Let `f, g : R -> R` be defined, respectively by `f(x) = x + 1, g(x) = 2x - 3`. Find `f + g, f g` and `f/g`.

Explanation

To solve the problem, we need to find the sum, difference, and quotient of the functions \( f \) and \( g \) defined as follows:

- \( f(x) = x + 1 \)
- \( g(x) = 2x - 3 \)

### Step 1: Find \( f + g \)

To find \( f + g \), we add the two functions together:

\[ f + g = f(x) + g(x) = (x + 1) + (2x - 3) \]

Now, combine like terms:

\[ f + g = x + 1 + 2x - 3 = 3x - 2 \]

### Step 2: Find \( f - g \)

Next, we find \( f - g \) by subtracting \( g \) from \( f \):

\[ f - g = f(x) - g(x) = (x + 1) - (2x - 3) \]

Distributing the negative sign:

\[ f - g = x + 1 - 2x + 3 \]

Now, combine like terms:

\[ f - g = -x + 4 \]

### Step 3: Find \( \frac{f}{g} \)

Now, we find \( \frac{f}{g} \) by dividing \( f \) by \( g \):

\[ \frac{f}{g} = \frac{f(x)}{g(x)} = \frac{x + 1}{2x - 3} \]

However, we need to consider the condition that \( g(x) \neq 0 \):

\[ 2x - 3 \neq 0 \implies x \neq \frac{3}{2} \]

### Final Results

1. \( f + g = 3x - 2 \)
2. \( f - g = -x + 4 \)
3. \( \frac{f}{g} = \frac{x + 1}{2x - 3}, \text{ where } x \neq \frac{3}{2} \)