F(x)=(βx+2)4 0β€xβ€3
Find the first derivative
Fβ²(x)=((βx+2)4)β²=β4(βx+2)3 Find the critical number(s)
Fβ²(x)=0=>β4(βx+2)3=0
x=2 Critical number: 2
F(0)=(β2+0)4=16
F(3)=(β2+3)4=1
F(2)=(β2+2)4=0The function F(x) has the absolute maximum with value of 16 on [0,3] at x=0:Point(0,16).
The function F(x) has the absolute minimum with value of on [0,3] at x=2:Point(2,0).
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