Find the least integer n such that f (x) is O(xn) for each of these functions. a) f (x) = 2x2 + x3 log x b) f (x) = 3x5 + (log x)4 c) f (x) = (x4 + x2 + 1)/(x4 + 1) d) f (x) = (x3 + 5 log x)/(x4 + 1)

In each scenario, we must identify the smallest integer (n) such that  implying there must be constants C and k such that.  

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a)4

b)5

c)0

d)0

Step-by-step explanation:

a) For x>0 we have that logx<=x, so f(x)<=2x^2+x^4

The largest power of x is a smallest n for which f(x) is O(x^n). Beacuse of f(x)<=2x^2+x^4, n is 4.

Now we have to find C and k for O(x^2).

We know that for x>0 it is x^2>x>2, so we have:

k=2 and

|f(x)|<= | 2x^2+x^4|<= |2x^2|+|x^4|=2x^2+x^4 <x^2*x^2+x^4=2x^4

it follows that C=2.

For a different k we will have other C.

b) The largest power of x in f(x) is smallest n of O(x^n), so n=5.

For x>0 we have logx<=x and for x>1 we have x^5>x^4.

SO for k=1, we have x>1, and :

It fallows that C=4.

c) , so n=0.

For k=0, we have x>0, and

It fallows that C=2.

d) Using x>0 we have .

When x>3 then  x^4+5x<x^4+1, so we have for k=3, n=0

|f(x)|<=1=x^0.

It fallows that C=1.

false

step-by-step explanation:

rulers and compasses arent required in order to bisect an angle because when bisecting angles, you're making a "half circle" and considering in order to bisect an angle it wouldn't be straight, you wouldn't need either of the two

7

step-by-step explanation: