`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 find the left-hand limit and the right-hand limit at that point and check if they are equal to each other and to \( f(2) \).

### Step 1: Define the function
The function is given as:
\[ f(x) = \begin{cases} 3x + 5 & \text{if } x > 2 \\ x^3 & \text{if } x \leq 2 \end{cases} \]

### Step 2: Find \( f(2) \)
Since \( x = 2 \) falls under the case \( x \leq 2 \), we use the second part of the function:
\[ f(2) = 2^3 = 8 \]

### Step 3: Calculate the left-hand limit as \( x \) approaches 2
The left-hand limit is defined as:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^3 \]
Since \( x \) is approaching 2 from the left, we substitute \( x = 2 \):
\[ \lim_{x \to 2^-} f(x) = 2^3 = 8 \]

### Step 4: Calculate the right-hand limit as \( x \) approaches 2
The right-hand limit is defined as:
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x + 5) \]
Since \( x \) is approaching 2 from the right, we substitute \( x = 2 \):
\[ \lim_{x \to 2^+} f(x) = 3(2) + 5 = 6 + 5 = 11 \]

### Step 5: Compare the limits and the function value
Now we compare the left-hand limit, right-hand limit, and the function value at \( x = 2 \):
- Left-hand limit: \( 8 \)
- Right-hand limit: \( 11 \)
- Function value: \( f(2) = 8 \)

Since the left-hand limit \( (8) \) is not equal to the right-hand limit \( (11) \), we conclude that:
\[ \text{Left-hand limit} \neq \text{Right-hand limit} \]

### Conclusion
Since the left-hand limit is not equal to the right-hand limit, the function \( f(x) \) is not continuous at \( x = 2 \).

Thus, the final answer is:
\[ \text{The function } f(x) \text{ is discontinuous at } x = 2. \]

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