How do you describe the end behavior for #f(x)=x^5-4x^3+x+1#?

Question

2304 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-19T16:32:19

See explanation for end trends, turning points and points of inflexion. Illustrative graph is inserted

graph{x^5-4x^3+x+1 [-3, 3, -29.54, 10.82]} #y=f(x)=x^5-4x^3+x+1#

#=x^5(1-4/x^2+1/x^4+1/x^6) to +-oo#, as #x to +-oo#.

#y'=5x^4-12x^2+1=5((x^2-6/5)^2-31/25)= 0#,

when #x^2=(6+-sqrt 31)/5 to x = +_sqrt((6+-sqrt 31)/5)#

#=+-1.521, +-0.2940#, giving local extrema.

#y''=20x^3-24x=20x(x^2-4/5)=0#, when #x = 0, +-2/sqrt5#, giving

possible points of inflexion, if f''' is not 0.

#y'''=12(5x^2-2)# and is not 0, when y'=0.