3. for ( f(x)=sqrt{x} ) and ( g(x)=6x + 7 ), find the following composite functions and state the domain of…

Explanation:

Step1: Find (f\circ g)

Composite function (f\circ g=f(g(x))). Substitute (g(x) = 6x + 7) into (f(x)): (f(g(x))=\sqrt{6x + 7}). For the domain, the expression under the square - root must be non - negative. So (6x+7\geq0), which gives (x\geq-\frac{7}{6}).

Step2: Find (g\circ f)

Composite function (g\circ f = g(f(x))). Substitute (f(x)=\sqrt{x}) into (g(x)): (g(f(x))=6\sqrt{x}+7). For the domain, the expression under the square - root must be non - negative. So (x\geq0).

Step3: Find (f\circ f)

Composite function (f\circ f=f(f(x))). Substitute (f(x)=\sqrt{x}) into (f(x)): (f(f(x))=\sqrt{\sqrt{x}}=\sqrt[4]{x}). For the domain, the expression under the fourth - root (which is equivalent to a square - root of a square - root) must be non - negative. So (x\geq0).

Step4: Find (g\circ g)

Composite function (g\circ g = g(g(x))). Substitute (g(x)=6x + 7) into (g(x)): (g(g(x))=6(6x + 7)+7=36x+42 + 7=36x+49). Since it is a linear function, the domain is all real numbers, (x\in(-\infty,\infty)).

Answer:

(a) (f\circ g=\sqrt{6x + 7}), domain: (\left[-\frac{7}{6},\infty\right)) (b) (g\circ f=6\sqrt{x}+7), domain: ([0,\infty)) (c) (f\circ f=\sqrt[4]{x}), domain: ([0,\infty)) (d) (g\circ g = 36x+49), domain: ((-\infty,\infty))