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If domain of $f(x)$ is $[-1,2]$ then what will be the domain of $f([x]-x^2+4)$ $?$ Here $[.]$ is for greatest integer function.

Attempt: since domain of $f(x)$ is $[-1,2]$ therefore for $f([x]-x^2+4)$

$-1\le[x]-x^2+4\le2$

$\Rightarrow x^2\le[x]+5$ and $x^2\ge[x]+2$

solving first inequality, as $x^2$ is always positive so $x\ge-5$

Now I can start taking intervals of $x$ and solve them but this brute force method is not taking me anywhere near to the correct answer. Can someone explain me how is this problem solved? Please give an elaborate solution.

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An $x\in{\mathbb R}$ is admissible if $-1\leq\lfloor x\rfloor-x^2+4\leq2$, or if $$2\leq x^2-\lfloor x\rfloor\leq 5\ .$$ Since for any $x$ one has $x-1<\lfloor x\rfloor\leq x$, an admissible $x$ necessarily has to fulfill $$1\leq x^2-x\leq 5\ .$$ The auxiliary function $g(x):=x^2-x$ is $\geq6$ if $x\leq-2$ or $x\geq3$, and $g(x)\leq0$ for $0\leq x\leq1$. It follows that admissible $x$ can only be found in the two intervals $\>]-2,0[\>$ and $\>]1,3[\>$.

This means that we have to treat separately the intervals $$J_1:=\>]-2,-1[\>,\quad J_2:=\>[-1,0[\>,\quad J_3:=\>]0,1[\>,\quad J_4:=\>[1,2[\>\ .$$ In each of these the term $\lfloor x\rfloor$ is constant, so that it should not be too difficult to locate the truly admissible $x$ in these intervals.

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Write the condition $$x^2\leq [x]+5$$ as $$x^2-[x]-5\leq 0. \quad (*)$$ Without the greatest integer function, this inequality is $$x^2-x-5\leq 0. \quad (**)$$ The roots of the quadratic are $$x=\frac{1\pm\sqrt{21}}{2},$$ so that the inequality $(**)$ holds when $$-1.79\approx\frac{1-\sqrt{21}}{2}\leq x\leq \frac{1+\sqrt{21}}{2}\approx 2.79.$$ Thus the upper bound on $x$ must satisfy $$x^2-2-5\leq 0,$$ so you get $x\leq \sqrt{7}$. Hopefully, you should be able to proceed from there.

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  • $\begingroup$ I am having difficulty on understanding how you deduced $(**)$ inequality. can you elaborate more? $\endgroup$ – nikola Jul 16 '16 at 11:31
  • $\begingroup$ I just rewrote $(*)$ without the greatest integer function. $\endgroup$ – smcc Jul 16 '16 at 11:35
  • $\begingroup$ thanks, smcc. i was having confusion earlier but now i can see how you approached to the solution. $\endgroup$ – nikola Jul 16 '16 at 12:20

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