Answer to Question #286172 in Calculus for Abdullah

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 $(\dfrac{\pi}{6}+2\pi n, \dfrac{5\pi}{6}+2\pi n), n\in \Z.$