f(x) = left[ frac{x^2 + 1}{x^2[|x|] + 1} righ | vedclass

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(B) Let $g(x) = \frac{x^2 + 1}{x^2[|x|] + 1}$.Case $1$: If $|x| < 1$,then $[|x|] = 0$. Thus,$g(x) = \frac{x^2 + 1}{0 + 1} = x^2 + 1$. Since $|x| < 1$,

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two points

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(B) Let $g(x) = \frac{x^2 + 1}{x^2[|x|] + 1}$.Case $1$: If $|x| Case $2$: If $|x| \ge 1$,then $[|x|] \ge 1$. Let $n = [|x|]$,where $n \in \{1, 2, 3, \dots\}$. Then $g(x) = \frac{x^2 + 1}{nx^2 + 1}$.As $x^2 	o \infty$,$g(x) 	o \frac{1}{n}$. Since $n \ge 1$,$0  0$. Thus,$[g(x)] = 0$ for $|x| \ge 1$ except when $g(x) = 1$,which happens at $x^2 = 1$ (i.e.,$x = \pm 1$).At $x = \pm 1$,$g(1) = \frac{1+1}{1(1)+1} = 1$,so $[g(1)] = 1$.Thus,$f(x) = 1$ for $|x|  1$,with $f(1) = f(-1) = 1$.The function jumps from $1$ to $0$ at $x = 1$ and $x = -1$. Hence,it is discontinuous at two points.

$f(x) = \left[ \frac{x^2 + 1}{x^2[|x|] + 1} \right]$ is discontinuous at (where $[.]$ denotes the greatest integer function):

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