Question

for $f(x) = x^2 + 8$ and $g(x) = \sqrt{x - 2}$, find the following composite functions and state the domain of each.\
(a) $f \circ g$ \quad (b) $g \circ f$ \quad (c) $f \circ f$ \quad (d) $g \circ g$\
\
(a) $(f \circ g)(x) = x + 6$ (simplify your answer.)\
select the correct choice below and fill in any answer boxes within your choice.\
\\(\boldsymbol{\circ}\\) a. the domain of $f \circ g$ is \\{x \\(\square\\)\\}.\
\quad (type an inequality. simplify your answer. use integers or fractions for any numbers in\
\quad the expression.)\
\\(\boldsymbol{\circ}\\) b. the domain of $f \circ g$ is all real numbers.

Explanation:

Step1: Recall the definition of composite function

To find \((f \circ g)(x)\), we substitute \(g(x)\) into \(f(x)\). Given \(f(x)=x^{2}+8\) and \(g(x)=\sqrt{x - 2}\), so \((f\circ g)(x)=f(g(x))=(\sqrt{x - 2})^{2}+8\).

Step2: Simplify the expression

Simplify \((\sqrt{x - 2})^{2}+8\). Since \((\sqrt{a})^{2}=a\) for \(a\geq0\), here \(a = x - 2\) (so \(x-2\geq0\) for the square root to be defined), then \((\sqrt{x - 2})^{2}+8=(x - 2)+8=x + 6\).

Step3: Determine the domain of \(f\circ g\)

The domain of a composite function \(f\circ g\) is the set of all \(x\) such that \(x\) is in the domain of \(g\) and \(g(x)\) is in the domain of \(f\).

• The domain of \(g(x)=\sqrt{x - 2}\) requires the expression inside the square root to be non - negative, so \(x-2\geq0\), which gives \(x\geq2\).
• The domain of \(f(x)=x^{2}+8\) is all real numbers, and since \(g(x)=\sqrt{x - 2}\) (for \(x\geq2\)) gives a real number, the domain of \(f\circ g\) is the domain of \(g\), which is \(x\geq2\).

Answer:

For the domain of \(f\circ g\), we have \(x\geq2\), so the inequality to fill in the box is \(x\geq2\)