How do you sketch the general shape of #f(x)=x^5-3x^3+2x+4# using end behavior?

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Answer 1 · 2017-01-21T02:25:33

See graph and explanation. The second graph reveals turning points and points of inflexion (POI).

graph{x^5-3x^3+2x^2+4 [-39.86, 39.8, -20.43, 19.43]}

#f = x^5(1-3/x^2+2/x^4+4/x^5) to +-oo#, as # x to +-oo#.

#f'=3x^4-6x^2+2#, giving turning points?POI, at its

zeros x = +-1.24 and +-0.51, nearly,

#f''= 12x(x^2-1)=0#, at #x = 0 and x = +-1#.

#f'''=12(3x^2-1) ne 0#, at these points.

So, the POI are at #x= 0, +-1/sqrt3#.

The second graph locates the turning points and POE that could not

be located in the first graph

.graph{x^5-3x^3+2x^2+4[-1.5, 1.5, -20.43, 19.43]}