The function `f(x)=(2x^(2)+7)/(x^(3)+3x^(2)-x-3)` is discontinuous for

Options

A. `x=1` only
B. `x=1, -1` only
C. `x=1, -1,-3` and other values of x
D. `x=1, -1, -3` only

Answer

To determine where the function \( f(x) = \frac{2x^2 + 7}{x^3 + 3x^2 - x - 3} \) is discontinuous, we need to analyze the denominator of the function. A function is discontinuous at points where its denominator equals zero, as the function is undefined at those points.

### Step-by-Step Solution:

1. **Identify the Denominator**:
The denominator of the function is \( g(x) = x^3 + 3x^2 - x - 3 \).

2. **Set the Denominator Equal to Zero**:
To find the points of discontinuity, we need to solve the equation:
\[ g(x) = 0 \implies x^3 + 3x^2 - x - 3 = 0 \]

3. **Factor the Polynomial**:
We can try to factor the polynomial. First, we can look for rational roots using the Rational Root Theorem, or we can use synthetic division or polynomial long division. Testing \( x = -3 \):
\[ g(-3) = (-3)^3 + 3(-3)^2 - (-3) - 3 = -27 + 27 + 3 - 3 = 0 \]
Thus, \( x + 3 \) is a factor of \( g(x) \).

4. **Perform Polynomial Division**:
Now we divide \( g(x) \) by \( x + 3 \):
\[ g(x) = (x + 3)(x^2 + 0x - 1) = (x + 3)(x^2 - 1) \]

5. **Factor Further**:
The quadratic \( x^2 - 1 \) can be factored further:
\[ x^2 - 1 = (x - 1)(x + 1) \]
Therefore, we have:
\[ g(x) = (x + 3)(x - 1)(x + 1) \]

6. **Set Each Factor to Zero**:
To find the points where the function is discontinuous, we set each factor equal to zero:
\[ x + 3 = 0 \implies x = -3 \]
\[ x - 1 = 0 \implies x = 1 \]
\[ x + 1 = 0 \implies x = -1 \]

7. **Conclusion**:
The function \( f(x) \) is discontinuous at the points:
\[ x = -3, \quad x = 1, \quad x = -1 \]

### Final Answer:
The function \( f(x) \) is discontinuous for \( x = -3, 1, -1 \).

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