Consider the function `f(x)={{:(x^(2)",",xgt2),(3x-2",",xle2):}`. Which one of the following statements is correct in respect of the above function?

Options

A. f(x) is derivable but not continuous at x=2.
B. f(x) if continuous but not derivalble at x=2.
C. f(x) is neither continuous nor derivable at x=2.
D. f(x) is continuous as well as derivable at x=2.

Explanation

To determine the correct statement about the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} x^2 & \text{if } x > 2 \\ 3x - 2 & \text{if } x \leq 2 \end{cases} \]

we need to check the continuity and differentiability of the function at \( x = 2 \).

### Step 1: Check Continuity at \( x = 2 \)

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

1. **Left-hand limit** as \( x \) approaches \( 2 \):
\[ \lim_{x \to 2^-} f(x) = \lim_{h \to 0} f(2 - h) = \lim_{h \to 0} (3(2 - h) - 2) = \lim_{h \to 0} (6 - 3h - 2) = 4 \]

2. **Right-hand limit** as \( x \) approaches \( 2 \):
\[ \lim_{x \to 2^+} f(x) = \lim_{h \to 0} f(2 + h) = \lim_{h \to 0} (2 + h)^2 = \lim_{h \to 0} (4 + 4h + h^2) = 4 \]

3. **Value of the function at \( x = 2 \)**:
\[ f(2) = 3(2) - 2 = 4 \]

Since the left-hand limit, right-hand limit, and the value of the function at \( x = 2 \) are all equal to \( 4 \), we conclude that:
\[ \text{The function is continuous at } x = 2. \]

### Step 2: Check Differentiability at \( x = 2 \)

Next, we need to check if the function is differentiable at \( x = 2 \) by finding the left-hand derivative and right-hand derivative.

1. **Left-hand derivative**:
\[ f'(x) = 3 \quad \text{for } x < 2 \]
\[ \lim_{h \to 0} \frac{f(2 - h) - f(2)}{-h} = \lim_{h \to 0} \frac{(3(2 - h) - 2) - 4}{-h} = \lim_{h \to 0} \frac{(6 - 3h - 2) - 4}{-h} = \lim_{h \to 0} \frac{-3h}{-h} = 3 \]

2. **Right-hand derivative**:
\[ f'(x) = 2x \quad \text{for } x > 2 \]
\[ \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} = \lim_{h \to 0} \frac{(2 + h)^2 - 4}{h} = \lim_{h \to 0} \frac{(4 + 4h + h^2) - 4}{h} = \lim_{h \to 0} \frac{4h + h^2}{h} = \lim_{h \to 0} (4 + h) = 4 \]

Since the left-hand derivative (3) does not equal the right-hand derivative (4), we conclude that:
\[ \text{The function is not differentiable at } x = 2. \]

### Conclusion

Based on the analysis, we find that:

- The function is continuous at \( x = 2 \).
- The function is not differentiable at \( x = 2 \).

Thus, the correct statement is:
**The function is continuous but not derivable at \( x = 2 \).**