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(B) The function is defined as $f(x) = \frac{1}{1 - e^{-\frac{x+1}{x-2}}}$.First,the function is undefined when the denominator is zero,i.e.,$x - 2 =

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$2$

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(B) The function is defined as $f(x) = \frac{1}{1 - e^{-\frac{x+1}{x-2}}}$.First,the function is undefined when the denominator is zero,i.e.,$x - 2 = 0$,which gives $x = 2$.Second,the function is discontinuous when the denominator of the entire expression is zero,i.e.,$1 - e^{-\frac{x+1}{x-2}} = 0$.This implies $e^{-\frac{x+1}{x-2}} = 1$,which means $-\frac{x+1}{x-2} = 0$.This occurs when the numerator is zero,i.e.,$x + 1 = 0$,which gives $x = -1$.Thus,the points of discontinuity are $x = -1$ and $x = 2$.Therefore,the number of points of discontinuity is $2$.

The number of points of discontinuity of the function $f(x) = \frac{1}{1 - e^{\frac{-x-1}{x-2}}}$ is

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