
Question content area top

Part 1

Let 

​f(x)equals=StartRoot 3 x minus 1 EndRoot3x−1 

and 

​g(x)equals=StartFraction 1 Over x EndFraction1x. 

Find 

​(fplus+​g)(x), 

​(fminus−​g)(x), 

​(fg)(x), and 

left parenthesis StartFraction f Over g EndFraction right parenthesis left parenthesis x right parenthesisfg(x). 

Give the domain of each.

Question content area bottom

Part 1

​(fplus+​g)(x)equals=enter your response here 

​(Simplify your​ answer.)

Part 2

​(fminus−​g)(x)equals=enter your response here 

​(Simplify your​ answer.)

Part 3

​(fg)(x)equals=enter your response here 

​(Simplify your​ answer.)

Part 4

left parenthesis StartFraction f Over g EndFraction right parenthesisfg​(x)equals=enter your response here 

​(Simplify your​ answer.)

Part 5

The domain of 

fplus+g 

is 

enter your response here.

​(Type your answer in interval​ notation.)

Part 6

The domain of 

fminus−g 

is 

enter your response here.

​(Type your answer in interval​ notation.)

Part 7

The domain of fg is 

enter your response here.

​(Type your answer in interval​ notation.)

Part 8

The domain of 

StartFraction f Over g EndFractionfg 

is 

enter your response here.

​(Type your answer in interval​ notation.)

Let's break this down step by step. Given the functions:

- \( f(x) = \sqrt{3x - 1} \)
- \( g(x) = \frac{1}{x} \)

We are asked to find:

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

Afterward, we will determine the domains of each expression.

---

### Part 1: \( (f + g)(x) \)

\[
(f + g)(x) = f(x) + g(x)
\]

\[
= \sqrt{3x - 1} + \frac{1}{x}
\]

The domain of this expression requires that:
- \( 3x - 1 \geq 0 \), i.e., \( x \geq \frac{1}{3} \), because the square root function is defined only for non-negative values.
- \( x \neq 0 \), because \( g(x) = \frac{1}{x} \) is undefined at \( x = 0 \).

Thus, the domain of \( (f + g)(x) \) is:

\[
x \in \left(\frac{1}{3}, \infty \right)
\]

---

### Part 2: \( (f - g)(x) \)

\[
(f - g)(x) = f(x) - g(x)
\]

\[
= \sqrt{3x - 1} - \frac{1}{x}
\]

The domain is the same as for \( (f + g)(x) \), since the conditions on \( f(x) \) and \( g(x) \) have not changed. Therefore, the domain of \( (f - g)(x) \) is:

\[
x \in \left(\frac{1}{3}, \infty \right)
\]

---

### Part 3: \( (fg)(x) \)

\[
(fg)(x) = f(x) \cdot g(x)
\]

\[
= \sqrt{3x - 1} \cdot \frac{1}{x} = \frac{\sqrt{3x - 1}}{x}
\]

The domain for \( (fg)(x) \) is still constrained by \( f(x) \) and \( g(x) \). Specifically:
- \( x \geq \frac{1}{3} \), to satisfy the square root.
- \( x \neq 0 \), since division by zero is undefined.

Thus, the domain of \( (fg)(x) \) is:

\[
x \in \left(\frac{1}{3}, \infty \right)
\]

---

### Part 4: \( \left(\frac{f}{g}\right)(x) \)

\[
\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{3x - 1}}{\frac{1}{x}} = x\sqrt{3x - 1}
\]

For this to be valid:
- \( x \geq \frac{1}{3} \), because of the square root.
- \( x \neq 0 \), since we are dividing by \( g(x) \).

Thus, the domain of \( \left(\frac{f}{g}\right)(x) \) is:

\[
x \in \left(\frac{1}{3}, \infty \right)
\]

---

### Domain Summary:
- The domain of \( (f + g)(x) \): \( \left(\frac{1}{3}, \infty\right) \)
- The domain of \( (f - g)(x) \): \( \left(\frac{1}{3}, \infty\right) \)
- The domain of \( (fg)(x) \): \( \left(\frac{1}{3}, \infty\right) \)
- The domain of \( \left(\frac{f}{g}\right)(x) \): \( \left(\frac{1}{3}, \infty\right) \)

Would you like further details on any part of the solution?

### Related Questions:
1. What happens if the expression inside the square root is negative?
2. How does the concept of domain differ for different operations on functions?
3. Can you describe the behavior of \( f(x) = \sqrt{3x - 1} \) as \( x \to \infty \)?
4. What would the graph of \( g(x) = \frac{1}{x} \) look like, and how does that affect the domain?
5. How would adding or subtracting constants affect the domain of these functions?

**Tip:** Always check the domains of individual functions before performing operations to ensure the result is valid.

Let f(x) = sqrt(3x - 1) and g(x) = 1/x. Find (f+g)(x), (f−g)(x), (fg)(x), and (f/g)(x). Give the domain of each.

Learn how to perform operations on the functions f(x) = sqrt(3x - 1) and g(x) = 1/x. This problem covers the sum, difference, product, and quotient of functions and includes domain restrictions based on square root and rational function rules. Suitable for students in Grades 9-12.

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