How do you describe the end behavior for f(x)=x^5-4x^3+x+1f(x)=x5−4x3+x+1?

1 Answer
A. S. Adikesavan
Dec 19, 2016

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

Explanation:

graph{x^5-4x^3+x+1 [-3, 3, -29.54, 10.82]} y=f(x)=x^5-4x^3+x+1y=f(x)=x5−4x3+x+1

=x^5(1-4/x^2+1/x^4+1/x^6) to +-oo=x5(1−4x2+1x4+1x6)→±∞, as x to +-oox→±∞.

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.

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