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

Question

anonymous · Answer

There are none.

anonymous · Answer

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 · Answer

Second derivative test yields y^'' = 24. set y^'' = 0 and we have a nonsensical statement.

anonymous · Answer

what about from x^4+50x^2+4?

anonymous · Answer

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 · Answer

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 · Answer

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.