Let f(x) = sgn (x) and g (x) = `x(x^(2)-5x+6)`. The function f(g(x)) is discontinuous at

Options

A. Infinitely many points
B. Exactly one point
C. Exactly three points
D. No point

Explanation

To solve the problem, we need to analyze the functions given: \( f(x) = \text{sgn}(x) \) and \( g(x) = x(x^2 - 5x + 6) \). We want to find where the function \( f(g(x)) \) is discontinuous.

### Step-by-step Solution:

1. **Understand the functions**:
- The signum function \( f(x) = \text{sgn}(x) \) is defined as:
\[ f(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases} \]
- The function \( g(x) = x(x^2 - 5x + 6) \) can be simplified.

2. **Factor \( g(x) \)**:
- First, factor the quadratic part:
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]
- Therefore, we can write:
\[ g(x) = x(x - 2)(x - 3) \]

3. **Identify the roots of \( g(x) \)**:
- The function \( g(x) \) has roots at \( x = 0, 2, 3 \). These points are where \( g(x) = 0 \).

4. **Determine the sign of \( g(x) \)**:
- Analyze the intervals determined by the roots:
- For \( x < 0 \): \( g(x) < 0 \) (since all factors are negative).
- For \( 0 < x < 2 \): \( g(x) > 0 \) (since \( x \) is positive and both \( (x - 2) \) and \( (x - 3) \) are negative).
- For \( 2 < x < 3 \): \( g(x) < 0 \) (since \( x \) is positive and \( (x - 2) \) is positive but \( (x - 3) \) is negative).
- For \( x > 3 \): \( g(x) > 0 \) (since all factors are positive).

5. **Evaluate \( f(g(x)) \)**:
- Now we can summarize the behavior of \( f(g(x)) \):
- For \( x < 0 \): \( g(x) < 0 \) → \( f(g(x)) = -1 \)
- For \( 0 < x < 2 \): \( g(x) > 0 \) → \( f(g(x)) = 1 \)
- For \( 2 < x < 3 \): \( g(x) < 0 \) → \( f(g(x)) = -1 \)
- For \( x > 3 \): \( g(x) > 0 \) → \( f(g(x)) = 1 \)

6. **Identify points of discontinuity**:
- The function \( f(g(x)) \) changes its value at the points \( x = 0, 2, 3 \):
- At \( x = 0 \): jumps from -1 to 1.
- At \( x = 2 \): jumps from 1 to -1.
- At \( x = 3 \): jumps from -1 to 1.

7. **Conclusion**:
- The function \( f(g(x)) \) is discontinuous at the points \( x = 0, 2, 3 \).

### Final Answer:
The function \( f(g(x)) \) is discontinuous at exactly 3 points: \( x = 0, 2, 3 \).