A-Level Functions: worked solution

Question

Functions are f: x → 4/(x−4)², x∈R, x≠4 and g: x → ln(1 + 1/x), x∈R, x>0. Show gf exists, then state its rule, domain and range. [4]

Step 1: find the range of f, because gf exists only if R_f ⊆ D_g. f(x) = 4/(x−4)², x ≠ 4. For every x ≠ 4 we have (x−4)² > 0, so 4/(x−4)² > 0. Hence f(x) > 0 always. As (x−4)² can be made as small as we like (x close to 4) f(x) can be made as large as we like, and as (x−4)² can be made as large as we like f(x) can be made as close to 0 as we like without reaching it. So R_f = (0, ∞). Step 2: compare with the domain of g. g(x) = ln(1 + 1/x) is defined for x > 0, so D_g = (0, ∞). Since R_f = (0, ∞) ⊆ (0, ∞) = D_g, the composite gf exists. Step 3: build the rule by substituting f(x) into g. gf(x) = g(f(x)) = ln(1 + 1/f(x)) 1/f(x) = 1 ÷ 4/(x−4)² = (x−4)²/4, so gf(x) = ln(1 + (x−4)²/4). Step 4: state the domain. The domain of a composite is the domain of the first function applied, so D_gf = D_f = R\{4}. Step 5: state the range by tracking the value through each stage. For x ∈ R\{4}: (x−4)² takes every value in (0, ∞), so (x−4)²/4 takes every value in (0, ∞), so 1 + (x−4)²/4 takes every value in (1, ∞), and since ln is increasing with ln 1 = 0, ln(1 + (x−4)²/4) takes every value in (0, ∞). So R_gf = (0, ∞).