Find all points of discontinuity of the function $f(t) = \frac{1}{{{t^2} + t - 2}}$, where $t = \frac{1}{{x - 1}}$

We have, $f(t) = \frac{1}{{{t^2} + t - 2}}$ and $t = \frac{1}{{x - 1}}$

therefore,$f(t) = \frac{1}{{\left( {\frac{1}{{{x^2} + 1 - 2x}}} \right) + \left( {\frac{1}{{x - 1}}} \right) - \frac{2}{1}}}$

$= \frac{1}{{\left( {\frac{{1 + x - 1 + \left[ { - 2{{(x - 1)}^2}} \right]}}{{\left( {{x^2} + 1 - 2x} \right)}}} \right)}}$

$= \frac{{{x^2} + 1 - 2x}}{{x - 2{x^2} - 2 + 4x}}$

$= \frac{{{x^2} + 1 - 2x}}{{ - 2{x^2} + 5x - 2}}$

$= \frac{{{{(x - 1)}^2}}}{{ - \left( {2{x^2} - 5x + 2} \right)}}$

$= \frac{{{{(x - 1)}^2}}}{{(2x - 1)(2 - x)}}$

So, $f(t)$ is discontinuous at $2x - 1 = 0 \Rightarrow x = 1/2$

and $2 - x = 0 \Rightarrow x = 2$

Therefore the given function is discontinuous at $x = 1/2$ $\&$ x = 2