`f(x)={{:(1+x, if x le 2),(5-x,ifx gt 2):}` at `x = 2`. Examine the differentiability

Explanation

To examine the differentiability of the function \( f(x) \) at \( x = 2 \), we will follow these steps:

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

### Step 2: Find the left-hand derivative at \( x = 2 \)
The left-hand derivative is calculated using the limit:
\[ f'(2^-) = \lim_{h \to 0} \frac{f(2 - h) - f(2)}{-h} \]
Since \( 2 - h \leq 2 \) for small \( h \), we use the first case of the function:
\[ f(2 - h) = 1 + (2 - h) = 3 - h \]
Now substituting into the limit:
\[ f'(2^-) = \lim_{h \to 0} \frac{(3 - h) - 3}{-h} = \lim_{h \to 0} \frac{-h}{-h} = \lim_{h \to 0} 1 = 1 \]

### Step 3: Find the right-hand derivative at \( x = 2 \)
The right-hand derivative is calculated using the limit:
\[ f'(2^+) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \]
Since \( 2 + h > 2 \) for small \( h \), we use the second case of the function:
\[ f(2 + h) = 5 - (2 + h) = 3 - h \]
Now substituting into the limit:
\[ f'(2^+) = \lim_{h \to 0} \frac{(3 - h) - 3}{h} = \lim_{h \to 0} \frac{-h}{h} = \lim_{h \to 0} -1 = -1 \]

### Step 4: Compare the left-hand and right-hand derivatives
We find:
\[ f'(2^-) = 1 \quad \text{and} \quad f'(2^+) = -1 \]
Since \( f'(2^-) \neq f'(2^+) \), the left-hand derivative is not equal to the right-hand derivative.

### Conclusion
Since the left-hand derivative and right-hand derivative at \( x = 2 \) are not equal, we conclude that the function \( f(x) \) is **not differentiable** at \( x = 2 \).

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