Question

f(x)= x^3 - 2x^2

relative min and max using 2nd derivative test

Answer 1

f'(x)=\[3x^2-4x\]
set it equal zero. you get x=0 or x=4/3

f"(x)=6x-4
f"(0)=-4<0 so (0,0) is a relative maximum
f"(4/3)=calculate this :) but it is positive so... (4/3,-32/27) is a relative minimum