Let `f:RtoR` (R is the set of real numbers) be defined as follows :
`f(x)={{:(2-x", for "1lexle2),(x-1/2x^(2)", for "xgt2):}`.
Examine the continuity and differentiability of f(x) at x = 2.

Answer

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

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

### Step 2: Calculate \( f(2) \)
To find \( f(2) \), we use the first piece of the function since \( 2 \) falls in the interval \( [1, 2] \):
\[ f(2) = 2 - 2 = 0 \]

### Step 3: Calculate the left-hand limit \( f(2^-) \)
Next, we calculate the left-hand limit as \( x \) approaches \( 2 \) from the left:
\[ f(2^-) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2 - x) = 2 - 2 = 0 \]

### Step 4: Calculate the right-hand limit \( f(2^+) \)
Now we calculate the right-hand limit as \( x \) approaches \( 2 \) from the right:
\[ f(2^+) = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \left(\frac{x - 1}{2} x^2\right) \]
Substituting \( x = 2 \):
\[ f(2^+) = \frac{2 - 1}{2} \cdot 2^2 = \frac{1}{2} \cdot 4 = 2 \]

### Step 5: Check continuity at \( x = 2 \)
For \( f(x) \) to be continuous at \( x = 2 \), we need:
\[ f(2^-) = f(2) = f(2^+) \]
We have:
- \( f(2^-) = 0 \)
- \( f(2) = 0 \)
- \( f(2^+) = 2 \)

Since \( f(2^-) \neq f(2^+) \), the function is **not continuous** at \( x = 2 \).

### Step 6: Check differentiability at \( x = 2 \)
A function must be continuous at a point to be differentiable there. Since we found that \( f(x) \) is not continuous at \( x = 2 \), it cannot be differentiable at that point.

### Conclusion
The function \( f(x) \) is neither continuous nor differentiable at \( x = 2 \).

---