The set of all points of discontinuity of f(x)=`(x-1)/(x^3+6x^2+11x+6)`

Options

A. `phi`
B. {-1}
C. {-1,-2,-3}
D. {1,2,3}

Answer

To find the set of all points of discontinuity of the function \( f(x) = \frac{x-1}{x^3 + 6x^2 + 11x + 6} \), we need to determine where the denominator is equal to zero, as these points will make the function undefined.

### Step-by-Step Solution:

1. **Identify the Denominator**:
The denominator of the function is \( b(x) = x^3 + 6x^2 + 11x + 6 \).

2. **Set the Denominator to Zero**:
We need to solve the equation:
\[ x^3 + 6x^2 + 11x + 6 = 0 \]

3. **Finding Roots**:
We can use the Rational Root Theorem to test possible rational roots. The possible rational roots of the polynomial can be the factors of the constant term (6) divided by the factors of the leading coefficient (1). Thus, the possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6 \).

4. **Testing Possible Roots**:
- Test \( x = -1 \):
\[ (-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0 \]
So, \( x = -1 \) is a root.

- Test \( x = -2 \):
\[ (-2)^3 + 6(-2)^2 + 11(-2) + 6 = -8 + 24 - 22 + 6 = 0 \]
So, \( x = -2 \) is also a root.

- Test \( x = -3 \):
\[ (-3)^3 + 6(-3)^2 + 11(-3) + 6 = -27 + 54 - 33 + 6 = 0 \]
So, \( x = -3 \) is also a root.

5. **Factoring the Polynomial**:
Since we found three roots, we can factor the polynomial as:
\[ b(x) = (x + 1)(x + 2)(x + 3) \]

6. **Finding Points of Discontinuity**:
The function \( f(x) \) will be discontinuous where the denominator is zero:
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]

7. **Conclusion**:
Therefore, the set of all points of discontinuity of the function \( f(x) \) is:
\[ \{-1, -2, -3\} \]