Find all points of discontinuity of the function

Given, $\quad f(t)=\frac{1}{t^{2}+t-2}$

$\Rightarrow f(t)=\frac{1}{\frac{1}{(x-1)^{2}}+\frac{1}{(x-1)}-2} \quad\left[\right.$ putting $\left.t=\frac{1}{x-1}\right]$

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

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

$=\frac{(x-1)^{2}}{-\left[2 x^{2}-4 x-x+2\right]}=\frac{(x-1)^{2}}{-[2 x(x-2)-1(x-2)]}$

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

Now,

if f(t) is discontinuous, then 2 – x = 0 ⇒ x = 2

And, 2x – 1 = 0 ⇒ x = ½

Therefore, the required points of discontinuity for the given function are 2 and ½.