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

Explanation

To determine the differentiability of the function \( f(x) \) at \( x = 2 \), we need to find the left-hand derivative and the right-hand derivative at that point. The function is defined as follows:

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

### Step 1: Calculate the Left-Hand Derivative at \( x = 2 \)

The left-hand derivative is given by the formula:

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

Since \( 2 - h \) will always be less than or equal to 2 for small \( h \), we use the first case of the function:

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

Now substituting these into the left-hand derivative formula:

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

### Step 2: Calculate the Right-Hand Derivative at \( x = 2 \)

The right-hand derivative is given by the formula:

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

For small \( h \), \( 2 + h \) will be greater than 2, so we use the second case of the function:

\[ f(2 + h) = 5 - (2 + h) = 3 - h \]
\[ f(2) = 3 \]

Now substituting these into the right-hand derivative formula:

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

### Step 3: Compare the Left-Hand and Right-Hand Derivatives

We have found:

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

Since the left-hand derivative is not equal to the right-hand derivative:

\[ f'(2^-) \neq f'(2^+) \]

### Conclusion

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