1. **State the problem:** We are given two functions: $$f(x) = \frac{x + 3}{\sqrt{x^2 - 1}}$$ $$g(x) = x^2 + 3$$ We need to determine whether the composition of functions satisfies: $$f \circ g(x) = g \circ f(x)$$ which means checking if: $$f(g(x)) = g(f(x))$$ 2. **Recall the composition of functions:** - The composition $f \circ g(x)$ means applying $g$ first, then $f$ to the result. - The composition $g \circ f(x)$ means applying $f$ first, then $g$ to the result. 3. **Calculate $f(g(x))$:** Substitute $g(x)$ into $f$: $$f(g(x)) = f(x^2 + 3) = \frac{(x^2 + 3) + 3}{\sqrt{(x^2 + 3)^2 - 1}} = \frac{x^2 + 6}{\sqrt{(x^2 + 3)^2 - 1}}$$ 4. **Calculate $g(f(x))$:** Substitute $f(x)$ into $g$: $$g(f(x)) = g\left(\frac{x + 3}{\sqrt{x^2 - 1}}\right) = \left(\frac{x + 3}{\sqrt{x^2 - 1}}\right)^2 + 3 = \frac{(x + 3)^2}{x^2 - 1} + 3$$ Simplify the right side: $$\frac{(x + 3)^2}{x^2 - 1} + 3 = \frac{(x + 3)^2 + 3(x^2 - 1)}{x^2 - 1}$$ Expand numerator: $$(x + 3)^2 + 3(x^2 - 1) = (x^2 + 6x + 9) + 3x^2 - 3 = 4x^2 + 6x + 6$$ So: $$g(f(x)) = \frac{4x^2 + 6x + 6}{x^2 - 1}$$ 5. **Compare $f(g(x))$ and $g(f(x))$:** - $f(g(x)) = \frac{x^2 + 6}{\sqrt{(x^2 + 3)^2 - 1}}$ - $g(f(x)) = \frac{4x^2 + 6x + 6}{x^2 - 1}$ These two expressions are clearly different functions because: - $f(g(x))$ has a square root in the denominator involving $(x^2 + 3)^2 - 1$. - $g(f(x))$ is a rational function with a polynomial numerator and denominator. 6. **Conclusion:** Since $f(g(x)) \neq g(f(x))$ for general $x$, the compositions are not equal. **Final answer:** $$f \circ g(x) \neq g \circ f(x)$$ Therefore, the compositions $f \circ g$ and $g \circ f$ are not equal functions.