Checking if the function is onto












0












$begingroup$


Consider a function $f$ defined over $R to R$



$y=x^2+2x$



Well it's inverse comes out to be



$f^{-1}(x)=sqrt {x+1} -1$



Well, this seems to be undefined for all values of co-domain $lt -1$



Am I correct in my reasoning?






functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
    $endgroup$
    – user247327
    Dec 25 '18 at 12:06


















0












$begingroup$


Consider a function $f$ defined over $R to R$



$y=x^2+2x$



Well it's inverse comes out to be



$f^{-1}(x)=sqrt {x+1} -1$



Well, this seems to be undefined for all values of co-domain $lt -1$



Am I correct in my reasoning?






functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
    $endgroup$
    – user247327
    Dec 25 '18 at 12:06
















0












0








0





$begingroup$


Consider a function $f$ defined over $R to R$



$y=x^2+2x$



Well it's inverse comes out to be



$f^{-1}(x)=sqrt {x+1} -1$



Well, this seems to be undefined for all values of co-domain $lt -1$



Am I correct in my reasoning?






functions







share|cite|improve this question









$endgroup$




Consider a function $f$ defined over $R to R$



$y=x^2+2x$



Well it's inverse comes out to be



$f^{-1}(x)=sqrt {x+1} -1$



Well, this seems to be undefined for all values of co-domain $lt -1$



Am I correct in my reasoning?






functions




functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 25 '18 at 11:57









user3767495user3767495

4078




4078












  • $begingroup$
    If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
    $endgroup$
    – user247327
    Dec 25 '18 at 12:06




















  • $begingroup$
    If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
    $endgroup$
    – user247327
    Dec 25 '18 at 12:06


















$begingroup$
If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
$endgroup$
– user247327
Dec 25 '18 at 12:06






$begingroup$
If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
$endgroup$
– user247327
Dec 25 '18 at 12:06












2 Answers
2






active

oldest

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2












$begingroup$

We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.



Can you proceed?






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.



      Can you proceed?






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.



        Can you proceed?






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.



          Can you proceed?






          share|cite|improve this answer









          $endgroup$



          We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.



          Can you proceed?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 12:03









          FredFred

          47.8k1849




          47.8k1849























              1












              $begingroup$

              If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.






                  share|cite|improve this answer









                  $endgroup$



                  If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 12:01









                  wendy.kriegerwendy.krieger

                  5,84511426




                  5,84511426






























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