Answer
$f + g = \sqrt{x + 4} + \sqrt{x - 1}$
$Domain_{f+g}: x\geq 1$
$f-g = \sqrt{x + 4} - \sqrt{x -1}$
$Domain_{f-g}: x\geq 1$
$f \times g = \sqrt{(x+4)(x-1)}$
$Domain_{fg} = (x\leq -4) OR (x \geq 1)$
$\frac{f}{g} = \frac{\sqrt{(x+4)(x-1)}}{x-1}$
$Domain_{\frac{f}{g}} = (x\leq -4) OR (x \gt 1)$
Work Step by Step
$$f(x) = \sqrt{x + 4}$$ $$g(x) = \sqrt{x - 1}$$
$f + g = \sqrt{x + 4} + \sqrt{x - 1}$
$Domain_{f+g}: x + 4 \geq 0; x-1\geq 0$
$Domain_{f+g}: x\geq -4; x\geq 1$
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:
$Domain_{\frac{f}{g}} = (x\leq -4) OR (x \gt 1)$
