There is a general method for finding ranges of function without resorting to calculus?











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I'm having trouble to find the range of more complicated functions such as



$ f(x) = frac{1}{sqrt{x + 1}} $



How one should proceed to tackled down these functions, specially the ones involving roots and quotients, without using tools from calculus? (limits, derivatives, etc.)



Thanks in advance.






algebra-precalculus functions







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  • Thank you very much.
    – Vinicius L. Deloi
    7 hours ago






  • 1




    Hint: Consider $sqrt {x+1}$ where $x lt -1$. Also consider what happens to $frac 1{sqrt{x+1}}$ when $x=-1$.
    – Raptor
    7 hours ago















up vote
0
down vote

favorite












I'm having trouble to find the range of more complicated functions such as



$ f(x) = frac{1}{sqrt{x + 1}} $



How one should proceed to tackled down these functions, specially the ones involving roots and quotients, without using tools from calculus? (limits, derivatives, etc.)



Thanks in advance.






algebra-precalculus functions







share|cite|improve this question
























  • Thank you very much.
    – Vinicius L. Deloi
    7 hours ago






  • 1




    Hint: Consider $sqrt {x+1}$ where $x lt -1$. Also consider what happens to $frac 1{sqrt{x+1}}$ when $x=-1$.
    – Raptor
    7 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm having trouble to find the range of more complicated functions such as



$ f(x) = frac{1}{sqrt{x + 1}} $



How one should proceed to tackled down these functions, specially the ones involving roots and quotients, without using tools from calculus? (limits, derivatives, etc.)



Thanks in advance.






algebra-precalculus functions







share|cite|improve this question















I'm having trouble to find the range of more complicated functions such as



$ f(x) = frac{1}{sqrt{x + 1}} $



How one should proceed to tackled down these functions, specially the ones involving roots and quotients, without using tools from calculus? (limits, derivatives, etc.)



Thanks in advance.






algebra-precalculus functions




algebra-precalculus functions






share|cite|improve this question















share|cite|improve this question













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edited 7 hours ago

























asked 7 hours ago









Vinicius L. Deloi

6071410




6071410












  • Thank you very much.
    – Vinicius L. Deloi
    7 hours ago






  • 1




    Hint: Consider $sqrt {x+1}$ where $x lt -1$. Also consider what happens to $frac 1{sqrt{x+1}}$ when $x=-1$.
    – Raptor
    7 hours ago


















  • Thank you very much.
    – Vinicius L. Deloi
    7 hours ago






  • 1




    Hint: Consider $sqrt {x+1}$ where $x lt -1$. Also consider what happens to $frac 1{sqrt{x+1}}$ when $x=-1$.
    – Raptor
    7 hours ago
















Thank you very much.
– Vinicius L. Deloi
7 hours ago




Thank you very much.
– Vinicius L. Deloi
7 hours ago




1




1




Hint: Consider $sqrt {x+1}$ where $x lt -1$. Also consider what happens to $frac 1{sqrt{x+1}}$ when $x=-1$.
– Raptor
7 hours ago




Hint: Consider $sqrt {x+1}$ where $x lt -1$. Also consider what happens to $frac 1{sqrt{x+1}}$ when $x=-1$.
– Raptor
7 hours ago










1 Answer
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We proceed using inequalities. For high school students it is more comprehensible if we formally use comparison with $infty.$




  • For the given function, start from its domain
    $$big(0<x+1<inftybig) Longrightarrow big(0<sqrt{x+1}<inftybig) Longrightarrow big(infty>{1over{sqrt{x+1}}}>0big), $$ so the range is $(0,infty).$


  • It becomes more interesting if e.g. $;g(x) = frac{3}{sqrt{(x + 1)}+2}$



$$begin{aligned}big(0<x+1<inftybig) Longrightarrow &;0<sqrt{x+1}<infty\ Longrightarrow &;2<sqrt{x+1}+2<infty\Longrightarrow &;{1over 2}>{1over{sqrt{(x+1)}+2}}>0\Longrightarrow &;{3over 2}>{1over{sqrt{(x+1)}+2}}>0end{aligned}$$ thus the range is $(0,{3over 2}).$






