Find domain of the function $y=frac{1}{sqrt{5+2lfloor -3xrfloor}}$











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I'm a starter in learning functions. Would you help me with this?



$$y=frac{1}{sqrt{5+2lfloor -3xrfloor}}$$



Tell me what inequalities i must solve?






calculus functions







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  • 4




    No problem. So, all you need is $5+2lfloor -3xrfloor>0$.
    – lulu
    Nov 25 at 20:23












  • Would you solve it in 3 or 2 lines of an instructing ANSWER please , bro?
    – user602338
    Nov 25 at 20:24










  • I'm not going to write out a full solution...but, honestly, it's almost there already. The inequality I wrote before is the same as $lfloor -3x rfloor >-2.5$ and since the floor is always an integer we must have $lfloor -3x rfloor ≥-2$
    – lulu
    Nov 25 at 20:26












  • Thanks. Bro. Would you rewrite the title! Sorry i'm a starter here to!
    – user602338
    Nov 25 at 20:28










  • I did edit the title (after you confirmed that my expression was what you had intended). Maybe you just need to refresh the screen.
    – lulu
    Nov 25 at 20:29















up vote
-1
down vote

favorite












I'm a starter in learning functions. Would you help me with this?



$$y=frac{1}{sqrt{5+2lfloor -3xrfloor}}$$



Tell me what inequalities i must solve?






calculus functions







share|cite|improve this question




















  • 4




    No problem. So, all you need is $5+2lfloor -3xrfloor>0$.
    – lulu
    Nov 25 at 20:23












  • Would you solve it in 3 or 2 lines of an instructing ANSWER please , bro?
    – user602338
    Nov 25 at 20:24










  • I'm not going to write out a full solution...but, honestly, it's almost there already. The inequality I wrote before is the same as $lfloor -3x rfloor >-2.5$ and since the floor is always an integer we must have $lfloor -3x rfloor ≥-2$
    – lulu
    Nov 25 at 20:26












  • Thanks. Bro. Would you rewrite the title! Sorry i'm a starter here to!
    – user602338
    Nov 25 at 20:28










  • I did edit the title (after you confirmed that my expression was what you had intended). Maybe you just need to refresh the screen.
    – lulu
    Nov 25 at 20:29













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I'm a starter in learning functions. Would you help me with this?



$$y=frac{1}{sqrt{5+2lfloor -3xrfloor}}$$



Tell me what inequalities i must solve?






calculus functions







share|cite|improve this question















I'm a starter in learning functions. Would you help me with this?



$$y=frac{1}{sqrt{5+2lfloor -3xrfloor}}$$



Tell me what inequalities i must solve?






calculus functions




calculus functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 20:21









lulu

38.4k24476




38.4k24476










asked Nov 25 at 20:19









user602338

1326




1326








  • 4




    No problem. So, all you need is $5+2lfloor -3xrfloor>0$.
    – lulu
    Nov 25 at 20:23












  • Would you solve it in 3 or 2 lines of an instructing ANSWER please , bro?
    – user602338
    Nov 25 at 20:24










  • I'm not going to write out a full solution...but, honestly, it's almost there already. The inequality I wrote before is the same as $lfloor -3x rfloor >-2.5$ and since the floor is always an integer we must have $lfloor -3x rfloor ≥-2$
    – lulu
    Nov 25 at 20:26












  • Thanks. Bro. Would you rewrite the title! Sorry i'm a starter here to!
    – user602338
    Nov 25 at 20:28










  • I did edit the title (after you confirmed that my expression was what you had intended). Maybe you just need to refresh the screen.
    – lulu
    Nov 25 at 20:29














  • 4




    No problem. So, all you need is $5+2lfloor -3xrfloor>0$.
    – lulu
    Nov 25 at 20:23












  • Would you solve it in 3 or 2 lines of an instructing ANSWER please , bro?
    – user602338
    Nov 25 at 20:24










  • I'm not going to write out a full solution...but, honestly, it's almost there already. The inequality I wrote before is the same as $lfloor -3x rfloor >-2.5$ and since the floor is always an integer we must have $lfloor -3x rfloor ≥-2$
    – lulu
    Nov 25 at 20:26












  • Thanks. Bro. Would you rewrite the title! Sorry i'm a starter here to!
    – user602338
    Nov 25 at 20:28










  • I did edit the title (after you confirmed that my expression was what you had intended). Maybe you just need to refresh the screen.
    – lulu
    Nov 25 at 20:29








4




4




No problem. So, all you need is $5+2lfloor -3xrfloor>0$.
– lulu
Nov 25 at 20:23






No problem. So, all you need is $5+2lfloor -3xrfloor>0$.
– lulu
Nov 25 at 20:23














Would you solve it in 3 or 2 lines of an instructing ANSWER please , bro?
– user602338
Nov 25 at 20:24




Would you solve it in 3 or 2 lines of an instructing ANSWER please , bro?
– user602338
Nov 25 at 20:24












I'm not going to write out a full solution...but, honestly, it's almost there already. The inequality I wrote before is the same as $lfloor -3x rfloor >-2.5$ and since the floor is always an integer we must have $lfloor -3x rfloor ≥-2$
– lulu
Nov 25 at 20:26






I'm not going to write out a full solution...but, honestly, it's almost there already. The inequality I wrote before is the same as $lfloor -3x rfloor >-2.5$ and since the floor is always an integer we must have $lfloor -3x rfloor ≥-2$
– lulu
Nov 25 at 20:26














Thanks. Bro. Would you rewrite the title! Sorry i'm a starter here to!
– user602338
Nov 25 at 20:28




Thanks. Bro. Would you rewrite the title! Sorry i'm a starter here to!
– user602338
Nov 25 at 20:28












I did edit the title (after you confirmed that my expression was what you had intended). Maybe you just need to refresh the screen.
– lulu
Nov 25 at 20:29




I did edit the title (after you confirmed that my expression was what you had intended). Maybe you just need to refresh the screen.
– lulu
Nov 25 at 20:29















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