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

Explanation

To examine the differentiability of the function \( f \) defined by

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

at \( x = 2 \), we need to check if the left-hand derivative and the right-hand derivative at \( x = 2 \) are equal.

### Step 1: Find the left-hand derivative at \( x = 2 \)

The left-hand derivative is defined as:

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

Since \( x \leq 2 \), we use the first case of the function:

\[ f(x) = 1 + x \]

Now, we need to find \( f(2) \):

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

Substituting into the derivative formula:

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

This simplifies to:

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

### Step 2: Find the right-hand derivative at \( x = 2 \)

The right-hand derivative is defined as:

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

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

\[ f(x) = 5 - x \]

Again, we know \( f(2) = 3 \). Now substituting into the derivative formula:

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

This can be rewritten as:

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

### Step 3: Compare the left-hand and right-hand derivatives

We have:

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

Since \( f'_{-}(2) \neq f'_{+}(2) \), the function \( f \) is not differentiable at \( x = 2 \).

### Conclusion

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

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