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. \( g(h(x)) \)

Evaluate the expression by following steps: - step0: Evaluate: \(3x+4\) - step1: Substitute: \(3\left(2x-5\right)+4\) - step2: Expand the expression: \(6x-15+4\) - step3: Add the numbers: \(6x-11\) Substitute \( x=3x+4 \) into the expression \( x^{-2} \). Evaluate the expression by following steps: - step0: Evaluate: \(x^{-2}\) - step1: Substitute: \(\left(3x+4\right)^{-2}\) - step2: Simplify: \(\frac{1}{\left(3x+4\right)^{2}}\) - step3: Calculate: \(\frac{1}{9x^{2}+24x+16}\) Substitute \( x=x^{-2} \) into the expression \( 2x-5 \). Evaluate the expression by following steps: - step0: Evaluate: \(2x-5\) - step1: Substitute: \(2x^{-2}-5\) - step2: Simplify: \(\frac{2}{x^{2}}-5\) - step3: Reduce fractions to a common denominator: \(\frac{2}{x^{2}}-\frac{5x^{2}}{x^{2}}\) - step4: Transform the expression: \(\frac{2-5x^{2}}{x^{2}}\) Let's perform the indicated operations step by step and state the domain for each composition. ### a. \( f(g(x)) \) 1. **Substituting \( g(x) \) into \( f(x) \)**: \[ f(g(x)) = f(x^{-2}) = 2(x^{-2}) - 5 = \frac{2 - 5x^{2}}{x^{2}} \] 2. **Domain**: The function \( g(x) = x^{-2} \) is defined for all \( x \) except \( x = 0 \). Therefore, the domain of \( f(g(x)) \) is: \[ x \in \mathbb{R} \setminus \{0\} \] ### b. \( h(f(x)) \) 1. **Substituting \( f(x) \) into \( h(x) \)**: \[ h(f(x)) = h(2x - 5) = 3(2x - 5) + 4 = 6x - 15 + 4 = 6x - 11 \] 2. **Domain**: The function \( f(x) = 2x - 5 \) is defined for all \( x \in \mathbb{R} \). Therefore, the domain of \( h(f(x)) \) is: \[ x \in \mathbb{R} \] ### c. \( g(h(x)) \) 1. **Substituting \( h(x) \) into \( g(x) \)**: \[ g(h(x)) = g(3x + 4) = (3x + 4)^{-2} = \frac{1}{(3x + 4)^{2}} \] 2. **Domain**: The function \( h(x) = 3x + 4 \) is defined for all \( x \in \mathbb{R} \). However, \( g(h(x)) \) is undefined when \( h(x) = 0 \), which occurs when \( 3x + 4 = 0 \) or \( x = -\frac{4}{3} \). Therefore, the domain of \( g(h(x)) \) is: \[ x \in \mathbb{R} \setminus \left\{-\frac{4}{3}\right\} \] ### Summary of Results - \( f(g(x)) = \frac{2 - 5x^{2}}{x^{2}} \) with domain \( x \in \mathbb{R} \setminus \{0\} \) - \( h(f(x)) = 6x - 11 \) with domain \( x \in \mathbb{R} \) - \( g(h(x)) = \frac{1}{(3x + 4)^{2}} \) with domain \( x \in \mathbb{R} \setminus \left\{-\frac{4}{3}\right\} \)