Let's work through the parts of the problem one by one.

Given:

$f(x) = \sqrt{5x}$ $g(x) = 7x - 4$

Part (a): Find $(f + g)(x)$

The sum of two functions is defined as: $(f + g)(x) = f(x) + g(x)$ Substitute the given functions: $(f + g)(x) = \sqrt{5x} + (7x - 4)$ So: $(f + g)(x) = \sqrt{5x} + 7x - 4$

Domain of $f + g$

For the domain of $f + g$, we must consider the domains of both functions $f(x)$ and $g(x)$.

• $f(x) = \sqrt{5x}$ is defined when $5x \geq 0$, i.e., $x \geq 0$.
• $g(x) = 7x - 4$ is a linear function and is defined for all real numbers.

Therefore, the domain of $f + g$ is $x \geq 0$.

Answer for part (a): $(f + g)(x) = \sqrt{5x} + 7x - 4$ Domain: $x \geq 0$.

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Part (b): Find $(f - g)(x)$

The difference of two functions is defined as: $(f - g)(x) = f(x) - g(x)$ Substitute the given functions: $(f - g)(x) = \sqrt{5x} - (7x - 4)$ Simplifying: $(f - g)(x) = \sqrt{5x} - 7x + 4$

Domain of $f - g$

The domain is still determined by $f(x)$, which is $x \geq 0$.

Answer for part (b): $(f - g)(x) = \sqrt{5x} - 7x + 4$ Domain: $x \geq 0$.

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Part (c): Find $(f \times g)(x)$

The product of two functions is defined as: $(f \times g)(x) = f(x) \times g(x)$ Substitute the given functions: $(f \times g)(x) = \sqrt{5x} \times (7x - 4)$ This simplifies to: $(f \times g)(x) = \sqrt{5x}(7x - 4)$ This is the simplified form, as the product involves a square root.

Domain of $f \times g$

As before, $f(x)$ imposes the condition $x \geq 0$, so the domain of $f \times g$ is $x \geq 0$.

Answer for part (c): $(f \times g)(x) = \sqrt{5x}(7x - 4)$ Domain: $x \geq 0$.

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Part (d): Find $\left( \frac{f}{g} \right)(x)$

The quotient of two functions is defined as: $\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)}$ Substitute the given functions: $\left( \frac{f}{g} \right)(x) = \frac{\sqrt{5x}}{7x - 4}$

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

For the domain, we need:

1. $f(x)$ is defined, i.e., $x \geq 0$.
2. $g(x) \neq 0$, i.e., $7x - 4 \neq 0$.

Solve for when $7x - 4 = 0$: $7x = 4 \quad \Rightarrow \quad x = \frac{4}{7}$ Thus, the domain is $x \geq 0$ and $x \neq \frac{4}{7}$.

Answer for part (d): $\left( \frac{f}{g} \right)(x) = \frac{\sqrt{5x}}{7x - 4}$ Domain: $x \geq 0$ and $x \neq \frac{4}{7}$.

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Summary of Answers:

1. $(f + g)(x) = \sqrt{5x} + 7x - 4$, Domain: $x \geq 0$
2. $(f - g)(x) = \sqrt{5x} - 7x + 4$, Domain: $x \geq 0$
3. $(f \times g)(x) = \sqrt{5x}(7x - 4)$, Domain: $x \geq 0$
4. $\left( \frac{f}{g} \right)(x) = \frac{\sqrt{5x}}{7x - 4}$, Domain: $x \geq 0$ and $x \neq \frac{4}{7}$

Would you like further details or have any questions about these calculations?

Here are 5 related questions you might find helpful:

1. How do you find the domain of a square root function?
2. What happens to the domain when you combine functions involving a square root and a linear term?
3. How do you simplify expressions involving square roots and linear functions?
4. What are common methods to solve domain restrictions involving denominators?
5. How does multiplication of functions impact the domain?

Tip: Always check for restrictions from both square roots and denominators when determining the domain of combined functions!