Answer:
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.