Find the Maximum and Minimum Values of $f(x) = \cos\sqrt{x+1} - \cos\sqrt{x}$ I saw today in a fb forum the following excersice:
Find the range of values of 
$f(x)=\cos\sqrt{x+1}-\cos\sqrt{x}$
I have tried to find the mix and max of the function but failed. Any ideas?

To do an analytic check of the range, and possibly other properties, note you can combine the $2 \cos$ terms into a multiple of one $\sin$ term using several identities from List of trigonometric identities. First, note that $\cos(\theta) = \sin(\frac{\pi}{2} - \theta)$, so you get
$$f(x) = \cos\sqrt{x+1}-\cos\sqrt{x} = \sin\left(\frac{\pi}{2} - \sqrt{x+1}\right) - \sin\left(\frac{\pi}{2} - \sqrt{x}\right) \tag{1}\label{eq1}$$
Next, you can use the Arbitrary phase shift formula for a linear sum of $2$ $\sin$ values which gives
$$a\sin y+b\sin(y+\theta )=c\sin(y+\varphi ) \tag{2}\label{eq2}$$
where
$$c={\sqrt {a^{2}+b^{2}+2ab\cos \theta }} \tag{3}\label{eq3}$$
and
$$\varphi =\operatorname {atan2} \left(b\,\sin \theta ,a+b\cos \theta \right) \tag{4}\label{eq4}$$
where $\operatorname {atan2}$ is the principal arc-tangent angle in $(-\pi,\pi]$ (more details are at atan2)). For the case of \eqref{eq1}, you have $a = 1$, $b = -1$, $y = \frac{\pi}{2} - \sqrt{x+1}$ and $\theta = \sqrt{x+1} - \sqrt{x}$. As such, \eqref{eq3} gives
$$c = \sqrt{2 - 2\cos\left(\sqrt{x+1} - \sqrt{x}\right)} \tag{5}\label{eq5}$$
Since $\sqrt{x+1} - \sqrt{x}$ is a strictly decreasing function, from $1$ towards $0$, for $x \ge 0$ and $\cos$ is also a decreasing function in $(0,1]$, this means that at $x = 0$ you have $c = \sqrt{2 - 2\cos(1)}$ is it's maximum and, since $\cos(0) = 1$, its infimum is $0$ since $c \to 0$ as $x \to \infty$.
You can do similar appropriate checks for the range of $\varphi$ values in \eqref{eq4} and then determine, overall, the range of possible values for \eqref{eq1}.
