`f(x) = (x^(2)+x-2)/(x^(2)-3x+2)` is discontinuous at x =

Options

A. 0,1
B. 1,2
C. `-1, -2`
D. `0,-1`

Answer

To determine the points at which the function \( f(x) = \frac{x^2 + x - 2}{x^2 - 3x + 2} \) is discontinuous, we need to analyze the denominator of the function, as discontinuities occur when the denominator is equal to zero.

### Step-by-step Solution:

1. **Identify the function**:
\[ f(x) = \frac{x^2 + x - 2}{x^2 - 3x + 2} \]

2. **Factor the numerator and denominator**:
- **Numerator**: \( x^2 + x - 2 \)
- We can factor this as:
\[ x^2 + x - 2 = (x + 2)(x - 1) \]
- **Denominator**: \( x^2 - 3x + 2 \)
- We can factor this as:
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

3. **Rewrite the function with the factored form**:
\[ f(x) = \frac{(x + 2)(x - 1)}{(x - 1)(x - 2)} \]

4. **Identify points of discontinuity**:
- The function \( f(x) \) is discontinuous where the denominator is equal to zero. Thus, we set the denominator equal to zero:
\[ (x - 1)(x - 2) = 0 \]
- This gives us the solutions:
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

5. **Conclusion**:
- The function \( f(x) \) is discontinuous at \( x = 1 \) and \( x = 2 \).

### Final Answer:
The function \( f(x) \) is discontinuous at \( x = 1 \) and \( x = 2 \).

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