# Science:Math Exam Resources/Courses/MATH100/December 2018/Question 07/Solution 1

We are being asked to show that the global maximum of $f(x)=\sin(x)+{\sqrt {3}}\cos(x)$ is at most equal to $2$ and its global minimum is at least equal to $-2$ . For this, note that the function is periodic with period $2\pi$ and therefore suffices to find the global maximum and the global minimum on the closed interval $[0,2\pi ]$ First determine the critical points in the interval by computing the derivative and setting it equal to 0:

$f'(x)=\cos(x)-{\sqrt {3}}\cdot \sin(x),$ so $f'(x)=0$ implies that $\tan(x)={\frac {1}{\sqrt {3}}}$ . The only solutions of thid equation in the interval $(0,2\pi )$ are $x={\frac {\pi }{6}}$ and $x=\pi +{\frac {\pi }{6}}$ . It remains to check the value of $f(x)$ at the endpoints of the interval and at the two critical points:

$f(0)={\sqrt {3}}{\text{, }}f\left({\frac {\pi }{6}}\right)={\frac {1}{2}}+{\sqrt {3}}\cdot {\frac {\sqrt {3}}{2}}=2.$ $f\left(\pi +{\frac {\pi }{6}}\right)=-{\frac {1}{2}}+{\sqrt {3}}\cdot {\frac {-{\sqrt {3}}}{2}}=-2{\text{ and }}f(2\pi )={\sqrt {3}}.$ Hence, $\color {blue}{\text{the global maximum is }}2{\text{ and the global minimum is }}-2$ , as desired.