Questions: The functions f and g are defined as f(x)=x-5, g(x)=√(x+6). a) Find the domain of f, g, f+g, f-g, fg, ff, f/g, and g/f. b) Find (f+g)(x), (f-g)(x), (fg)(x), (f)(x),(f/g)(x), and (g/f)(x). a) The domain of f is (-∞, ∞). (Type your answer in interval notation.) The domain of g is . (Type your answer in interval notation.)

Solution Steps

To solve the given problem, we need to determine the domains of the functions $f$ and $g$, as well as their combinations. Then, we will find the expressions for the combined functions.

1. Domain of $f(x) = x - 5$: Since $f(x)$ is a linear function, its domain is all real numbers, $(-\infty, \infty)$.
2. Domain of $g(x) = \sqrt{x + 6}$: The expression inside the square root must be non-negative, so $x + 6 \geq 0$. Therefore, the domain is $[-6, \infty)$.

For the combined functions:

• $f + g$: The domain is the intersection of the domains of $f$ and $g$, which is $[-6, \infty)$.
• $f - g$: The domain is the same as $f + g$, which is $[-6, \infty)$.
• $f \cdot g$: The domain is the same as $f + g$, which is $[-6, \infty)$.
• $\frac{f}{g}$: The domain is $[-6, \infty)$ excluding points where $g(x) = 0$. Since $g(x) = \sqrt{x + 6}$, $g(x) = 0$ when $x = -6$. Thus, the domain is $(-6, \infty)$.
• $\frac{g}{f}$: The domain is $[-6, \infty)$ excluding points where $f(x) = 0$. Since $f(x) = x - 5$, $f(x) = 0$ when $x = 5$. Thus, the domain is $[-6, 5) \cup (5, \infty)$.

1. $(f + g)(x)$: This is $f(x) + g(x)$.
2. $(f - g)(x)$: This is $f(x) - g(x)$.
3. $(f \cdot g)(x)$: This is $f(x) \cdot g(x)$.
4. $\left(\frac{f}{g}\right)(x)$: This is $\frac{f(x)}{g(x)}$.
5. $\left(\frac{g}{f}\right)(x)$: This is $\frac{g(x)}{f(x)}$.

The function $f(x) = x - 5$ is a linear function. Linear functions are defined for all real numbers.

$$\text{Domain of } f: (-\infty, \infty)$$

The function $g(x) = \sqrt{x + 6}$ is defined when the expression inside the square root is non-negative.

$$x + 6 \geq 0 \implies x \geq -6$$

$$\text{Domain of } g: [-6, \infty)$$

• $f + g$: The domain is the intersection of the domains of $f$ and $g$.

$$\text{Domain of } f + g: [-6, \infty)$$

• $f - g$: The domain is the same as $f + g$.

$$\text{Domain of } f - g: [-6, \infty)$$

• $f \cdot g$: The domain is the same as $f + g$.

$$\text{Domain of } f \cdot g: [-6, \infty)$$

• $\frac{f}{g}$: The domain is $[-6, \infty)$ excluding points where $g(x) = 0$. Since $g(x) = \sqrt{x + 6}$, $g(x) = 0$ when $x = -6$.

$$\text{Domain of } \frac{f}{g}: (-6, \infty)$$

• $\frac{g}{f}$: The domain is $[-6, \infty)$ excluding points where $f(x) = 0$. Since $f(x) = x - 5$, $f(x) = 0$ when $x = 5$.

$$\text{Domain of } \frac{g}{f}: [-6, 5) \cup (5, \infty)$$

• $(f + g)(x)$:

$$(f + g)(x) = f(x) + g(x) = x - 5 + \sqrt{x + 6}$$

• $(f - g)(x)$:

$$(f - g)(x) = f(x) - g(x) = x - 5 - \sqrt{x + 6}$$

• $(f \cdot g)(x)$:

$$(f \cdot g)(x) = f(x) \cdot g(x) = (x - 5) \sqrt{x + 6}$$

• $\left(\frac{f}{g}\right)(x)$:

$$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x - 5}{\sqrt{x + 6}}$$

• $\left(\frac{g}{f}\right)(x)$:

$$\left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)} = \frac{\sqrt{x + 6}}{x - 5}$$

Final Answer