Point of inflexion?

Question

f(x)=x^2 - (27/x^2)
Second derivative gives a point of inflexion at (3,6). dy/dx is positive after 0. Without a change in concavity how can there exist a point of inflexion?

Answer 1

f''(x) is what gives you the concavity,
so while f'(x)>0 for x>0,
the second derivative changes signs

Answer 2

The second derivative is f"(x)= 2 - [ 162/(x^4) ]
if f"(x)=0 then you equation should end up in x^4=81 or just to make it easier get the square root to get x^2=9.
therefore x= +3 or -3.

Don't forget that (+)^2 or (-)^2 will give you a +.
The changes of concavity are given by the inflection points. (3;6) is one of them and the other is (-3;6) so in all this segment of the curve there will be no changes of concavity.

Also something you can't miss is the fact that x can't be 0. So x=0 is an asymptote but it still doesn't change the concavity.

Answer 3

f(x) = 2x^3 +3x^2 -36x +5 dy/dx = 12x² + 6x - 36 dy/dx = 0 x = 2 , -3 d²y/dx² = 24x + 6 d²y/dx² = 0 x = - a million/4 try the two area of x = - a million/4 say - a million/2 and 0 whilst x = - a million/2 d²y/dx² = - 6 whilst x = 0 d²y/dx² = 6 try exhibits that it is going from - ve to + ve so factor of inflection

Answer 4

The first derivative does not have to change sign at an inflection point. The second derivative does.