How can I find the range of the function $f(x) =\frac{2 - x \cos x}{2 + x \cos x}$? How can I find the range of the function $f(x) =\frac{2 - x \cos x}{2 + x \cos x}$?
I was trying to see if the function is monotone or not. But the calculation is becoming so tedious.
Can anyone please help me by giving some hint?
 A: $$f(x)=\frac{4}{2+x\cos{x}}-1$$
Consider $g(x)=x\cos{x}$.
This is function has a range $\mathbb{R}$, which you can see by a bit of imagination. The $x$ part of the function increases linearly, while $\cos{x}$ will keep varying this linear input sinusoidally. It will be like a $x-t$ graph of Simple Harmonic motion ($x=A\sin{\omega t}$), but with a twist that the amplitude ($A$) keeps increasing linearly with time. So at $x \to \infty$ the function will vary rapidly from $+\infty$ to $-\infty$. Thus its range comes out to be $(-\infty,\infty)$ 
So from this you can calculate the range of $f(x)$ by replacing $x\cos x$ with another function of same range ($y=x$ for eg.)
So the range of $f(x)$ will be the range of $h(x)$, where :
$$h(x)=\frac{4}{2+x}-1$$
This is a rectangular hyperbola (shifted), of the form $xy=k....(k=constant)$ which has a rather commonly known graph (and hence range).
Thus the range is $\mathbb{R}-\{-1\}$
NOTE: Graph of $y=x\cos{x}$ if interested (Desmos)
A: Consider  $f(t) =\frac{2-t}{2+t}$. We have $f’(t) = \frac{-2}{(2+t)^2} \lt 0$ and so $f$ is monotone decreasing. Also, it is easy to see that the range of $f$ is $\mathbb R -\{-1\}$ since $\frac{2-t}{2+t} = c \ne -1$ will always have a unique solution for $t$.
Now, clearly $x\cos x$ is continuous and thus has range $\mathbb R$, so replacing $t$ by $x\cos x$ wouldn’t change the range of $f$. Hence, the range of $\frac{2-x\cos x}{2+x\cos x} $ is $\mathbb R -\{-1\}$.
A: Let $r$ denote a number in the range of $f$. We have
$$\begin{align}
r={2-x\cos x\over2+x\cos x}&\iff r(2+x\cos x)=2-x\cos x\\
&\iff (r+1)x\cos x=2(1-r)\\
&\iff x\cos x=\displaystyle{2(1-r)\over 1+r}
\end{align}$$
Now the final right hand side does not make sense for $r=-1$, so $-1$ is not in the range. But for every other value of $r$, the equation $x\cos x=2(1-r)/(1+r)$ is satisfied by some value of $x$. (On each interval of the form $(2n-1)\pi\le x\le2n\pi$, with $n\in\mathbb{Z}$, the function $x\cos x$ takes on all values between $-(2n-1)\pi$ and $2n\pi$, so every equation of the form $x\cos x=R$ is satisfied, in fact, by infinitely many values of $x$.) The range of $f$ is therefore $\mathbb{R}\backslash\{-1\}$.
