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

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MATH100 December 2018

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Question 07
Show that for each real number $x$ , we have $-2\leq \sin(x)+{\sqrt {3}}\cos(x)\leq 2$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Find the absolute maximum and minimum of the function.

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Solution
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.