anonymous
  • anonymous
y=x^3 has a point of inflection at x=o. I know it is true but how i prove it with an statement?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
You test the point by, take the Second Derivato on it.. and then you can test it by using, a point before 0 and after.. For exempel test with the point -1 and 1
anonymous
  • anonymous
you get a parabola
anonymous
  • anonymous
and the parabola touches the origin..is this fine?

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anonymous
  • anonymous
@nabaz thankyou. it means i have to solve it.
anonymous
  • anonymous
No problem, is it a parabola...???
anonymous
  • anonymous
yeah , of course, u need proof?
anonymous
  • anonymous
i thouht it was a Curve ..
anonymous
  • anonymous
@thinker thankyou
anonymous
  • anonymous
if so, u have the answer wid u...just try substituting x=-1,1,0,-2,2,...... and you'll get corresponding y values, with which u can get a curve...and that curve is a parabola
anonymous
  • anonymous
now , is this fine?? @nabaz & @abhinn
anonymous
  • anonymous
yeah, i last thing... u mean this is a parabola.. http://en.wikipedia.org/wiki/File:X_cubed_plot.svg
anonymous
  • anonymous
Copie the text.. could get the hole adress..
anonymous
  • anonymous
no no....not that one...:).try this http://en.wikipedia.org/wiki/File:Parabola.svg this is a parabola...both of go to this link n see
anonymous
  • anonymous
*both of u
anonymous
  • anonymous
its like a "U" shape
anonymous
  • anonymous
Yes but this was a Y = x^3 and its not a parabole
anonymous
  • anonymous
Yes, but x3 is like a S with is lying downwards.
anonymous
  • anonymous
yeah...y=x^3 is a symmetric curve, but parabola looks like a "U" ...
anonymous
  • anonymous
haha.. my bad, i thought u meant that x^3 was a parabola
anonymous
  • anonymous
its ok:) nabaz
anonymous
  • anonymous
i wud like to give u both a hint
anonymous
  • anonymous
u der?
anonymous
  • anonymous
hint for what?
anonymous
  • anonymous
to know if an equation is parabola or what..only if u need it:)
anonymous
  • anonymous
yes, i know i think. ever x^2 x^4 x^6 .. ( not odd nrs )
anonymous
  • anonymous
exactly :)...good

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