Given the functions $f\left(x\right)=\frac{1}{x^2-2}$ and $g\left(x\right)=\sqrt{x+2}$, find $(f\circ g)\left(x\right)$ and $(g\circ f)\left(x\right)$.

• A. $(f\(\circ\) g)(x)=\(\frac{1}{x}\)$ ; $(g\(\circ\) f)(x)=\(\sqrt{\frac{2x^2-3}{x^2-2}\)}$
• B. $(f\(\circ\) g)(x)=\(\frac{1}{x}\)$ ; $(g\(\circ\) f)(x)=\(\sqrt{\frac{3}{x^2-2}\)}$
• C. $(f∘g)(x)=x$ ; $(g\(\circ\) f)(x)=\(\sqrt{x^2-2}\)$
• D. $(f∘g)(x)=x$ ; $(g\(\circ\) f)(x)=\(\frac{1}{\sqrt{x+2}\)-2}$

Answer: (A)

Verified step by step guidance

1. Recall that the composition of functions (f \(\circ\) g)(x) means you substitute g(x) into the function f. So, start by writing (f \(\circ\) g)(x) = f(g(x)).
2. Given f(x) = \(\frac{1}{x^2 - 2}\) and g(x) = \(\sqrt{x + 2}\), substitute g(x) into f: (f \(\circ\) g)(x) = \(\frac{1}{(\sqrt{x + 2}\))^2 - 2}.
3. Simplify the expression inside the denominator: (\(\sqrt{x + 2}\))^2 simplifies to x + 2, so the denominator becomes (x + 2) - 2.
4. Simplify the denominator further: (x + 2) - 2 = x, so (f \(\circ\) g)(x) = \(\frac{1}{x}\).
5. Next, find (g \(\circ\) f)(x) by substituting f(x) into g: (g \(\circ\) f)(x) = g(f(x)) = \(\sqrt{f(x) + 2}\) = \(\sqrt{\frac{1}{x^2 - 2}\) + 2}. To simplify, write 2 as \(\frac{2(x^2 - 2)}{x^2 - 2}\) and combine the fractions under the square root.