Question

14. Determine whether x3 is O(g(x)) for each of these functions g(x). a) g(x) = x2 b) g(x) = x3 c) g(x) = x2 + x3 d) g(x) = x2 + x4 e) g(x) = 3x f ) g(x) = x3/2

Answer 1

To check if x³ is O(g(x)) for given functions, we compare the growth rates. x³ is O(g(x)) for g(x) = x³, g(x) = x² + x³, g(x) = x² + x´, and g(x) = x³/2, but not for g(x) = x² or g(x) = 3x.

Step-by-step explanation:

The question asks us to determine whether x³ is O(g(x)) for various functions g(x). This is a common problem in analysis and computational complexity where we establish upper bounds for functions using Big O notation.

Determine whether x³ is O(g(x)):

1. g(x) = x²: x³ is not O(x²) because as x goes to infinity, x³ grows faster than x², so there is no constant c such that x³ ≤ cx² for all x.
2. g(x) = x³: x³ is O(x³) as we can choose c = 1, so x³ ≤ cx³ for all x.
3. g(x) = x² + x³: x³ is O(x² + x³) since x³ is less than or equal to x² + x³ for all x ≥ 0, thus c = 1 suffices.
4. g(x) = x² + x´: x³ is O(x² + x´) because x³ is eventually less than any positive multiple of x² + x´ as x goes to infinity.
5. g(x) = 3x: x³ is not O(3x) since as x goes to infinity, x³ grows faster than any linear function and there is no constant c that would make x³ ≤ c3x for all x.
6. g(x) = x³/2: x³ is O(x³/2) because we can choose c = 2, and so x³ ≤ cx³/2 for all x.

Answer 2

Answer and explanation with steps are in the following attachments

Step-by-step explanation: