F(x)=3xˇ5-20xˇ3
f´(x)=15xˇ4-60xˇ2=15xˇ2(xˇ2-4)=15xˇ2(x+2)(x-2)=0
x1=0,x2=-2,x3=2
f´´(x)=60xˇ3-120x=60x(xˇ2 - 2)
f´´(x1)=f´´(0)=0
f´´(x2)=f´´(-2)=60.(-2).2=-240 max
f´´(x3)=f´´(2)=60.2.2=240 min
f(-2)=3.(-2)ˇ5-20.(-2)ˇ3=3.(-32)-20(-8)=-96+160=64
Otvet: max v točke M(-2,64)