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Show that x log(x)+x^1/2 is O(x2).

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To show that the function f(x) = x * log(x) + x^(1/2) is O(x^2), we need to find a positive constant c and a value of x₀ such that for all x > x₀, f(x) ≤ c * x^2.

Let's break it down into two parts:

x * log(x) is O(x^2):
First, consider the term x * log(x). We can use this rule to show that the limit of (x * log(x)) / x^2 as x approaches infinity is 0. Therefore, x * log(x) is O(x^2).
x^(1/2) is O(x^2):
The term x^(1/2) is a polynomial of degree 1/2, which is less than 2. Therefore, it is also O(x^2).
Now, we've shown that both x * log(x) and x^(1/2) are O(x^2). Since O notation is additive, their sum, which is f(x), is still O(x^2).

So, we can conclude that f(x) = x * log(x) + x^(1/2) is indeed O(x^2).
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