Question

consider the function on the given interval defined below.

$f(x)=3x - 6\cos(x), -\pi,\pi$

find the local and absolute maxima and minima points. if there are multiple points in a single category, list the points in a comma - separated list. if no such point exists, enter dne.

local maxima:

local minima:

absolute maxima:

absolute minima:

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=3x - 6\cos(x)\) is \(f^\prime(x)=3 + 6\sin(x)\).

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 \([-\pi,\pi]\), \(x =-\frac{\pi}{6},\frac{7\pi}{6}\). But \(\frac{7\pi}{6}
otin[-\pi,\pi]\), so the critical point is \(x =-\frac{\pi}{6}\).

Step3: Use the second - derivative test

The second - derivative \(f^{\prime\prime}(x)=6\cos(x)\).
When \(x =-\frac{\pi}{6}\), \(f^{\prime\prime}(-\frac{\pi}{6})=6\cos(-\frac{\pi}{6}) = 6\times\frac{\sqrt{3}}{2}=3\sqrt{3}>0\). So \(x =-\frac{\pi}{6}\) is a local minimum.

Step4: Evaluate the function at the endpoints and the critical point

• \(f(-\pi)=3(-\pi)-6\cos(-\pi)=-3\pi + 6\approx - 3\times3.14+6=-9.42 + 6=-3.42\)
• \(f(-\frac{\pi}{6})=3(-\frac{\pi}{6})-6\cos(-\frac{\pi}{6})=-\frac{\pi}{2}-6\times\frac{\sqrt{3}}{2}\approx-1.57-5.196=-6.766\)
• \(f(\pi)=3\pi-6\cos(\pi)=3\pi + 6\approx3\times3.14+6 = 9.42+6=15.42\)

Answer:

Local maxima: DNE
Local minima: \((-\frac{\pi}{6},-\frac{\pi}{2}-3\sqrt{3})\)
Absolute maxima: \((\pi,3\pi + 6)\)
Absolute minima: \((-\frac{\pi}{6},-\frac{\pi}{2}-3\sqrt{3})\)