Question

Find the local and absolute extreme values of the function on the given interval. Also

specify the intervals where function is increasing or decreasing

𝑓(𝑥) = 𝑥 + 2𝑐𝑜𝑠𝑥

Answer 1

$$f(x)=x+2\cos x$$

Domain: $(-\infin, \infin)$

Find the fist derivative with respect to $x$

$$f'(x)=(x+2\cos x)'=1-2\sin x$$

Find the critical number(s)

$$f'(x)=0=>1-2\sin x=0$$

$$x=(-1)^n\dfrac{\pi}{6}+\pi n, n\in \Z$$

Critical numbers: $(-1)^n\dfrac{\pi}{6}+\pi n, n\in \Z$

Find the second derivative with respect to $x$

$$f''(x)=(1-2\sin x)'=-\cos x$$

$$f''(\dfrac{\pi}{6}+2\pi n)=-\dfrac{\sqrt{3}}{2}<0$$

$$f''(\dfrac{5\pi}{6}+2\pi n)=\dfrac{\sqrt{3}}{2}>0$$

i) If $x\in [-\pi, \pi]$

$$f(-\pi)=-\pi+2\cos (-\pi)=-\pi-2$$

$$f(\pi)=\pi+2\cos (\pi)=\pi-2$$

$$f(\dfrac{\pi}{6})=\dfrac{\pi}{6}+\sqrt{3}$$

$$f(\dfrac{5\pi}{6})=\dfrac{5\pi}{6}-\sqrt{3}$$

The function $f(x)$ has a local maximum on $[-\pi, \pi]$ with value of $\dfrac{\pi}{6}+\sqrt{3}$ at $x=\dfrac{\pi}{6}.$

The function $f(x)$ has a local minimum on $[-\pi, \pi]$ with value of $\dfrac{5\pi}{6}-\sqrt{3}$ at $x=\dfrac{5\pi}{6}.$

The function $f(x)$ has the absolute maximum on $[-\pi, \pi]$ with value of $\dfrac{\pi}{6}+\sqrt{3}$ at $x=\dfrac{\pi}{6}.$

The function $f(x)$ has the absolute minimum on $[-\pi, \pi]$ with value of $-\pi-2$ at $x=-\pi.$

The function $f(x)$ increases on $(-\pi, \dfrac{\pi}{6})\cup(\dfrac{5\pi}{6}, \pi).$

The function $f(x)$ decreases on $(\dfrac{\pi}{6}, \dfrac{5\pi}{6}).$

ii)

The function $f(x)$ increases on $(\dfrac{5\pi}{6}+2\pi n, \dfrac{13\pi}{6}+2\pi n), n\in \Z.$

The function $f(x)$ decreases on $(\dfrac{\pi}{6}+2\pi n, \dfrac{5\pi}{6}+2\pi n), n\in \Z.$