Range of f(x) = $\frac{\sqrt3\,\sin x}{2 + \cos x}$ Can you give any idea about the range of the following function?
$$f(x) = \frac{\sqrt{3}\,\sin x}{2 + \cos x}$$

Yes:


*

*$-1 \leq \sin x \leq 1 \implies -\sqrt{3} \leq \sqrt{3}\sin x \leq \sqrt{3}$

*$-1 \leq \cos x \leq 1 \implies 1 \leq 2+\cos x \leq 3$


Hence: $$-\sqrt{3} \leq \frac{\sqrt{3}\sin x}{2+\cos x} \leq \sqrt{3}$$

Of course, you can do better than this using $f'(x)$ in order to calculate min/max points of $f(x)$:


*

*$f'(x)=\frac{\sqrt{3}(1+2\cos x)}{(2+\cos x)^2}$

*$f'(x)=0 \implies \cos x=-\frac{1}{2} \implies x=\frac{2}{3}\pi+2\pi k$


And then using $f''(x)$ in order to be sure that they are indeed min/max (and not curve) points:


*

*$f''(x)=\frac{4\sqrt{3}(\sin\frac{x}{2})^2\sin x}{(2+\cos x)^3}$

*$f''(x)=0 \implies x=\pi k \implies x \neq \frac{2}{3}\pi+2\pi k$


Finally, you can calculate the values of $f(x)$ at these points:


*

*$f(x)=\frac{\sqrt{3}\sin x}{2+\cos x}$

*$x=\frac{2}{3}\pi+2\pi k \implies -1 \leq f(x) \leq 1$

