How do you find all points of inflection given #y=x^3-2x^2+1#?

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Answer 1 · 2017-04-17T22:25:31

#(2/3,0.41)#

Inflection points occur when the second derivative is equal to #0#

#dy/dx=3x^2-4x#

#(d^2y)/(dx^2)=6x-4#

Let #(d^2y)/(dx^2)=0#

#0=6x-4#

#6x=4#

#x=4/6=2/3#

Solve for y-cordinate,

#y=(2/3)^3-2(2/3)^2+1#

#y=8/27-2(4/9)+1#

#y=8/27-8/9+1#

#y=11/27# or #0.41#

Therefore the point of inflection for the function #y=x^3-2x^2+1# is #(2/3,0.41)#