ≤ 0 and only the NEGATIVE square root is possible: x = 3/2 − √(25/4 − y) So g⁻¹(x) = 3/2 − √(25/4 − x). The domain of g⁻¹ is the range of g. On x ≤ 3/2 the bracket (x − 3/2)² takes every value from 0 upwards, so 25/4 − (x − 3/2)² takes every value up to and including 25/4: D_{g⁻¹} = R_g = (−∞, 25/4] (iii) The composite g⁻¹f⁻¹ exists only if R_{f⁻¹} ⊆ D_{g⁻¹}. The range of an inverse is the domain of the original function, and with a = 2 we have D_f = [2, ∞), so R_{f⁻¹} = D_f = [2, ∞) From part (ii), D_{g⁻¹} = (−∞, 25/4] [2, ∞) is not contained in (−∞, 25/4] (for instance 7 lies in R_{f⁻¹} but 7 > 25/4, so g⁻¹(7) is undefined). Therefore g⁻¹f⁻¹ does not exist.