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Let $f(x) = [x^3-3]$ where x is the greatest integer function. Then find the number of points in the interval (1,2) where this function is discontinuous ?

Please suggest , how to proceed in this question.

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2 Answers 2

HINT: I’m going to use the more modern notation $\lfloor x\rfloor$ for the greatest integer (or floor) function instead of $[x]$. We’re interested in points of discontinuity of the function $f(x)=\lfloor x^3-3\rfloor$ on the interval $(1,2)$. The floor function is constant on intervals between consecutive integers and jumps at each integer, so it has a discontinuity at each integer. Thus, $f(x)$ will have a discontinuity at each $x\in(1,2)$ at which $x^3-3$ is an integer. So for what real numbers $x\in(1,2)$ is it true that $x^3-3$ is an integer?

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Function $x\longmapsto x^3-3$ is continuous and monotonically increasing in your interval. Hence its range is the interval $(-2,5)$ and each value is attained once. Then the range of $f$ is $[-2,4]\cap\mathbb{Z}$, which contains 7 elements, producing 6 points of discontinuity in $(1,2)$

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I can write the given function as $[x^3]-3$ So the values in the given interval can start from : (1.1111. 1.222.1.333,1.4444,1.5555,1.6666,1.7777,1.888,1.9999) which gives the result : (-2,-2,-1,0,0,1,2,3,4) Can we do something of this.... –  sultan Apr 30 '13 at 8:31
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