Given the functions $f\left(x\right)=x^2$ and $g\left(x\right)=\sqrt{x-8}$, find $(f\circ g)\left(x\right)$ and determine its domain.

• A. $(f∘g)(x)=x-8$ ; $Dom:(-\(\infty\),\(\infty\))$
• B. $(f\(\circ\) g)(x)=\(\sqrt{x^2-8}\)$ ; $Dom:(-∞,∞)$
• C. $(f∘g)(x)=x-8$ ; $Dom:[8,∞)$
• D. $(f\(\circ\) g)(x)=\(\sqrt{x^2-8}\)$ ; $Dom:[8,∞)$

Answer: (C)

Verified step by step guidance

1. Recall that the composition of functions \( (f \circ g)(x) \) means applying \( g \) first, then applying \( f \) to the result of \( g(x) \). So, \( (f \circ g)(x) = f(g(x)) \).
2. Given \( f(x) = x^2 \) and \( g(x) = \sqrt{x - 8} \), substitute \( g(x) \) into \( f \) to get \( (f \circ g)(x) = f(\sqrt{x - 8}) = (\sqrt{x - 8})^2 \).
3. Simplify the expression \( (\sqrt{x - 8})^2 \) by recognizing that squaring a square root cancels out, leaving \( x - 8 \). So, \( (f \circ g)(x) = x - 8 \).
4. Next, determine the domain of \( (f \circ g)(x) \). Since \( g(x) = \sqrt{x - 8} \), the expression inside the square root must be non-negative: \( x - 8 \geq 0 \).
5. Solve the inequality \( x - 8 \geq 0 \) to find \( x \geq 8 \). Therefore, the domain of \( (f \circ g)(x) \) is all real numbers \( x \) such that \( x \geq 8 \), or in interval notation, \( [8, \infty) \).