Question

Find all point of discontinuity of the function 

\[f\left( t \right) = \frac{1}{t^2 + t - 2}, \text{ where }  t = \frac{1}{x - 1}\]

Answer 1

\[f\left( t \right) = \frac{1}{t^2 + t - 2}\]
\[\text{ Now, let  u }  = \frac{1}{x - 1}\]
\[ \therefore f\left( u \right) = \frac{1}{u^2 + 2u - u - 2} = \frac{1}{u^2 + u - 2} = \frac{1}{\left( u + 2 \right)\left( u - 1 \right)}\]
\[So, f\left( u \right) \text{is not defined at} u = - 2 \text{ and }  u = 1\]
\[\text{ If u = - 2, then } \]
\[ - 2 = \frac{1}{x - 1}\]
\[ \Rightarrow 2x = 1\]
\[ \Rightarrow x = \frac{1}{2}\]
\[\text{ If u = 1, then} \]
\[1 = \frac{1}{x - 1}\]
\[ \Rightarrow x = 2\]

Hence, the function is discontinuous at 

\[x = \frac{1}{2}, 2\]