Determine the range of $f(x)=\frac{\sin x}{x}$ I am having trouble understanding the solution to this question.
''Determine the range of the following function:
$f(x)$ = $(1$  $if$  $x=0)$ or (${\sin x\over x}$ if $x$$\neq$$0$)
where the domain is the set $E$=($-\infty$,0)$\cup$(0,$\infty$).
Answer:
The answer for the range is $R=(c,1]$ where $c=-\cos x_0$, such that $x_0$ is the smallest positive solution of $x=\tan x$. ''
Could someone please explain how we get this range?
Thanks
 A: note that $f(0) = 1$ and $f$ is even. so we only need to worry about the global minimum on $0 \le x < \infty$
the critical numbers of $f$ defined by $f(x) = \dfrac{\sin x}{x}$ are given by 
$ f^\prime(x)= \dfrac{x\cos x - \sin x}{x^2} = 0$ the positive critical numbers are the positive solutions $x \cos x - \sin x = 0$ which is equivalent to $\tan x = x$ this has a solution in $(\pi, 3/2 \pi) + k\pi$ where $k$ is nonnegative integer. the critical numbers corresponding to the even values of $k$ gives local min and odd values of $k$ give local max. the first local min is also the global min. let us call that $x_0$ which satisfies $\cos x_0 = \dfrac{\sin x_0}{x_0}$ and $\pi < x_0 < 3\pi/2.$  the range of $f$ is $[\cos x_0, 1]$
A: The function $ sinc (x)= \dfrac{\sin x }{x}$ has maximum value 1 and minimum that can be found by Chain Rule when we have 
$$\frac{\sin x }{x}=\frac{\cos x }{1}$$
or $$ \tan x = x $$ 
whose solution $x_{min}$ has to be  found graphically or numerically by iteration. Its corresponding $y_{min}$ negative value is evaluated. Then,
the range is $ 1>y> y_{min},$
because the graph shows the functon clinging to intermediate values of $y$.
