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Question

Given the function $$\displaystyle f\left ( x \right )=\dfrac {1}{x-1}.$$ Find the number of points of discontinuity of the composite function $$\displaystyle f\left [ f\left \{ f\left ( x \right ) \right \} \right ].$$


Solution

The point $$x=1$$ is clearly a point of discontinuity of the function $$\displaystyle y=f\left ( x \right )=\frac{1}{1-x}.$$ 
If $$\displaystyle x\neq 1,$$ then $$\displaystyle v\left ( x \right )=f\left [ f\left ( x \right ) \right ]=f\left ( \frac{1}{1-x} \right )=\frac{1}{1-\left [ 1/\left ( 1-x \right ) \right ]}=\frac{x-1}{x}$$ 
Hence, the point $$x=0$$ is a discontinuity of the function $$v$$.
If $$\displaystyle x\neq 0,x\neq 1,then w\left ( x \right )=f\left [ f\left \{ f\left ( x \right ) \right \} \right ]=f\left [ f\left ( \frac{1}{1-x} \right ) \right ]=f\left ( \frac{x-1}{x} \right )=\frac{1}{1-\left ( x-1 \right )/x}=x.$$
 Hence w is clearly continuous everywhere.
Thus,the points of discontinuity of the composite function $$\displaystyle f\left [ f\left \{ f\left ( x \right ) \right \} \right ] are  x= 0.x=1$$ and the composite function $$\displaystyle f\left \{ f\left ( x \right ) \right \}$$ has a discontinuity at $$\displaystyle x=1$$ only. 

Physics

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