The smallest integer n such that n3 - 11n2 + 32n - 28 > 0 is
Correct Answer: 8
Solution and Explanation
To find the smallest integer \( n \) such that \( n^3 - 11n^2 + 32n - 28 > 0 \), we follow these steps:
- Evaluate the polynomial for initial values close to the suspected range.
- Check for the smallest integer satisfying the inequality.
First, substitute incremental integer values starting from \( n = 8 \):
| n | f(n) = \( n^3 - 11n^2 + 32n - 28 \) | f(n) > 0? |
|---|---|---|
| 8 | \( 8^3 - 11 \times 8^2 + 32 \times 8 - 28 = 512 - 704 + 256 - 28 = 36 \) | Yes |
| 7 | \( 7^3 - 11 \times 7^2 + 32 \times 7 - 28 = 343 - 539 + 224 - 28 = 0 \) | No, equals 0 |
- The polynomial evaluates to 36 (which is > 0) when \( n = 8 \). At \( n = 7 \), it evaluates to exactly 0, which is not greater than 0.
- Thus, the smallest integer \( n \) such that \( f(n) > 0 \) is \( n = 8 \).
The computed value \( n = 8 \) falls within the provided range: \([8, 8]\).
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