Range of the function `f (x) = (1)/( 3x +2)` is

Options

A. R
B. `R -{0}`
C. ` (0,oo)`
D. `R- {-(2)/(3)}`

Answer

To find the range of the function \( f(x) = \frac{1}{3x + 2} \), we will follow these steps:

### Step 1: Set up the equation
We start with the function:
\[ y = \frac{1}{3x + 2} \]

### Step 2: Rearrange the equation
To find the values of \( y \), we can rearrange the equation to express \( x \) in terms of \( y \):
\[ 3x + 2 = \frac{1}{y} \]

### Step 3: Solve for \( x \)
Now, we solve for \( x \):
\[ 3x = \frac{1}{y} - 2 \]
\[ x = \frac{1}{3y} - \frac{2}{3} \]

### Step 4: Identify restrictions on \( y \)
From the rearranged equation, we see that \( y \) cannot be equal to 0 because it would make the denominator of the original function undefined. Therefore, \( y \) can take any real number value except for 0.

### Step 5: Write the range
Thus, the range of the function \( f(x) \) is:
\[ y \in \mathbb{R} \setminus \{0\} \]
This means \( y \) can be any real number except 0.

### Final Answer
The range of the function \( f(x) = \frac{1}{3x + 2} \) is \( \mathbb{R} \setminus \{0\} \).

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