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 expressions for \( f + g \), \( f - g \), and \( \frac{f}{g} \) given the functions:

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

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

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

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

Now, simplify the expression:

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

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

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

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

Now, simplify the expression:

\[ f - g = x + 1 - 2x + 3 = -x + 4 \]

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

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

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

### Step 4: State the domain restriction

Since \( g(x) = 2x - 3 \) is in the denominator, we need to ensure it is not equal to zero:

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

### Final Results

Thus, the results are:

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

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