The function f(x) = begin{cases} sgn([x]) & x | vedclass

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(B) Let us analyze the function $f(x)$ for different intervals.For $x 
otin I$,$f(x) = sgn([x])$.If $x \in (0, 1)$,$[x] = 0$,so $f(x) = sgn(0) = 0$.I

Vedclass · Accepted Answer

Discontinuous at every integer

Vedclass · Answer

(B) Let us analyze the function $f(x)$ for different intervals.For $x 
otin I$,$f(x) = sgn([x])$.If $x \in (0, 1)$,$[x] = 0$,so $f(x) = sgn(0) = 0$.If $x \in (1, 2)$,$[x] = 1$,so $f(x) = sgn(1) = 1$.If $x \in (-1, 0)$,$[x] = -1$,so $f(x) = sgn(-1) = -1$.For $x \in I$,$f(x) = [sgn(x)]$.If $x = 0$,$f(0) = [sgn(0)] = [0] = 0$.If $x > 0$ and $x \in I$,$f(x) = [sgn(x)] = [1] = 1$.If $x Combining these,the function is discontinuous at every integer $n \in I$ because the left-hand limit and right-hand limit do not match the function value at integers. Specifically,for any integer $n$,$\lim_{x 	o n^-} f(x) = sgn(n-1)$ and $\lim_{x 	o n^+} f(x) = sgn(n)$. These are not equal to $f(n) = [sgn(n)]$. Thus,the function is discontinuous at every integer.

The function $f(x) = \begin{cases} sgn([x]) & x \notin I \\ [sgn(x)] & x \in I \end{cases}$ is (where $sgn()$ denotes the signum function and $[.]$ denotes the greatest integer function):

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