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:

$f^{\prime }(x)=\cos (x)-\sqrt{3}\cdot \sin (x),$

so $f^{\prime }(x)=0$ 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:

$f(0)=\sqrt{3}\text{, }f\left(\frac{\pi }{6}\right)=\frac{1}{2}+\sqrt{3}\cdot \frac{\sqrt{3}}{2}=2.$

$f(\pi +\frac{\pi }{6})=-\frac{1}{2}+\sqrt{3}\cdot \frac{-\sqrt{3}}{2}=-2\text{ and }f(2\pi )=\sqrt{3}.$

Hence, $\text{the global maximum is }2\text{ and the global minimum is }-2$, as desired.