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anonymous
 5 years ago
help please! inflection points from 12x^2100
anonymous
 5 years ago
help please! inflection points from 12x^2100

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0Second derivative test yields y^'' = 24. set y^'' = 0 and we have a nonsensical statement.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what about from x^4+50x^2+4?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The process is always the same. Get to the second derivative. Set it =0. Solve for x. And that's your point of inflection.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Be 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
 5 years ago
Best ResponseYou've already chosen the best response.0if 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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