A function `f(x) = {:{(1+x",",xle2),(5-x,","xgt2):}` is

Options

A. not continuous at x = 2
B. differentiable x = 2
C. continuous but not differentiable at x = 2
D. None of the above

Explanation

To solve the problem, we need to analyze the piecewise function defined as:

\[ f(x) = \begin{cases} 1 + x & \text{if } x \leq 2 \\ 5 - x & \text{if } x > 2 \end{cases} \]

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

To determine if the function is continuous at \( x = 2 \), we need to check the following:

1. \( f(2) \)
2. \( \lim_{x \to 2^-} f(x) \)
3. \( \lim_{x \to 2^+} f(x) \)

**Calculating \( f(2) \):**

Since \( x = 2 \) falls under the first case:

\[ f(2) = 1 + 2 = 3 \]

**Calculating \( \lim_{x \to 2^-} f(x) \):**

For \( x < 2 \), we use the first case of the function:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (1 + x) = 1 + 2 = 3 \]

**Calculating \( \lim_{x \to 2^+} f(x) \):**

For \( x > 2 \), we use the second case of the function:

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5 - x) = 5 - 2 = 3 \]

**Conclusion for Continuity:**

Since:

\[ f(2) = 3, \quad \lim_{x \to 2^-} f(x) = 3, \quad \lim_{x \to 2^+} f(x) = 3 \]

Thus, \( f(2) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \), which means \( f(x) \) is continuous at \( x = 2 \).

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

To check differentiability, we need to find the left-hand derivative and the right-hand derivative at \( x = 2 \).

**Calculating the Left-Hand Derivative:**

The left-hand derivative at \( x = 2 \) is given by:

\[ f'(2^-) = \lim_{h \to 0^-} \frac{f(2 + h) - f(2)}{h} \]

Using \( f(x) = 1 + x \) for \( x < 2 \):

\[ f'(2^-) = \lim_{h \to 0^-} \frac{(1 + (2 + h)) - 3}{h} = \lim_{h \to 0^-} \frac{(3 + h) - 3}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1 \]

**Calculating the Right-Hand Derivative:**

The right-hand derivative at \( x = 2 \) is given by:

\[ f'(2^+) = \lim_{h \to 0^+} \frac{f(2 + h) - f(2)}{h} \]

Using \( f(x) = 5 - x \) for \( x > 2 \):

\[ f'(2^+) = \lim_{h \to 0^+} \frac{(5 - (2 + h)) - 3}{h} = \lim_{h \to 0^+} \frac{(3 - h) - 3}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1 \]

**Conclusion for Differentiability:**

Since:

\[ f'(2^-) = 1 \quad \text{and} \quad f'(2^+) = -1 \]

These two derivatives are not equal, which means \( f(x) \) is not differentiable at \( x = 2 \).

### Final Conclusion

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

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