Show that 8x^3 + 12x + 100logx=O(x^3).
To show that the function 8x^3 + 12x + 100logx = O(x^3) as x approaches infinity, we need to find a constant C and a value x0 such that for all x greater than x0, the absolute value of the function is less than or equal to C times x^3.
First, let's simplify the equation by considering the dominant term as x approaches infinity, which is 8x^3. Ignoring the smaller terms (12x and 100logx) will not affect the result since we are interested in the behavior of the function for large values of x.
So, we have:
8x^3 = O(x^3)
This equation is trivially true since the left side is equal to the right side.
Alternatively, we can use the formal definition of big O notation. For the given function, we need to find a constant C and a value x0 such that:
|8x^3 + 12x + 100log(x)| ≤ C * |x^3| for x > x0
Since the constant term and the logarithmic term are smaller compared to the dominant term (8x^3) when x is large, we can choose C to be a value greater than or equal to 8. Let's say C = 8.
Now, we need to find a value x0 such that the inequality holds for x > x0. Let's start by analyzing the absolute value of the function.
|8x^3 + 12x + 100log(x)| = |8x^3| ≤ 8|x^3| for x > 1
So, if we take x0 = 1, the inequality holds true because |8x^3| is always less than or equal to 8|x^3|.
Therefore, we can conclude that 8x^3 + 12x + 100logx = O(x^3) as x approaches infinity.