Question 397766
{{{df/dx  =4x^3 + 3x^2}}} ==>  {{{d^2f/dx^2 = 12x^2 + 6x }}}.
Set the 2nd derivative to 0:

 {{{d^2f/dx^2 = 12x^2 + 6x  = 6x(2x + 1) = 0}}} ==> x = 0, -1/2.
When x = -1, {{{d^2f/dx^2 > 0}}},
when x = -1/4, {{{d^2f/dx^2 < 0}}}, and 
when x = 1, {{{d^2f/dx^2 > 0}}}.
Therefore there are changes of concavity in the intervals ({{{-infinity}}}, -1/2), (-1/2, 0), and (0, {{{infinity}}}).  To find the y-coordinates, substitute x = 0 and x = -1/2 in to original function.

The answer is c.