N no_arg Member Joined Mar 25, 2004 Messages 67 Gender Undisclosed HSC N/A Jul 1, 2008 #1 Here's an odd one! Define f to be the piecemeal function f(x)=x^2 for x>=0 and f(x) =-(x^2) for x<=0 The graph of f is then sort of like x^3 Does f have a point of inflection at the origin?
Here's an odd one! Define f to be the piecemeal function f(x)=x^2 for x>=0 and f(x) =-(x^2) for x<=0 The graph of f is then sort of like x^3 Does f have a point of inflection at the origin?
C conics2008 Active Member Joined Mar 26, 2008 Messages 1,225 Gender Male HSC 2005 Jul 1, 2008 #2 the graph looks SIMILAR to x^3 but NOOOOOOOOOOOOOOO because x^2 dy/dx = 2x d^2y/dx^2=2 hence no point of inflexion ??
the graph looks SIMILAR to x^3 but NOOOOOOOOOOOOOOO because x^2 dy/dx = 2x d^2y/dx^2=2 hence no point of inflexion ??
N no_arg Member Joined Mar 25, 2004 Messages 67 Gender Undisclosed HSC N/A Jul 1, 2008 #3 Is only a change in concavity required?
R ronnknee Live to eat Joined Aug 2, 2006 Messages 473 Gender Male HSC 2008 Jul 6, 2008 #4 My Math teacher always tells us f''(x) = 0 is necessary but not sufficient proof for a point of inflexion. It must also change in concavity. In this case, yes it does change in concavity but f''(x) does not equal to 0. Therefore there is no point of inflexion Last edited: Jul 6, 2008
My Math teacher always tells us f''(x) = 0 is necessary but not sufficient proof for a point of inflexion. It must also change in concavity. In this case, yes it does change in concavity but f''(x) does not equal to 0. Therefore there is no point of inflexion
