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(B) To find the points of discontinuity for $(f \circ g)(x)$,we check the points where $g(x)$ is discontinuous,the points where $f(u)$ is discontinuou

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two points

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(B) To find the points of discontinuity for $(f \circ g)(x)$,we check the points where $g(x)$ is discontinuous,the points where $f(u)$ is discontinuous (where $u = g(x)$),and the points where $g(x)$ equals the value at which $f$ is discontinuous.$1$. $g(x)$ is continuous everywhere since $\lim_{x 	o 0^-} g(x) = e^0 - 0 = 1$ and $g(0) = (0-1)^2 - 1 = 0$. Wait,$g(x)$ is discontinuous at $x=0$ because $\lim_{x 	o 0^-} g(x) = 1$ and $g(0) = 0$. So,$x=0$ is a point of discontinuity for $f \circ g$.$2$. $f(u)$ is discontinuous at $u=0$ (since $[u]$ is discontinuous at integers and $|1-u|$ is continuous,but at $u=0$,$\lim_{u 	o 0^-} [u] = -1$ and $f(0) = |1-0| = 1$).$3$. We check $g(x) = 0$. For $x $4$. At $x=2$: $\lim_{x 	o 2^+} (f \circ g)(x) = f(\lim_{x 	o 2^+} g(x)) = f(0^+) = |1-0| = 1$. $\lim_{x 	o 2^-} (f \circ g)(x) = f(\lim_{x 	o 2^-} g(x)) = f(0^-) = [-0] = 0$. Since the limits are not equal,$x=2$ is a point of discontinuity.$5$. At $x=0$: $\lim_{x 	o 0^-} (f \circ g)(x) = f(\lim_{x 	o 0^-} g(x)) = f(1) = |1-1| = 0$. $\lim_{x 	o 0^+} (f \circ g)(x) = f(g(0)) = f(0) = 1$. Since $0 
eq 1$,$x=0$ is a point of discontinuity.Thus,the function is discontinuous at $x=0$ and $x=2$. There are two points of discontinuity.

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