f'(x)=(2x³-6x²-18x+7)'=6x²-12x-18=6(x²-2x-3)
f'(x)=0=>6(x²-2x-3)=0=>x²-2x-3=0=>ecuatia are solutiile -1 si 3.
x |-infinit_______-1______3______________+infinit
f'(x)|+++++++++++0------------0+++++++++++++
f(x)| ccccccccccccc17ddddd-47ccccccccccc
c=crescatoare
d=descrescatoare
x=-1 punct de maxim local
x=3 punct de minim local
f'(x)=-8x³-16x=-8x(x²+2)
f'(x)=0=>-8x(x²+2)=0=>x=0
x |-infinit_______0____________________+infinit
f'(x)|+++++++++++0----------------------------------
f(x)| cccccccccccc9ddddddddddddddddddddd
x=0 punct de maxim global.
