find the absolute extrema of the function on the closed interval.\n$y = 5\\cos x$, $0, 2\\pi$\nminimum…

find the absolute extrema of the function on the closed interval.\n$y = 5\\cos x$, $0, 2\\pi$\nminimum $(x,y)=()$\nmaximum $(x,y)=()$ (smaller $x$-value)\n$(x,y)=()$ (larger $x$-value)

find the absolute extrema of the function on the closed interval.\n$y = 5\\cos x$, $0, 2\\pi$\nminimum $(x,y)=()$\nmaximum $(x,y)=()$ (smaller $x$-value)\n$(x,y)=()$ (larger $x$-value)

Answer

Explanation:

Step1: Find the derivative of the function

The derivative of (y = 5\cos x) is (y'=-5\sin x).

Step2: Find the critical points

Set (y' = 0), so (-5\sin x=0). Then (\sin x = 0) in the interval ([0,2\pi]). The solutions are (x = 0,\pi,2\pi).

Step3: Evaluate the function at the critical points and endpoints

  • When (x = 0), (y=5\cos(0)=5).
  • When (x=\pi), (y = 5\cos(\pi)=- 5).
  • When (x = 2\pi), (y=5\cos(2\pi)=5).

Answer:

minimum ((x,y)=(\pi,-5)) maximum ((x,y)=(0,5)) (smaller (x) - value) ((x,y)=(2\pi,5)) (larger (x) - value)