## College Algebra (6th Edition)

$f + g = \sqrt{x + 4} + \sqrt{x - 1}$ $Domain_{f+g}: x\geq 1$ $f-g = \sqrt{x + 4} - \sqrt{x -1}$ $Domain_{f-g}: x\geq 1$ $f \times g = \sqrt{(x+4)(x-1)}$ $Domain_{fg} = (x\leq -4) OR (x \geq 1)$ $\frac{f}{g} = \frac{\sqrt{(x+4)(x-1)}}{x-1}$ $Domain_{\frac{f}{g}} = (x\leq -4) OR (x \gt 1)$
$$f(x) = \sqrt{x + 4}$$ $$g(x) = \sqrt{x - 1}$$ $f + g = \sqrt{x + 4} + \sqrt{x - 1}$ $Domain_{f+g}: x + 4 \geq 0; x-1\geq 0$ $Domain_{f+g}: x\geq -4; x\geq 1$ Therefore, $Domain_{f+g}: x\geq 1$ $f-g = \sqrt{x + 4} - \sqrt{x -1}$ $Domain_{f-g}: x + 4 \geq 0; x-1\geq 0$ $Domain_{f-g}: x\geq -4; x\geq 1$ Therefore, $Domain_{f-g}: x\geq 1$ $f \times g = \sqrt{x + 4} \times \sqrt{x-1}$ $f \times g = \sqrt{(x+4)(x-1)}$ **To find the domain of $fg$, since we already have the radical function in terms of its factors, we need to find the values of $x$ that, when applied in both factors, always yields $\geq 0$ once they're multiplied. The values that satisfy this condition are: $Domain_{fg} = (x\leq -4) OR (x \geq 1)$ $\frac{f}{g} = \frac{\sqrt{x+4}}{\sqrt{x-1}}$ $\frac{f}{g} = \frac{\sqrt{(x+4)(x-1)}}{x-1}$ **To find its domain, we can apply the same logic as the previous exercise and reach a preliminary conclusion of $Domain_{\frac{f}{g}} = (x\leq -4) OR (x \geq 1)$. However, since $x-1$ in the denominator implies that $x\ne 1$ or otherwise the function would be undefined, we can conclude that the domain is as follows: $Domain_{\frac{f}{g}} = (x\leq -4) OR (x \gt 1)$