For the function  x³ + 2x² - 4x the derivative is equal to 3x² + 4x - 4 and the second derivative is equal to 6x + 4. If we equal the derivative functions to zero we would have  3x² + 4x - 4=0 is (3x-2)(x+2)=0 x will be equal to -2 and 2/3, these are the critical points. When we plug each one to the second derivative we will find 6(-2)+4=-8 and 6(2/3)+4=8. -8<0 which makes x=-2 a local maximum and 8>0 which makes x=2/3 a local minimum. The answer is D.