A-Level Functions: worked solution

Question

With f (as defined) and g(x)=3+eˣ, and knowing fg exists, determine the exact k satisfying fg(k)=a/7.

Step 1 - decide WHICH branch of the piecewise function f is used. fg(k) means f(g(k)), so the number being fed into f is g(k) = 3 + e^k. For every real k, e^k > 0. Therefore g(k) = 3 + e^k > 3. Since 3 > 2, the input to f always lands in the x > 2 part of the domain, so f acts by the rule f(x) = a/x. (This is also exactly why the composite fg exists: the whole range of g sits inside the domain of f.) Step 2 - write down fg(k) using that branch. fg(k) = f(3 + e^k) = a / (3 + e^k) Step 3 - set it equal to the value required. a / (3 + e^k) = a/7 Step 4 - a is a non-zero constant, so divide both sides by a. 1 / (3 + e^k) = 1/7 Two fractions with equal numerators are equal only when their denominators are equal, so 3 + e^k = 7 Step 5 - solve for k. e^k = 7 - 3 = 4 Take natural logarithms of both sides: k = ln 4 Check: g(ln 4) = 3 + e^(ln 4) = 3 + 4 = 7, which is greater than 2, so the branch used was the right one, and f(7) = a/7 as required.