What are the points of inflection, if any, of #f(x) = x^6 + 3x^5 - (15/2)x^4 - 40x^3 - 60x^2 + 8x + 5 #?

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Answer 1 · 2017-11-25T05:24:18

Points of inflection occur at #x=-1,-2# and #2#.

At point of inflection #(d^2f(x))/(dx^2)=0#

As #f(x) = x^6+3x^5-(15/2)x^4-40x^3-60x^2+8x+5#

#(df)/(dx)=6x^5+15x^4-30x^3-120x^2-120x+8#

and #(d^2f(x))/(dx^2)=30x^4+60x^3-90x^2-240x-120#

Hence points of inflection are given by #30x^4+60x^3-90x^2-240x-120=0#

or #x^4+2x^3-3x^2-8x-4=0#

Using factor theorem we have zeros at #-1,-2# and #2#

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

Hence, points of inflection occur at #x=-1,-2# and #2#.

Graph not drrawn to scale (shrunk vertically).
graph{x^6+3x^5-(15/2)x^4-40x^3-60x^2+8x+5 [-4, 4, -200, 200]}