If `f(x)={((3x^(2)+2x)/(x),"when " x ne 0),(k,"when " x=0):}}`, what must the value of k be equal to in order for f(x) to be a continuous function?

Options

A. `-(3)/(2)`
B. `-(2)/(3)`
C. 0
D. 2

Explanation

To determine the value of \( k \) for which the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit and the right-hand limit at that point are equal to the value of the function at that point. Here’s how we can solve this step by step:

### Step 1: Understand the function definition
The function is defined as:
\[ f(x) = \begin{cases} \frac{3x^2 + 2x}{x} & \text{when } x \neq 0 \\ k & \text{when } x = 0 \end{cases} \]

### Step 2: Find the limit as \( x \) approaches 0
To find the limits, we need to evaluate:
\[ \lim_{x \to 0} f(x) \text{ when } x \neq 0 \]
This means we need to simplify the expression \( \frac{3x^2 + 2x}{x} \):
\[ \lim_{x \to 0} \frac{3x^2 + 2x}{x} = \lim_{x \to 0} \left(3x + 2\right) \]

### Step 3: Simplify the limit
Now, we can separate the terms:
\[ \lim_{x \to 0} (3x + 2) = \lim_{x \to 0} 3x + \lim_{x \to 0} 2 \]
Calculating these limits gives:
\[ \lim_{x \to 0} 3x = 0 \quad \text{and} \quad \lim_{x \to 0} 2 = 2 \]
Thus,
\[ \lim_{x \to 0} (3x + 2) = 0 + 2 = 2 \]

### Step 4: Set the limit equal to \( k \)
For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ \lim_{x \to 0} f(x) = f(0) \]
This means:
\[ 2 = k \]

### Conclusion
Thus, the value of \( k \) must be:
\[ \boxed{2} \]

---