Question

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

Answer 1

Use the concept of continuity of composite functions Given: $f\left(t\right)=\frac{1}{t^2+t-2}$, where $t=\left(\frac{1}{x-1}\right)$.
Clearly, at $x=1$, 𝑡 is not defined.
$\therefore$ $x=1$ is not in the domain of $F\left(x\right)$.
Hence $f\left(x\right)$ is discontinuous at $x=1$.
Consider $F\left(x\right)=\frac{1}{{\left(\frac{1}{x-1}\right)}^2+\left(\frac{1}{x-1}\right)-2}$
$=\frac{(x-1{)}^2}{(1+(x-1)-2(x-1{)}^2}$ $=\frac{(x-1{)}^2}{-(2x^2-5x+2)}=\frac{(x-1{)}^2}{(2x-1)(2-x)}$
∵ The quotient of two continuous functions
i.e.$\frac{h\left(x\right)}{g\left(x\right)}$ is discontinuous at $x=a$ if $g\left(a\right)=0$
So, $f\left(x\right)$ is discontinuous at $2x-1=0$ $\Rightarrow x=\frac{1}{2}$ and $2-x=0\Rightarrow x=2$
Hence the function $f\left(x\right)$ is discontinuous at $x=1$, $\frac{1}{2}$ and $2$