determine the location and value of the absolute maximum and absolute minimum for the given function. f(x) = (-x+2)⁴ , where 0≤x≤3 this answer is incomple please help me to complete it and provided...

The given function is:

f(x)=(-x+2)^4, 0\le x\le3f(x)=(−x+2)4,0≤x≤3

let c be the critical point. To find c

f'(x)=0\\\implies4(-x+2)^3(-1)=0\\\implies(-x+2)^3=0\\\implies x=2,2,2f′(x)=0⟹4(−x+2)3(−1)=0⟹(−x+2)3=0⟹x=2,2,2

The critical points are: 2,2,2

Now,

f''(2)=0\\f(2)=0f′′(2)=0f(2)=0

So the extreme value of the given function is 0 and the location is at (0,0)