Question

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

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)