A-Level Functions: worked solution

Question

Given f(x) = 1 + 3/(e^x - 2) for x > ln 2 with f inverse (x) = ln(2 + 3/(x - 1)), and g(x) = 2(x - 1)^2 + 2 for x <= 0 with g(x) = 2 + |2 - x| for 0 < x <= 3, find the exact k satisfying f inverse g(k) = ln(5/2).

Work from the outside in: first find what g(k) must be, then solve for k. Write y = g(k). The condition f inverse (g(k)) = ln(5/2) becomes f inverse (y) = ln(5/2), and the given formula turns this into ln(2 + 3/(y - 1)) = ln(5/2). ln is a one-one function, so the insides must be equal: 2 + 3/(y - 1) = 5/2. 3/(y - 1) = 5/2 - 2 = 1/2. Taking reciprocals of both sides, (y - 1)/3 = 2, so y - 1 = 6 and y = 7. So the requirement is g(k) = 7. g is defined in two pieces, so test each branch separately and check whether the solution lies in that branch's domain. Branch 1, k <= 0, where g(k) = 2(k - 1)² + 2: 2(k - 1)² + 2 = 7 2(k - 1)² = 5 (k - 1)² = 5/2 k - 1 = + sqrt(5/2) or k - 1 = - sqrt(5/2). Now sqrt(5/2) = sqrt5/sqrt2 = sqrt10/2, so k = 1 + sqrt10/2 or k = 1 - sqrt10/2. Since sqrt10 = 3.162..., the first gives k = 2.58, which is NOT <= 0, so it is rejected; the second gives k = -0.581, which does satisfy k <= 0, so it is accepted. Branch 2, 0 < k <= 3, where g(k) = 2 + |2 - k|: 2 + |2 - k| = 7, so |2 - k| = 5. Then 2 - k = 5 or 2 - k = -5, giving k = -3 or k = 7. Neither of these lies in 0 < k <= 3, so this branch gives no solution. (Also check that 7 is a legitimate input to f inverse: for x > ln 2 we have e^x > 2, so e^x - 2 > 0 and f(x) = 1 + 3/(e^x - 2) > 1. The range of f is therefore x > 1, and 7 > 1.) So the only solution is k = 1 - sqrt10 / 2.