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.

step-by-step explanation:

let

we know that

the y-intercept is the y-coordinate of the point when the value of x is equal to zero

so

the y-coordinate of the point a is the y-intercept

the answer part a) is

the value of y-intercept is

part b) find the slope m

we know that

the value of the slope is equal to

substitute the values

therefore

the answer part b) is

the value of the slope is equal to

the general answer is

the slope is

the value of y-intercept is

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497