1. **State the problem:** We want to analyze the function $g(x) = \arccos\left(\frac{4x}{4 + x^2}\right)$.\n\n2. **Recall the domain of arccos:** The function $\arccos(y)$ is defined only for $y$ in the interval $[-1,1]$. So we must ensure that the expression inside the arccos, $\frac{4x}{4 + x^2}$, lies within $[-1,1]$.\n\n3. **Analyze the inner function:** Let $f(x) = \frac{4x}{4 + x^2}$. We want to find the domain where $-1 \leq f(x) \leq 1$.\n\n4. **Check the range of $f(x)$:**\n- The denominator $4 + x^2$ is always positive.\n- To find extrema, differentiate $f(x)$:\n$$f'(x) = \frac{4(4 + x^2) - 4x(2x)}{(4 + x^2)^2} = \frac{16 + 4x^2 - 8x^2}{(4 + x^2)^2} = \frac{16 - 4x^2}{(4 + x^2)^2}.$$\n\n5. **Set derivative to zero to find critical points:**\n$$16 - 4x^2 = 0 \implies 4x^2 = 16 \implies x^2 = 4 \implies x = \pm 2.$$\n\n6. **Evaluate $f(x)$ at critical points:**\n$$f(2) = \frac{4 \cdot 2}{4 + 4} = \frac{8}{8} = 1,$$\n$$f(-2) = \frac{4 \cdot (-2)}{4 + 4} = \frac{-8}{8} = -1.$$\n\n7. **Check limits at infinity:**\n$$\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \frac{4x}{4 + x^2} = 0.$$\n\n8. **Conclusion on range:** The function $f(x)$ attains values between $-1$ and $1$ for all real $x$, so the domain of $g(x)$ is all real numbers.\n\n9. **Final expression:**\n$$g(x) = \arccos\left(\frac{4x}{4 + x^2}\right).$$\n\nThis function is defined for all real $x$ and outputs values in $[0, \pi]$ because of the arccos range.