Number of integral values of x in the domain of `f(x)=sqrt(-[x]^2+3[x]-2)` is

Answer

To find the number of integral values of \( x \) in the domain of the function \( f(x) = \sqrt{-[x]^2 + 3[x] - 2} \), we need to ensure that the expression inside the square root is non-negative. Let's break this down step by step.

### Step 1: Set up the inequality

We need to satisfy the condition:
\[ -[x]^2 + 3[x] - 2 \geq 0 \]
This means we want:
\[ -[x]^2 + 3[x] - 2 \geq 0 \]

### Step 2: Rearranging the inequality

To make it easier to solve, we can multiply the entire inequality by -1 (which flips the inequality sign):
\[ [x]^2 - 3[x] + 2 \leq 0 \]

### Step 3: Factor the quadratic expression

Now we need to factor the quadratic expression \( [x]^2 - 3[x] + 2 \):
\[ [x]^2 - 3[x] + 2 = ([x] - 1)([x] - 2) \]
Thus, we have:
\[ ([x] - 1)([x] - 2) \leq 0 \]

### Step 4: Determine the critical points

The critical points from the factors are:
- \( [x] - 1 = 0 \) → \( [x] = 1 \) → \( x \) in the interval \( [1, 2) \)
- \( [x] - 2 = 0 \) → \( [x] = 2 \) → \( x \) in the interval \( [2, 3) \)

### Step 5: Analyze the intervals

To solve the inequality \( ([x] - 1)([x] - 2) \leq 0 \), we can test the intervals defined by the critical points:
1. For \( [x] < 1 \) (e.g., \( [x] = 0 \)): The product is positive.
2. For \( 1 \leq [x] < 2 \): The product is non-positive (valid).
3. For \( [x] \geq 2 \): The product is positive.

Thus, the valid interval for \( [x] \) is:
\[ 1 \leq [x] \leq 2 \]

### Step 6: Determine integral values of \( x \)

The values of \( [x] \) can be:
- \( [x] = 1 \) corresponds to \( x \) in the interval \( [1, 2) \) (includes 1, excludes 2).
- \( [x] = 2 \) corresponds to \( x \) in the interval \( [2, 3) \) (includes 2, excludes 3).

Thus, the integral values of \( x \) in the domain are:
- From \( [1, 2) \): \( 1 \)
- From \( [2, 3) \): \( 2 \)

### Conclusion

The integral values of \( x \) in the domain of \( f(x) \) are \( 1 \) and \( 2 \). Therefore, the number of integral values of \( x \) is:
\[ \boxed{2} \]