Let `f (x)=` signum (x) and `g (x) =x (x ^(2) -10x+21),` then the number of points of discontinuity of `f [g (x)]` is

Answer

To solve the problem, we need to find the number of points of discontinuity of the function \( f[g(x)] \), where \( f(x) = \text{signum}(x) \) and \( g(x) = x(x^2 - 10x + 21) \).

### Step 1: Analyze the function \( g(x) \)

First, we need to simplify \( g(x) \):
\[ g(x) = x(x^2 - 10x + 21) \]
The quadratic \( x^2 - 10x + 21 \) can be factored:
\[ x^2 - 10x + 21 = (x - 3)(x - 7) \]
Thus, we can rewrite \( g(x) \):
\[ g(x) = x(x - 3)(x - 7) \]

### Step 2: Find the roots of \( g(x) \)

Next, we find the values of \( x \) for which \( g(x) = 0 \):
\[ g(x) = 0 \implies x(x - 3)(x - 7) = 0 \]
The roots are:
\[ x = 0, \quad x = 3, \quad x = 7 \]

### Step 3: Determine the sign of \( g(x) \)

Now, we need to analyze the sign of \( g(x) \) in the intervals defined by these roots:
- For \( x < 0 \): \( g(x) < 0 \)
- For \( 0 < x < 3 \): \( g(x) > 0 \)
- For \( 3 < x < 7 \): \( g(x) < 0 \)
- For \( x > 7 \): \( g(x) > 0 \)

### Step 4: Analyze the function \( f(g(x)) \)

The function \( f(x) = \text{signum}(x) \) is defined as:
\[ f(x) = \begin{cases} -1 & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 & \text{if } x > 0 \end{cases} \]

### Step 5: Determine the points of discontinuity in \( f[g(x)] \)

The function \( f[g(x)] \) will be discontinuous at points where \( g(x) = 0 \) or where \( g(x) \) changes sign. The points where \( g(x) = 0 \) are \( x = 0, 3, 7 \).

1. At \( x = 0 \): \( g(x) \) changes from negative to positive.
2. At \( x = 3 \): \( g(x) \) changes from positive to negative.
3. At \( x = 7 \): \( g(x) \) changes from negative to positive.

### Conclusion

Thus, the number of points of discontinuity of \( f[g(x)] \) is:
\[ \text{Number of points of discontinuity} = 3 \]