anonymous
  • anonymous
help please! inflection points from 12x^2-100
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
There are none.
anonymous
  • anonymous
You find d^2y/dx^2 and find that its 24, which is >0, so concavity doesn't change and hence there are no points of inflexion
anonymous
  • anonymous
Second derivative test yields y^'' = 24. set y^'' = 0 and we have a nonsensical statement.

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anonymous
  • anonymous
what about from x^4+50x^2+4?
anonymous
  • anonymous
The process is always the same. Get to the second derivative. Set it =0. Solve for x. And that's your point of inflection.
anonymous
  • anonymous
Be careful here, the solution of x from solving the second derivative equal 0 not necessarily give the inflection point, for example f(x) = x^4
amistre64
  • amistre64
if this is the original: y=12x^2 - 100 then it is an big "U" graph and it has no inflection points. it never changes concavity. Inflection only occurs where the concavity chnges from up to down between to "bowls". 12x^2 is just one big bowl, so it has no inflection points.

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