Check the continuity of f(x) = `{{:(x^(2)/2, if 0le x le 1),(2x^(2)-3x+3/2, if 1 lt x le 2):}` at `x = 1`

Explanation

To check the continuity of the function \( f(x) \) at \( x = 1 \), we need to verify three conditions:

1. The left-hand limit as \( x \) approaches 1 exists.
2. The right-hand limit as \( x \) approaches 1 exists.
3. The value of the function at \( x = 1 \) is equal to both the left-hand limit and the right-hand limit.

Given:
\[ f(x) = \begin{cases} \frac{x^2}{2} & \text{if } 0 \leq x \leq 1 \\ 2x^2 - 3x + \frac{3}{2} & \text{if } 1 < x \leq 2 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit
The left-hand limit as \( x \) approaches 1 is given by:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2}{2} \]
Substituting \( x = 1 \):
\[ = \frac{1^2}{2} = \frac{1}{2} \]

### Step 2: Calculate the Right-Hand Limit
The right-hand limit as \( x \) approaches 1 is given by:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \left(2x^2 - 3x + \frac{3}{2}\right) \]
Substituting \( x = 1 \):
\[ = 2(1^2) - 3(1) + \frac{3}{2} = 2 - 3 + \frac{3}{2} = -1 + \frac{3}{2} = \frac{1}{2} \]

### Step 3: Calculate the Value of the Function at \( x = 1 \)
Now, we find the value of the function at \( x = 1 \):
\[ f(1) = \frac{1^2}{2} = \frac{1}{2} \]

### Conclusion
Now we compare the limits and the function value:
- Left-hand limit: \( \frac{1}{2} \)
- Right-hand limit: \( \frac{1}{2} \)
- Function value at \( x = 1 \): \( \frac{1}{2} \)

Since:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = \frac{1}{2} \]
All three conditions for continuity are satisfied.

Thus, we conclude that \( f(x) \) is continuous at \( x = 1 \).