Answer to Question #206093 in Calculus for Seebi

Question

Determine the location and values of the absolute maximum and absolute 

minimum for the given function: 

F(x) = (−x + 2)

Power 4

, 𝑤ℎ𝑒𝑟𝑒 0 ≤ x ≤ 3

Answer 1

$$F(x)=(-x+2)^4$$

$0\leq x\leq 3$

Find the first derivative

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

Find the critical number(s)

$$F'(x)=0=>-4(-x+2)^3=0$$

$$x=2$$

Critical number: $2$

$$F(0)=(-2+0)^4=16$$

$$F(3)=(-2+3)^4=1$$

$$F(2)=(-2+2)^4=0$$

The 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 $x=2: Point(2,0).$