consider the function on the given interval defined below.\n\n$f(x)=3x - 6\\cos(x), -\\pi,\\pi$\n\nfind the…

Explanation:

Step1: Find the derivative of the function

The derivative of (f(x)=3x - 6\cos(x)) is (f^\prime(x)=3 + 6\sin(x)) using the sum rule ((u + v)^\prime=u^\prime+v^\prime) (where (u = 3x), (u^\prime=3) and (v=-6\cos(x)), (v^\prime = 6\sin(x)))

Step2: Find the critical points

Set (f^\prime(x)=0), so (3+6\sin(x)=0). Then (\sin(x)=-\frac{1}{2}). On the interval ([-\pi,\pi]), (x =-\frac{\pi}{6}) and (x =-\frac{5\pi}{6})

Step3: Use the second - derivative test

The second - derivative (f^{\prime\prime}(x)=6\cos(x))

• For (x =-\frac{\pi}{6}): (f^{\prime\prime}(-\frac{\pi}{6})=6\cos(-\frac{\pi}{6}) = 6\times\frac{\sqrt{3}}{2}=3\sqrt{3}>0). So (x =-\frac{\pi}{6}) is a local minimum. (f(-\frac{\pi}{6})=3\times(-\frac{\pi}{6})-6\cos(-\frac{\pi}{6})=-\frac{\pi}{2}-6\times\frac{\sqrt{3}}{2}=-\frac{\pi}{2}-3\sqrt{3})
• For (x =-\frac{5\pi}{6}): (f^{\prime\prime}(-\frac{5\pi}{6})=6\cos(-\frac{5\pi}{6})=6\times(-\frac{\sqrt{3}}{2})=-3\sqrt{3}<0). So (x =-\frac{5\pi}{6}) is a local maximum. (f(-\frac{5\pi}{6})=3\times(-\frac{5\pi}{6})-6\cos(-\frac{5\pi}{6})=-\frac{5\pi}{2}-6\times(-\frac{\sqrt{3}}{2})=-\frac{5\pi}{2}+3\sqrt{3})

Step4: Evaluate the function at the endpoints

• When (x =-\pi): (f(-\pi)=3\times(-\pi)-6\cos(-\pi)=-3\pi + 6)
• When (x=\pi): (f(\pi)=3\pi-6\cos(\pi)=3\pi + 6)

Compare the values: (f(-\frac{5\pi}{6})=-\frac{5\pi}{2}+3\sqrt{3}\approx-\frac{5\times3.14}{2}+3\times1.732=-7.85 + 5.196=-2.654) (f(-\pi)=-3\pi + 6\approx-9.42+6=-3.42) (f(-\frac{\pi}{6})=-\frac{\pi}{2}-3\sqrt{3}\approx-1.57-5.196=-6.766) (f(\pi)=3\pi + 6\approx9.42 + 6=15.42)

Answer:

Local maxima: (-\frac{5\pi}{6}) Local minima: (-\frac{\pi}{6}) Absolute maxima: (\pi) Absolute minima: (-\frac{\pi}{6})