Answer to Question #206764 in Calculus for Shaboo

Question

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

Answer 1

given:

$f(x)=(-x+2)^4, \\where\space 0\le x\le3$

differentiate with respect to x

$f'(x)=4(-x+2)^3$

now

$f'(x)=0\\\implies-4(-x+2)^3=0\\\implies-x+2=0\\\implies x=2$

now

$f(0)=(0+2)^4=16\\f(1)=(-1+2)^4=1\\f(2)=(-2+2)^4=0\\f(3)=(-3+2)^4=1$

Therefore, the absolute maximum is at x=0 and the value is 16. the absolute minimum is at x=2 and the value is 0