`{{:(x^(2)/2, if 0le x le 1),(2x^(2)-3x+3/2, if l lt x le 2):}` at `x = 1`

Explanation

To determine the continuity of the piecewise function at \( x = 1 \), we will follow these steps:

### Step 1: Define the function
The function \( f(x) \) is given as:
\[ f(x) = \begin{cases} \frac{x^2}{2} & \text{if } 0 \leq x \leq 1 \\ 2x^2 - 3x + \frac{3}{2} & \text{if } 1 < x \leq 2 \end{cases} \]

### Step 2: Find the left-hand limit as \( x \) approaches 1
To find the left-hand limit, we compute:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1} \frac{x^2}{2} \]
Substituting \( x = 1 \):
\[ \lim_{x \to 1^-} f(x) = \frac{1^2}{2} = \frac{1}{2} \]

### Step 3: Find the right-hand limit as \( x \) approaches 1
To find the right-hand limit, we compute:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1} \left( 2x^2 - 3x + \frac{3}{2} \right) \]
Substituting \( x = 1 \):
\[ \lim_{x \to 1^+} f(x) = 2(1^2) - 3(1) + \frac{3}{2} = 2 - 3 + \frac{3}{2} \]
Calculating this:
\[ = -1 + \frac{3}{2} = \frac{-2 + 3}{2} = \frac{1}{2} \]

### Step 4: Compare the left-hand and right-hand limits
We have:
\[ \lim_{x \to 1^-} f(x) = \frac{1}{2} \]
\[ \lim_{x \to 1^+} f(x) = \frac{1}{2} \]
Since both limits are equal, we conclude:
\[ \lim_{x \to 1} f(x) = \frac{1}{2} \]

### Step 5: Check the value of the function at \( x = 1 \)
Now, we need to check the value of the function at \( x = 1 \):
\[ f(1) = \frac{1^2}{2} = \frac{1}{2} \]

### Step 6: Conclude continuity
Since:
\[ \lim_{x \to 1} f(x) = f(1) = \frac{1}{2} \]
The function \( f(x) \) is continuous at \( x = 1 \).

### Final Answer
The function is continuous at \( x = 1 \).

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