find the absolute extrema of the function on the closed interval.\n\n$f(x)=8 - x, \\quad -2,4$\n\nminimum…

find the absolute extrema of the function on the closed interval.\n\n$f(x)=8 - x, \\quad -2,4$\n\nminimum $(x,y)=\\left(\\begin{array}{l}\\square\\end{array}\\right)$\n\nmaximum $(x,y)=\\left(\\begin{array}{l}\\square\\end{array}\\right)$

find the absolute extrema of the function on the closed interval.\n\n$f(x)=8 - x, \\quad -2,4$\n\nminimum $(x,y)=\\left(\\begin{array}{l}\\square\\end{array}\\right)$\n\nmaximum $(x,y)=\\left(\\begin{array}{l}\\square\\end{array}\\right)$

Answer

Explanation:

Step1: Find the derivative

The derivative of (f(x)=8 - x) is (f^\prime(x)=-1). Since (f^\prime(x)) is a constant ((- 1\neq0)), there are no critical points in the open interval ((-2,4)).

Step2: Evaluate the function at the endpoints

  • For (x=-2): (y = f(-2)=8-(-2)=10)
  • For (x = 4): (y=f(4)=8 - 4=4)

Answer:

minimum ((x,y)=(4,4)) maximum ((x,y)=(-2,10))