How do you find all points of inflection given #y=x^4-3x^2#?

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Answer 1 · 2017-08-10T05:51:42

Inflection points at #x = +- 1/sqrt(2)#.

Find the second derivative.

#y' = 4x^3 - 6x#

#y'' = 12x^2 - 6#

Inflection points occur when #y'' = 0#.

#0 = 12x^2 - 6#

#0 = 6(2x^2 - 1)#

#x = +- 1/sqrt(2)#

If we select test points, we see that the sign of the second derivative does indeed change at #x = -1/sqrt(2)# and #x = +1/sqrt(2)#.

Therefore, #x = -1/sqrt(2)# and #x = + 1/sqrt(2)# are both inflection points.

Hopefully this helps!

Answer 2 · 2017-08-10T06:07:32

Point of infection

#(0.7, -1.23)#
#(-0.7, -1.23)#

Given -

#y=x^4-3x^2#

#dy/dx=4x^3-6x#

#(d^2y)/(dx^2)=12x^2-6#

#(d^2y)/(dx^2)=0=>12x^2-6=0#

To find the point of inflection, set the second derivative equal to zero

#x^2=6/12=1/2=0.5#
#x=+-sqrt0.5=0.7#

At #x=0.7#
At #x=-0.7#

#y=0.7^4-3(0.7)^2#

#y=0.2401-1.47=-1.23#
#y=-1.23#

Point of infection

#(0.7, -1.23)#

#y=0.2401-1.47=-1.23#
#y=-0.5096#
#y=-1.23#
Point of infection
#(-0.7, -1.23)#