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Question

At what value of x does the graph of #y=(1/x^2 + 1/x^3)# have a point of inflection?

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Answer 1 · 2018-04-15T20:52:28

No inflexion

#y=1/x^2 + 1/x^3#

#y' =- 2/x^3 - 3/x^4 = - 1/x^3 (2 + 3/x)#

#y' = 0 implies x = color(blue)(- 3/2)#

#y'' =6/x^4 + 12/x^5 = 6/x^4 (1 + 2/x ) #

#y'' = 0 implies x = color(red)(- 2) color(blue)(ne - 3/2)#