Let `S` be the set of all `x` such that `x^4-10 x^2+9lt=0` . The maximum value of `f(x)=x^3-3x\ ` on `S ,` is 16 (b) 17 (c) 18 (d) 19

Answer

To solve the problem, we need to find the maximum value of the function \( f(x) = x^3 - 3x \) on the set \( S \), where \( S \) is defined by the inequality \( x^4 - 10x^2 + 9 \leq 0 \).

### Step 1: Solve the inequality \( x^4 - 10x^2 + 9 \leq 0 \)

Let \( y = x^2 \). Then, we can rewrite the inequality as:
\[ y^2 - 10y + 9 \leq 0 \]
Next, we will factor the quadratic:
\[ y^2 - 10y + 9 = (y - 1)(y - 9) \leq 0 \]
This implies:
\[ (y - 1)(y - 9) \leq 0 \]
The roots of the equation \( y^2 - 10y + 9 = 0 \) are \( y = 1 \) and \( y = 9 \).

### Step 2: Determine the intervals for \( y \)

The critical points divide the number line into intervals:
- \( (-\infty, 1) \)
- \( (1, 9) \)
- \( (9, \infty) \)

We will test a point in each interval to determine where the product is less than or equal to zero.

- For \( y < 1 \) (e.g., \( y = 0 \)): \( (0 - 1)(0 - 9) = 9 > 0 \)
- For \( 1 < y < 9 \) (e.g., \( y = 5 \)): \( (5 - 1)(5 - 9) = -16 < 0 \)
- For \( y > 9 \) (e.g., \( y = 10 \)): \( (10 - 1)(10 - 9) = 9 > 0 \)

Thus, the solution to the inequality is:
\[ 1 \leq y \leq 9 \]
Since \( y = x^2 \), we have:
\[ 1 \leq x^2 \leq 9 \]
Taking square roots gives:
\[ -3 \leq x \leq -1 \quad \text{or} \quad 1 \leq x \leq 3 \]
So, the set \( S \) is:
\[ S = [-3, -1] \cup [1, 3] \]

### Step 3: Find the maximum value of \( f(x) \) on \( S \)

Next, we will evaluate \( f(x) \) at the endpoints and critical points within the intervals of \( S \).

1. **Endpoints of the intervals**:
- \( f(-3) = (-3)^3 - 3(-3) = -27 + 9 = -18 \)
- \( f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2 \)
- \( f(1) = (1)^3 - 3(1) = 1 - 3 = -2 \)
- \( f(3) = (3)^3 - 3(3) = 27 - 9 = 18 \)

2. **Critical points**:
We find the derivative:
\[ f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1) \]
Setting \( f'(x) = 0 \) gives critical points \( x = -1 \) and \( x = 1 \). Both points are already evaluated.

### Step 4: Compare the values

Now we compare the values we calculated:
- \( f(-3) = -18 \)
- \( f(-1) = 2 \)
- \( f(1) = -2 \)
- \( f(3) = 18 \)

The maximum value of \( f(x) \) on the set \( S \) is:
\[ \max(-18, 2, -2, 18) = 18 \]

### Conclusion

The maximum value of \( f(x) \) on the set \( S \) is \( 18 \).