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    We proceed using inequalities. For high school students it is more comprehensible if we formally use comparison with $infty.$




    • For the given function, start from its domain
      $$big(0<x+1<inftybig) Longrightarrow big(0<sqrt{x+1}<inftybig) Longrightarrow big(infty>{1over{sqrt{x+1}}}>0big), $$ so the range is $(0,infty).$


    • It becomes more interesting if e.g. $;g(x) = frac{3}{sqrt{(x + 1)}+2}$



    $$begin{aligned}big(0<x+1<inftybig) Longrightarrow &;0<sqrt{x+1}<infty\ Longrightarrow &;2<sqrt{x+1}+2<infty\Longrightarrow &;{1over 2}>{1over{sqrt{(x+1)}+2}}>0\Longrightarrow &;{3over 2}>{1over{sqrt{(x+1)}+2}}>0end{aligned}$$ thus the range is $(0,{3over 2}).$






    share|cite|improve this answer

























      up vote
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      down vote













      We proceed using inequalities. For high school students it is more comprehensible if we formally use comparison with $infty.$




      • For the given function, start from its domain
        $$big(0<x+1<inftybig) Longrightarrow big(0<sqrt{x+1}<inftybig) Longrightarrow big(infty>{1over{sqrt{x+1}}}>0big), $$ so the range is $(0,infty).$


      • It becomes more interesting if e.g. $;g(x) = frac{3}{sqrt{(x + 1)}+2}$



      $$begin{aligned}big(0<x+1<inftybig) Longrightarrow &;0<sqrt{x+1}<infty\ Longrightarrow &;2<sqrt{x+1}+2<infty\Longrightarrow &;{1over 2}>{1over{sqrt{(x+1)}+2}}>0\Longrightarrow &;{3over 2}>{1over{sqrt{(x+1)}+2}}>0end{aligned}$$ thus the range is $(0,{3over 2}).$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        We proceed using inequalities. For high school students it is more comprehensible if we formally use comparison with $infty.$




        • For the given function, start from its domain
          $$big(0<x+1<inftybig) Longrightarrow big(0<sqrt{x+1}<inftybig) Longrightarrow big(infty>{1over{sqrt{x+1}}}>0big), $$ so the range is $(0,infty).$


        • It becomes more interesting if e.g. $;g(x) = frac{3}{sqrt{(x + 1)}+2}$



        $$begin{aligned}big(0<x+1<inftybig) Longrightarrow &;0<sqrt{x+1}<infty\ Longrightarrow &;2<sqrt{x+1}+2<infty\Longrightarrow &;{1over 2}>{1over{sqrt{(x+1)}+2}}>0\Longrightarrow &;{3over 2}>{1over{sqrt{(x+1)}+2}}>0end{aligned}$$ thus the range is $(0,{3over 2}).$






        share|cite|improve this answer












        We proceed using inequalities. For high school students it is more comprehensible if we formally use comparison with $infty.$




        • For the given function, start from its domain
          $$big(0<x+1<inftybig) Longrightarrow big(0<sqrt{x+1}<inftybig) Longrightarrow big(infty>{1over{sqrt{x+1}}}>0big), $$ so the range is $(0,infty).$


        • It becomes more interesting if e.g. $;g(x) = frac{3}{sqrt{(x + 1)}+2}$



        $$begin{aligned}big(0<x+1<inftybig) Longrightarrow &;0<sqrt{x+1}<infty\ Longrightarrow &;2<sqrt{x+1}+2<infty\Longrightarrow &;{1over 2}>{1over{sqrt{(x+1)}+2}}>0\Longrightarrow &;{3over 2}>{1over{sqrt{(x+1)}+2}}>0end{aligned}$$ thus the range is $(0,{3over 2}).$







        share|cite|improve this answer












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        answered 18 mins ago









        user376343

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