We are given the functions:

• $f(x) = \sqrt{25 - x^2}$
• $g(x) = 5 + x$

We need to find:

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

We will also find the domain of each function and operation. Let's go through each step.

1. $f + g$

$(f + g)(x) = f(x) + g(x) = \sqrt{25 - x^2} + (5 + x)$

Domain of $f + g$

• For $f(x) = \sqrt{25 - x^2}$, we need $25 - x^2 \geq 0$, meaning $x^2 \leq 25$, so $-5 \leq x \leq 5$.
• $g(x) = 5 + x$ has no restrictions, its domain is all real numbers.

Thus, the domain of $f + g$ is $[-5, 5]$.

2. $f - g$

$(f - g)(x) = f(x) - g(x) = \sqrt{25 - x^2} - (5 + x)$

Domain of $f - g$

• The domain of $f - g$ is the same as the domain of $f(x)$, which is $[-5, 5]$.

3. $fg$

$(fg)(x) = f(x) \cdot g(x) = \sqrt{25 - x^2} \cdot (5 + x)$

Domain of $fg$

• The domain is again constrained by $f(x)$, so the domain of $fg$ is $[-5, 5]$.

4. $\frac{f}{g}$

$\left( \frac{f}{g} \right)(x) = \frac{\sqrt{25 - x^2}}{5 + x}$

Domain of $\frac{f}{g}$

• In addition to $-5 \leq x \leq 5$, we need to exclude points where $g(x) = 0$, which happens when $5 + x = 0$, or $x = -5$.

Thus, the domain of $\frac{f}{g}$ is $(-5, 5]$.

Summary of Domains:

1. $f + g$: Domain = $[-5, 5]$
2. $f - g$: Domain = $[-5, 5]$
3. $fg$: Domain = $[-5, 5]$
4. $\frac{f}{g}$: Domain = $(-5, 5]$

Would you like more details on any step, or have any specific questions?

Related Questions:

1. What are the restrictions on $f(x) = \sqrt{25 - x^2}$ and why?
2. How do you find the domain of a square root function in general?
3. How does the domain of a quotient function differ from other operations?
4. Can you solve for specific values of $f + g$ at $x = 0$ or other points?
5. What happens to the domain of $f(x)$ when its form changes to a cubic root instead of a square root?

Tip:

Always remember that the square root function requires the radicand to be non-negative, which affects the domain significantly.