Question

Find a point of discontinuity of the function $f\left(t\right)=\frac{1}{t^2+t-2}$, where $t=\frac{1}{x-1}$

Answer 1

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

Clearly $t=\frac{1}{x-1}$ is discontinuous and undefined at $x=1$,

For $x\ne 1$, we have

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

For $t=-2$,

$t=\frac{1}{x-1}\Rightarrow x=\frac{1}{2}$

For $t=1$,

$t=\frac{1}{x-1}\Rightarrow x=2$

Hence $f\left(x\right)$ is discontinuous at $x=\frac{1}{2},x=1$ and $x=2$.