Given the functions $f(x) = x + 1$ and $g(x) = 3x - 4$, let's evaluate the following operations:

1. $(f + g)(x)$: $(f + g)(x) = f(x) + g(x) = (x + 1) + (3x - 4) = x + 1 + 3x - 4 = 4x - 3$

2. $(f - g)(x)$: $(f - g)(x) = f(x) - g(x) = (x + 1) - (3x - 4) = x + 1 - 3x + 4 = -2x + 5$

3. $(f \times g)(x)$: $(f \times g)(x) = f(x) \cdot g(x) = (x + 1)(3x - 4) = 3x^2 - 4x + 3x - 4 = 3x^2 - x - 4$

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

Summary:

1. $(f + g)(x) = 4x - 3$
2. $(f - g)(x) = -2x + 5$
3. $(f \times g)(x) = 3x^2 - x - 4$
4. $\left(\frac{f}{g}\right)(x) = \frac{x + 1}{3x - 4}$

Would you like more details or have any questions?

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Here are 5 related questions to deepen understanding:

1. What is the value of $(f + g)(2)$?
2. What happens when you substitute $x = 0$ in $\left(\frac{f}{g}\right)(x)$?
3. How do the graphs of $f(x)$ and $g(x)$ compare?
4. Can you factorize $(f \times g)(x)$?
5. What is the domain of $\left(\frac{f}{g}\right)(x)$?

Tip: Always check for restrictions in the domain when dividing functions. In $\frac{f(x)}{g(x)}$, ensure the denominator $g(x) \neq 0$.