Question

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

Answer 1

$$\begin{gathered} f\left(t\right)=\frac{1}{t^2+t-2} \\ \mathrm{Now},\mathrm{let}u=\frac{1}{x-1} \\ \therefore f\left(u\right)=\frac{1}{u^2+2u-u-2}=\frac{1}{u^2+u-2}=\frac{1}{\left(u+2\right)\left(u-1\right)} \\ \mathrm{So},f\left(u\right)\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{at}u=-2\mathrm{and}u=1 \\ \mathrm{If}u=-2,\mathrm{then} \\ -2=\frac{1}{x-1} \\ \Rightarrow 2x=1 \\ \Rightarrow x=\frac{1}{2} \\ \mathrm{If}u=1,\mathrm{then} \\ 1=\frac{1}{x-1} \\ \Rightarrow x=2 \end{gathered}$$
Hence, the function is discontinuous at $x=\frac{1}{2},2$