`f(x)={{:(3x+5, if x ge 2),(x^(3), if x le 2):}`at `x = 2`

Options

A. `f(x)` is dicontinuous at `x = 2`.
B. `f(x)` is continuous at `x = 2`.
C. Can not be determined
D. None of these

Explanation

To determine whether the function \( f(x) \) is continuous at \( x = 2 \), we need to check the left-hand limit, the right-hand limit, and the value of the function at that point.

The function is defined as:
\[ f(x) = \begin{cases} 3x + 5 & \text{if } x > 2 \\ x^3 & \text{if } x \leq 2 \end{cases} \]

### Step 1: Find the left-hand limit as \( x \) approaches 2.
The left-hand limit is given by:
\[ \lim_{x \to 2^-} f(x) \]
Since we are considering values of \( x \) that are less than or equal to 2, we use the definition of \( f(x) \) for \( x \leq 2 \):
\[ f(x) = x^3 \]
Now, we calculate the limit:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2} x^3 = 2^3 = 8 \]

### Step 2: Find the right-hand limit as \( x \) approaches 2.
The right-hand limit is given by:
\[ \lim_{x \to 2^+} f(x) \]
For values of \( x \) greater than 2, we use the definition of \( f(x) \) for \( x > 2 \):
\[ f(x) = 3x + 5 \]
Now, we calculate the limit:
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2} (3x + 5) = 3(2) + 5 = 6 + 5 = 11 \]

### Step 3: Check the value of the function at \( x = 2 \).
Since \( f(x) \) is defined for \( x \leq 2 \) as \( x^3 \):
\[ f(2) = 2^3 = 8 \]

### Step 4: Compare the limits and the function value.
We have:
- Left-hand limit: \( \lim_{x \to 2^-} f(x) = 8 \)
- Right-hand limit: \( \lim_{x \to 2^+} f(x) = 11 \)
- Value of the function at \( x = 2 \): \( f(2) = 8 \)

Since the left-hand limit (8) is not equal to the right-hand limit (11), we conclude that:
\[ \text{The function } f(x) \text{ is discontinuous at } x = 2. \]

### Final Answer:
The function \( f(x) \) is discontinuous at \( x = 2 \).

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