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anonymous

  • 5 years ago

help please! inflection points from 12x^2-100

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  1. anonymous
    • 5 years ago
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    There are none.

  2. anonymous
    • 5 years ago
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    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

  3. anonymous
    • 5 years ago
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    Second derivative test yields y^'' = 24. set y^'' = 0 and we have a nonsensical statement.

  4. anonymous
    • 5 years ago
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    what about from x^4+50x^2+4?

  5. anonymous
    • 5 years ago
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    The process is always the same. Get to the second derivative. Set it =0. Solve for x. And that's your point of inflection.

  6. anonymous
    • 5 years ago
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    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

  7. amistre64
    • 5 years ago
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    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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