Answer to Question #206291 in Calculus for Hashir

Question

Determine the location and values of the absolute maximum and absolute 

minimum for the given function: 

𝑓(𝑥) = (−𝑥 + 2)

ସ

, 𝑤ℎ𝑒𝑟𝑒 0 ≤ 𝑥 ≤ 3

Answer 1

$$F(x)=(-x+2)^3$$

$0\leq x\leq 3$

Find the first derivative

$$F'(x)=((-x+2)^3)'=-3(-x+2)^2$$

Find the critical number(s)

$$F'(x)=0=>-3(-x+2)^2=0$$

$$x=2$$

Critical number: $2$

$$F(0)=(-2+0)^3=-8$$

$$F(3)=(-2+3)^3=1$$

$$F(2)=(-2+2)^3=0$$

The 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 $-8$ on $[0, 3]$ at $x=0: Point(0,-8).$