# Thread: Find maximum and minimum value of

1. ## Find maximum and minimum value of

Find maximum and minimum value of

$\displaystyle f(x)=|1+2\cos x|+|1+2\sin x|, x\in R$

2. Originally Posted by pankaj
Find maximum and minimum value of

$\displaystyle f(x)=|1+2\cos x|+|1+2\sin x|, x\in R$
you only need to consider the interval $\displaystyle [0, 2\pi].$ now $\displaystyle \cos x = \frac{-1}{2}$ at $\displaystyle x = \frac{2 \pi}{3},\ \frac{4 \pi}{3}$ and $\displaystyle \sin x = \frac{-1}{2}$ at $\displaystyle x=\frac{7 \pi}{6}, \ \frac{11 \pi}{6}.$ thus:

$\displaystyle |1 + 2\cos x|=1 + 2 \cos x, \ |1 + 2\sin x|=1+2 \sin x$ for any $\displaystyle x \in [0, 2 \pi/3] \cup [11 \pi/6, 2\pi],$

$\displaystyle |1 + 2\cos x|=-1 - 2 \cos x, \ |1 + 2\sin x|=1+2 \sin x$ for any $\displaystyle x \in [2 \pi/3, 7\pi/6],$

$\displaystyle |1 + 2\cos x|=-1 - 2 \cos x, \ |1 + 2\sin x|=-1-2 \sin x$ for any $\displaystyle x \in [7 \pi/6, 4 \pi/3],$ and finally:

$\displaystyle |1 + 2\cos x|=1 + 2 \cos x, \ |1 + 2\sin x|=-1-2 \sin x$ for any $\displaystyle x \in [4 \pi/3, 11 \pi/6].$

now use the indentity $\displaystyle \sin x \pm \cos x = \sqrt{2} \sin(x \pm \pi/4)$ and the rest is easy. the maximum is quickly seen to be $\displaystyle 2(1+\sqrt{2}).$