2. for ( f(x)=\frac{8}{x - 5} ) and ( g(x)=\frac{3}{x} ), find the following composite functions and state…

Explanation:

Step1: Find (f\circ g)

Composite function (f\circ g=f(g(x))). Substitute (g(x)=\frac{3}{x}) into (f(x)): (f(g(x))=\frac{8}{\frac{3}{x}-5}=\frac{8x}{3 - 5x}) For the domain, (g(x)) has (x\neq0), and (\frac{3}{x}-5\neq0) (since denominator of (f) can't be (0)). Solving (\frac{3}{x}-5 = 0) gives (x=\frac{3}{5}). So domain is (x\neq0,\frac{3}{5}).

Step2: Find (g\circ f)

Composite function (g\circ f = g(f(x))). Substitute (f(x)=\frac{8}{x - 5}) into (g(x)): (g(f(x))=\frac{3}{\frac{8}{x - 5}}=\frac{3(x - 5)}{8}) For the domain, (f(x)) has (x\neq5), and since (g(f(x))) is a linear - like function (after simplification) with no additional restrictions from (g) (as long as (f(x)) is defined), domain is (x\neq5).

Step3: Find (f\circ f)

Composite function (f\circ f=f(f(x))). Substitute (f(x)=\frac{8}{x - 5}) into (f(x)): (f(f(x))=\frac{8}{\frac{8}{x - 5}-5}=\frac{8(x - 5)}{8-5(x - 5)}=\frac{8(x - 5)}{33 - 5x}) For the domain, (f(x)) has (x\neq5), and (\frac{8}{x - 5}-5\neq0). Solving (\frac{8}{x - 5}-5 = 0) gives (x=\frac{33}{5}). So domain is (x\neq5,\frac{33}{5}).

Step4: Find (g\circ g)

Composite function (g\circ g = g(g(x))). Substitute (g(x)=\frac{3}{x}) into (g(x)): (g(g(x))=\frac{3}{\frac{3}{x}}=x) For the domain, (g(x)) has (x\neq0), and since (g(g(x))) is (x) (but we still need to consider the domain of the inner function). So domain is (x\neq0).

Answer:

(a) (f\circ g=\frac{8x}{3 - 5x}), domain: (\left{x|x\neq0,x\neq\frac{3}{5}\right}) (b) (g\circ f=\frac{3(x - 5)}{8}), domain: (\left{x|x\neq5\right}) (c) (f\circ f=\frac{8(x - 5)}{33 - 5x}), domain: (\left{x|x\neq5,x\neq\frac{33}{5}\right}) (d) (g\circ g=x), domain: (\left{x|x\neq0\right})