How do I sketch y = 4 x^3 - 4x ?

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Answer 1 · 2018-08-07T01:54:06

See details and graph depicting all the aspects.

#y = 4 x (x - 1 ) ( x + 1 ) = 0, at x = -1. 0 and 1

#rArr# the graph passes through #( -1, 0 ), ( 0, 0 ) and ( 1, 0 )#

#y ( - x ) = - 4 x^3 + 4 x = - y ( x ) rArr# the graph is symmetrical

about the origin.

#y = (x^3 ( 4 - 4/x^2 )) to +- oo#, as #x to +-oo#, respectively.

#y' = 4 ( 3 x^2 - 1 ) = 0#, at #x +- 1/sqrt 3#

#y'' = 24x >< 0at x = +-1/sqrt3#

And so, at #( 1/sqrt3, - 8/sqrt 3 )#, the curve has

local minimum #= - 8/sqrt2#.

At #( - 1/ sqrt3, 8/sqrt3 )#, it has the local maximum #= 8/sqrt3#

#y, y', y'' = 0 and y''' = 24 ne 0, at ( 0, 0 )#.

There is no asymptote.

So. the origin is the ( tangent-crossing-curve )point of inflexion.
graph{(y- 4x(x^2-1))=0}