1. **Problem Statement:** Find the domain of the function $$f(x) = \frac{\sqrt{x+3}}{\sqrt{x^2-16}}$$ 2. **Formula and Rules:** - The domain of a function includes all values of $x$ for which the function is defined. - The square root function $\sqrt{y}$ is defined only for $y \geq 0$. - The denominator cannot be zero. 3. **Step 1: Domain restrictions from numerator** $$x+3 \geq 0 \implies x \geq -3$$ 4. **Step 2: Domain restrictions from denominator** $$x^2 - 16 > 0$$ This is because the denominator cannot be zero or negative (square root of negative is undefined, and division by zero is undefined). 5. **Step 3: Solve inequality for denominator** $$x^2 - 16 > 0$$ $$ (x-4)(x+4) > 0$$ 6. **Step 4: Analyze intervals for denominator** - For $x < -4$, both $(x-4)$ and $(x+4)$ are negative, so product is positive. - For $-4 < x < 4$, one factor is negative, one positive, product negative. - For $x > 4$, both factors positive, product positive. So, $$x < -4 \quad \text{or} \quad x > 4$$ 7. **Step 5: Combine numerator and denominator restrictions** - Numerator: $x \geq -3$ - Denominator: $x < -4$ or $x > 4$ The intersection is only where both conditions hold: - $x \geq -3$ and $x < -4$ is impossible. - $x \geq -3$ and $x > 4$ is $x > 4$. 8. **Step 6: Final domain** $$\boxed{(4, \infty)}$$ This means the function is defined for all real numbers greater than 4.