The function f(t) = frac{1}{t^2 + t - 2},wher | vedclass

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(B) Given the function $f(t) = \frac{1}{t^2 + t - 2}$.Factoring the denominator,we get $f(t) = \frac{1}{(t + 2)(t - 1)}$.The function $f(t)$ is discon

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$2, \frac{1}{2}$

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(B) Given the function $f(t) = \frac{1}{t^2 + t - 2}$.Factoring the denominator,we get $f(t) = \frac{1}{(t + 2)(t - 1)}$.The function $f(t)$ is discontinuous where the denominator is zero,i.e.,at $t = -2$ and $t = 1$.Now,substitute $t = \frac{1}{x - 1}$ to find the values of $x$:For $t = -2$:$\frac{1}{x - 1} = -2 \implies x - 1 = -\frac{1}{2} \implies x = 1 - \frac{1}{2} = \frac{1}{2}$.For $t = 1$:$\frac{1}{x - 1} = 1 \implies x - 1 = 1 \implies x = 2$.Additionally,the expression $t = \frac{1}{x - 1}$ is discontinuous at $x = 1$.However,checking the options provided,the points of discontinuity for the composite function $f(x)$ are $x = \frac{1}{2}$ and $x = 2$.

The function $f(t) = \frac{1}{t^2 + t - 2}$,where $t = \frac{1}{x - 1}$,is discontinuous at

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