Domain of $f(x)$ is $[-1,2]$; find domain of $f([x]-x^2+4)$ [closed]

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The question goes like this
Domain of f(x) is [-1,2] find domain of f([x]-x²+4) [•] is greatest integer function.



I know that I have to do this
$-1leq [x]-x^2+4leq 2$
But what to do next. I can't even factorise it.




algebra-precalculus functions




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closed as off-topic by amWhy, José Carlos Santos, Chris Custer, Leucippus, Parcly Taxel Jul 26 at 1:10


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Chris Custer, Leucippus, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
















    up vote
    -1
    down vote

    favorite












    The question goes like this
    Domain of f(x) is [-1,2] find domain of f([x]-x²+4) [•] is greatest integer function.



    I know that I have to do this
    $-1leq [x]-x^2+4leq 2$
    But what to do next. I can't even factorise it.




    algebra-precalculus functions




    share|cite|improve this question













    closed as off-topic by amWhy, José Carlos Santos, Chris Custer, Leucippus, Parcly Taxel Jul 26 at 1:10


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Chris Custer, Leucippus, Parcly Taxel
    If this question can be reworded to fit the rules in the help center, please edit the question.














      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      The question goes like this
      Domain of f(x) is [-1,2] find domain of f([x]-x²+4) [•] is greatest integer function.



      I know that I have to do this
      $-1leq [x]-x^2+4leq 2$
      But what to do next. I can't even factorise it.




      algebra-precalculus functions




      share|cite|improve this question













      The question goes like this
      Domain of f(x) is [-1,2] find domain of f([x]-x²+4) [•] is greatest integer function.



      I know that I have to do this
      $-1leq [x]-x^2+4leq 2$
      But what to do next. I can't even factorise it.





      algebra-precalculus functions





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 25 at 21:40









      amWhy

      189k25219431




      189k25219431









      asked Jul 25 at 6:05









      user199925

      194




      194




      closed as off-topic by amWhy, José Carlos Santos, Chris Custer, Leucippus, Parcly Taxel Jul 26 at 1:10


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Chris Custer, Leucippus, Parcly Taxel
      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by amWhy, José Carlos Santos, Chris Custer, Leucippus, Parcly Taxel Jul 26 at 1:10


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Chris Custer, Leucippus, Parcly Taxel
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          1 Answer
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          Let $n le x <n+1$. It is easy to see that for $|n| ge 3, [x]-x^2+4 <-1$.



          Hence you have consider only the values $n=-2,-1,0,1,2$. Now try to solve the inequalities for these cases.






          share|cite|improve this answer



















          • 1




            We are defining $n = [x]$ (that is what $n le x < n+1$ says). $x$ cannot be 2.8 or 2.9 if $|n| ge 3$.
            – Ted
            Jul 25 at 7:15










          • Sorry guys for the trouble. I m really a blockhead. Now I understand everything that he wrote.
            – user199925
            Jul 25 at 13:34

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote













          Let $n le x <n+1$. It is easy to see that for $|n| ge 3, [x]-x^2+4 <-1$.



          Hence you have consider only the values $n=-2,-1,0,1,2$. Now try to solve the inequalities for these cases.






          share|cite|improve this answer



















          • 1




            We are defining $n = [x]$ (that is what $n le x < n+1$ says). $x$ cannot be 2.8 or 2.9 if $|n| ge 3$.
            – Ted
            Jul 25 at 7:15










          • Sorry guys for the trouble. I m really a blockhead. Now I understand everything that he wrote.
            – user199925
            Jul 25 at 13:34














          up vote
          2
          down vote













          Let $n le x <n+1$. It is easy to see that for $|n| ge 3, [x]-x^2+4 <-1$.



          Hence you have consider only the values $n=-2,-1,0,1,2$. Now try to solve the inequalities for these cases.






          share|cite|improve this answer



















          • 1




            We are defining $n = [x]$ (that is what $n le x < n+1$ says). $x$ cannot be 2.8 or 2.9 if $|n| ge 3$.
            – Ted
            Jul 25 at 7:15










          • Sorry guys for the trouble. I m really a blockhead. Now I understand everything that he wrote.
            – user199925
            Jul 25 at 13:34












          up vote
          2
          down vote










          up vote
          2
          down vote









          Let $n le x <n+1$. It is easy to see that for $|n| ge 3, [x]-x^2+4 <-1$.



          Hence you have consider only the values $n=-2,-1,0,1,2$. Now try to solve the inequalities for these cases.






          share|cite|improve this answer















          Let $n le x <n+1$. It is easy to see that for $|n| ge 3, [x]-x^2+4 <-1$.



          Hence you have consider only the values $n=-2,-1,0,1,2$. Now try to solve the inequalities for these cases.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 25 at 21:41









          amWhy

          189k25219431




          189k25219431











          answered Jul 25 at 6:29









          Kavi Rama Murthy

          20.1k2829




          20.1k2829







          • 1




            We are defining $n = [x]$ (that is what $n le x < n+1$ says). $x$ cannot be 2.8 or 2.9 if $|n| ge 3$.
            – Ted
            Jul 25 at 7:15










          • Sorry guys for the trouble. I m really a blockhead. Now I understand everything that he wrote.
            – user199925
            Jul 25 at 13:34












          • 1




            We are defining $n = [x]$ (that is what $n le x < n+1$ says). $x$ cannot be 2.8 or 2.9 if $|n| ge 3$.
            – Ted
            Jul 25 at 7:15










          • Sorry guys for the trouble. I m really a blockhead. Now I understand everything that he wrote.
            – user199925
            Jul 25 at 13:34







          1




          1




          We are defining $n = [x]$ (that is what $n le x < n+1$ says). $x$ cannot be 2.8 or 2.9 if $|n| ge 3$.
          – Ted
          Jul 25 at 7:15




          We are defining $n = [x]$ (that is what $n le x < n+1$ says). $x$ cannot be 2.8 or 2.9 if $|n| ge 3$.
          – Ted
          Jul 25 at 7:15












          Sorry guys for the trouble. I m really a blockhead. Now I understand everything that he wrote.
          – user199925
          Jul 25 at 13:34




          Sorry guys for the trouble. I m really a blockhead. Now I understand everything that he wrote.
          – user199925
          Jul 25 at 13:34


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