The function $f(x) = \frac{{4 - {x^2}}}{{4x - {x^3}}}$ is

We have, $f(x) = \frac{{4 - {x^2}}}{{4x - {x^3}}} = \frac{{\left( {4 - {x^2}} \right)}}{{x\left( {4 - {x^2}} \right)}}$

$= \frac{{\left( {4 - {x^2}} \right)}}{{x\left( {{2^2} - {x^2}} \right)}} = \frac{{4 - {x^2}}}{{x(2 + x)(2 - x)}}$

Clearly, $f(x)$ is discontinuous at exactly three points $x = 0,x = - 2$ and $x = 2$.