Question #206093

Determine the location and values of the absolute maximum and absolute 

minimum for the given function: 

F(x) = (βˆ’x + 2)

Power 4

, π‘€β„Žπ‘’π‘Ÿπ‘’ 0 ≀ x ≀ 3


1
Expert's answer
2021-06-14T13:57:39-0400
F(x)=(βˆ’x+2)4F(x)=(-x+2)^4

0≀x≀30\leq x\leq 3

Find the first derivative


Fβ€²(x)=((βˆ’x+2)4)β€²=βˆ’4(βˆ’x+2)3F'(x)=((-x+2)^4)'=-4(-x+2)^3

Find the critical number(s)


Fβ€²(x)=0=>βˆ’4(βˆ’x+2)3=0F'(x)=0=>-4(-x+2)^3=0

x=2x=2

Critical number: 22


F(0)=(βˆ’2+0)4=16F(0)=(-2+0)^4=16


F(3)=(βˆ’2+3)4=1F(3)=(-2+3)^4=1

F(2)=(βˆ’2+2)4=0F(2)=(-2+2)^4=0

The function F(x)F(x) has the absolute maximum with value of 1616 on [0,3][0, 3] at x=0:Point(0,16).x=0: Point(0,16).


The function F(x)F(x) has the absolute minimum with value of on [0,3][0, 3] at x=2:Point(2,0).x=2: Point(2,0).



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