F(x)=(βx+2)3 0β€xβ€3
Find the first derivative
Fβ²(x)=((βx+2)3)β²=β3(βx+2)2 Find the critical number(s)
Fβ²(x)=0=>β3(βx+2)2=0
x=2 Critical number: 2
F(0)=(β2+0)3=β8
F(3)=(β2+3)3=1
F(2)=(β2+2)3=0The function F(x) has the absolute maximum with value of 1 on [0,3] at x=3:Point(3,1).
The function F(x) has the absolute minimum with value of β8 on [0,3] at x=0:Point(0,β8).
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