10. Perform the indicated operation and state the domain. \( f(x)=2 x-5, g(x)=x^{-2} h(x)=3 x+4 \) a. \( f(g(x)) \) b. \( h(f(x)) \) c. \( g(h(x)) \)

We are given the functions:   f(x) = 2x – 5   g(x) = x^(–2) (which is the same as 1/x²)   h(x) = 3x + 4 Now, let’s perform each of the indicated compositions and state the domain. ───────────────────────────── a. Find f(g(x)) Step 1: Substitute g(x) into f(x):   f(g(x)) = 2·(g(x)) – 5 = 2·(x^(–2)) – 5 = 2/x² – 5 Step 2: Domain Consideration: g(x) = 1/x² is defined only when x² ≠ 0, i.e., x ≠ 0. Thus, the domain of f(g(x)) is all real numbers except x = 0. ───────────────────────────── b. Find h(f(x)) Step 1: Substitute f(x) into h(x):   h(f(x)) = 3·(f(x)) + 4 = 3·(2x – 5) + 4 = 6x – 15 + 4 = 6x – 11 Step 2: Domain Consideration: The function f(x) is linear (defined for all x), and h(x) is also linear (defined for all x). Thus, the domain of h(f(x)) is all real numbers. ───────────────────────────── c. Find g(h(x)) Step 1: Substitute h(x) into g(x):   g(h(x)) = (h(x))^(–2) = [3x + 4]^(–2) = 1/(3x + 4)² Step 2: Domain Consideration: For g(h(x)) to be defined, the expression (3x + 4)² must not be zero. Solve 3x + 4 = 0 → x = –4/3. Thus, the domain of g(h(x)) is all real numbers except x = –4/3. ───────────────────────────── Summary: a. f(g(x)) = 2/x² – 5, domain: x ≠ 0 b. h(f(x)) = 6x – 11, domain: all real numbers c. g(h(x)) = 1/(3x + 4)², domain: x ≠ –4/3