Recall the formal definition for Big-Oh: We say that f(x) = O(g(x) if there exists positive constants c and x0 such that f(x) =x0.

Question

Let g(x) =5x^2 + 25x + 10

f(x) = 2500x + 200

We can see that f(x) = O(g(x))

Recall the formal definition for Big-Oh: We say that f(x) = O(g(x) if there exists positive constants c and x0 such that f(x) <= cg(x) for all x >=x0. Specify integer values for c and x0 such that f(x) <= cg(x).

There are many values for c and x0 that you may choose. You should specify the answer which minimizes the value of x0.

Summary

Big O C = 100 X0 = 0 are the values of ‘c’ and ‘x0.’
For c and x0, there are numerous possibilities. The preceding are chosen in such a way that x0 is given the smallest possible value.
The definition of Big-O yields these values for c and x0.

Explanation

Big- O definition

IF g and f be the functions from a set of natural numbers to itself

The function ‘of is and to be C(a), if there is a constant C>0  and a natural number x₀ to such that:

F(n)<=cg(n) for all n>=x₀

[f(x)<=cg(x) for all x>=x₀]

In the question:

Given f(t) = 8500x + 900

g(x) = 5x+ 5x+10

The above are chosen such that x0 gets its aim from the definition of Big O.

f(x)=0(901) iff f(x) <= cg(x) +1>²

(500 x +200)<= (5x+a5x+10)

(8500x+800) <= 50x²³²+a5cx+10c of C=100 900- 500x+ 2500x+1000 (0=4500 (0) + 000 = ROC = f(0) 300 = 500 (0)² + 12500 (0) +1000 100 900-1000 = (g(0) f(x) < .=cg(x)  forall x>=x₀

Valid

Therefore, the values of ‘c’ and ‘x₀’ would be:-

C=100

X₀=0

Taken in such a way that x₀ gets the minimum most possible value.

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