16. \( f(x)=-\frac{3}{2} x-4, \quad g(x)=-\frac{2 x+8}{3} \)

We are given   f(x) = –(3/2)x – 4    and  g(x) = –(2x + 8)/3. A natural thing to try is to compute the compositions f(g(x)) and g(f(x)). Let’s do that step by step. Step 1. Compute f(g(x)):   f(g(x)) = –(3/2)[g(x)] – 4 Substitute g(x) = –(2x + 8)/3:   f(g(x)) = –(3/2)·[–(2x + 8)/3] – 4 Notice that the negatives cancel:   = (3/2)·((2x + 8)/3) – 4 Now the factor 3 in the numerator and denominator cancel:   = (2x + 8)/2 – 4 Factor the 1/2:   = x + 4 – 4 = x. So, f(g(x)) = x. Step 2. Compute g(f(x)):   g(f(x)) = –[2·f(x) + 8] / 3 Substitute f(x) = –(3/2)x – 4:   = –[2(–(3/2)x – 4) + 8] / 3 Simplify inside the bracket:   2(–(3/2)x) = –3x  and  2(–4) = –8,   thus: –3x – 8 + 8 = –3x. So we get:   g(f(x)) = –(–3x)/3 = 3x/3 = x. Thus, g(f(x)) = x. Since both compositions yield x, we conclude that f and g are inverses of each other. Answer:   f(g(x)) = x  and  g(f(x)) = x,   so g(x) is the inverse of f(x), and we can write f⁻¹(x) = g(x) = –(2x + 8)/3.