How do you describe the end behavior of #f(x)=x^10-x^9+5x^8#?

1 Answer
Jan 17, 2017

Answer:

#f >=0#. As # x to +-oo, f to oo# See the graph.
The rate of rise of a function to #oo uarr#, in this order :
#ln x (x x^2 x^3 ...x^10...) e^x#

Explanation:

#f=x^8(x^2-x+5)=x^8((x-1/2)^2+19/4)>=0#

Also,

#f=x^10( 1-1/x+5/x^2 ) to oo#, as #x to +-oo#.

The rate of rise of a function to #oo uarr#, in this order :

ln x (x x^2 x^3 ,,,x^10...) e^x

As the derivatives of f, up to the order 10, are 0 at x = 0, the graph is

horizontal, at the 10-tuple point O. I would like to call this 10 as order

of #flatitude#.

Definition: If the derivatives of f are the same, up to and inclusive of order m at x = a, with #f^((m+1))(a)# different, the order of #flatitude# of the point x = a is m.

graph{x^8(x^2-x+5) [-5, 5, -2.5, 2.5]}