Finding the absolute minimum of $f(x)=\frac{1}{x}\sin x$

Question

For the function $f(x)=\frac{1}{x}\sin x$, the domain is $\mathbb{R}-\left\{0\right\}$, but how do I analytically determine the range of the function? Since it's not $[-1,1]$.

Graph:

Answer 1

Hint: Let $f(x)={\displaystyle \sin x\over x}$. Note that $f(x)$ is an even function (its graph is symmetric about the $y$-axis). Note also that $$|f(x)|<{1\over 2\pi} \qquad \mbox{ for } |x|>2\pi.$$ Now it remains to analyze only a couple of local extrema points with $|x|\le2\pi$ and confirm that the one with $x\approx4.49$ is indeed the global minimum, with $f(x)<-{1\over2\pi}$. (You will get two minimum points which are symmetric about the $y$-axis.)