Range of f(x) = $\frac{\sqrt3\,\sin x}{2 + \cos x}$

Question

Can you give any idea about the range of the following function?

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

Answer 1

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

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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:

• $f(x)=\frac{\sqrt{3}\sin x}{2+\cos x}$
• $x=\frac{2}{3}\pi+2\pi k \implies -1 \leq f(x) \leq 1$

Answer 2

First notice that your function is continuous and defined everywhere, since $2+cos x \ge 0 \ \forall x$. At this point, I do recommend you check out the plot.

The derivative of the function is $(\sqrt{3} (2 \cos(x)+1))/(\cos(x)+2)^2$, and it has roots $x = 2/3 (3 \pi n \pm \pi)$, for $n \in \mathbb Z$. This means that the extremal values of $\sqrt{3} \sin(x)/(2+\cos(x))$ are $\pm 1$.

Answer 3

Consider $f^2(x) = \dfrac{3 - 3cos^2x}{cos^2x + 4cosx + 4}$. Now let $t = cosx$, then look at $f(t) = \dfrac{3 - 3t^2}{t^2 + 4t + 4}$. We find $f_{max}$. $f'(t) = \dfrac{-6(2t + 1)}{t + 2} = 0$ when $t = \dfrac{-1}{2}$. So $f(-1) = 0$, $f(\frac{-1}{2}) = 1$, and $f(1) = 0$. So $f_{max}^2 = 1$. This means that: $f^2(x) \le 1$, and so: $-1 \le f(x) \le 1$. So the range of $f$ is: $[-1, 1]$