What's the domain of $f(x) = sqrt{x^2 - 4x + 5}$? [closed]












-1












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What's the domain of $f(x) = sqrt{x^2 - 4x + 5}$ ?



Thanks






algebra-precalculus







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closed as off-topic by José Carlos Santos, Hans Lundmark, Claude Leibovici, Adrian Keister, Abcd Jan 29 at 16:12


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Hans Lundmark, Claude Leibovici, Adrian Keister, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    I would say all real numbers
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 29 at 7:11










  • $begingroup$
    @Dr.SonnhardGraubner, I think you are right
    $endgroup$
    – BadAtGeometry
    Jan 29 at 7:13








  • 1




    $begingroup$
    ... and how to show this? Hint $x^2 -4x +5 = x^2 -2times 2x + 2^2 +1 = (x-2)^2 +1$
    $endgroup$
    – Matti P.
    Jan 29 at 7:15
















-1












$begingroup$


What's the domain of $f(x) = sqrt{x^2 - 4x + 5}$ ?



Thanks






algebra-precalculus







share|cite|improve this question











$endgroup$



closed as off-topic by José Carlos Santos, Hans Lundmark, Claude Leibovici, Adrian Keister, Abcd Jan 29 at 16:12


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Hans Lundmark, Claude Leibovici, Adrian Keister, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    I would say all real numbers
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 29 at 7:11










  • $begingroup$
    @Dr.SonnhardGraubner, I think you are right
    $endgroup$
    – BadAtGeometry
    Jan 29 at 7:13








  • 1




    $begingroup$
    ... and how to show this? Hint $x^2 -4x +5 = x^2 -2times 2x + 2^2 +1 = (x-2)^2 +1$
    $endgroup$
    – Matti P.
    Jan 29 at 7:15














-1












-1








-1





$begingroup$


What's the domain of $f(x) = sqrt{x^2 - 4x + 5}$ ?



Thanks






algebra-precalculus







share|cite|improve this question











$endgroup$




What's the domain of $f(x) = sqrt{x^2 - 4x + 5}$ ?



Thanks






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 7:11









Chinnapparaj R

5,8602928




5,8602928










asked Jan 29 at 7:10









Sully Sully

21




21




closed as off-topic by José Carlos Santos, Hans Lundmark, Claude Leibovici, Adrian Keister, Abcd Jan 29 at 16:12


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Hans Lundmark, Claude Leibovici, Adrian Keister, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by José Carlos Santos, Hans Lundmark, Claude Leibovici, Adrian Keister, Abcd Jan 29 at 16:12


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Hans Lundmark, Claude Leibovici, Adrian Keister, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    I would say all real numbers
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 29 at 7:11










  • $begingroup$
    @Dr.SonnhardGraubner, I think you are right
    $endgroup$
    – BadAtGeometry
    Jan 29 at 7:13








  • 1




    $begingroup$
    ... and how to show this? Hint $x^2 -4x +5 = x^2 -2times 2x + 2^2 +1 = (x-2)^2 +1$
    $endgroup$
    – Matti P.
    Jan 29 at 7:15














  • 2




    $begingroup$
    I would say all real numbers
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 29 at 7:11










  • $begingroup$
    @Dr.SonnhardGraubner, I think you are right
    $endgroup$
    – BadAtGeometry
    Jan 29 at 7:13








  • 1




    $begingroup$
    ... and how to show this? Hint $x^2 -4x +5 = x^2 -2times 2x + 2^2 +1 = (x-2)^2 +1$
    $endgroup$
    – Matti P.
    Jan 29 at 7:15








2




2




$begingroup$
I would say all real numbers
$endgroup$
– Dr. Sonnhard Graubner
Jan 29 at 7:11




$begingroup$
I would say all real numbers
$endgroup$
– Dr. Sonnhard Graubner
Jan 29 at 7:11












$begingroup$
@Dr.SonnhardGraubner, I think you are right
$endgroup$
– BadAtGeometry
Jan 29 at 7:13






$begingroup$
@Dr.SonnhardGraubner, I think you are right
$endgroup$
– BadAtGeometry
Jan 29 at 7:13






1




1




$begingroup$
... and how to show this? Hint $x^2 -4x +5 = x^2 -2times 2x + 2^2 +1 = (x-2)^2 +1$
$endgroup$
– Matti P.
Jan 29 at 7:15




$begingroup$
... and how to show this? Hint $x^2 -4x +5 = x^2 -2times 2x + 2^2 +1 = (x-2)^2 +1$
$endgroup$
– Matti P.
Jan 29 at 7:15










2 Answers
2






active

oldest

votes


















5












$begingroup$

Assuming that you are considering this as a function from real numbers to real numbers, you can rearrange to write:$$f(x)=sqrt{(x-2)^2+1}$$
So the radicand is positive no matter what real number $x$ is, and therefore the domain is all real numbers.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    The domain is all real numbers. See this graph.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      a graph is not a proof
      $endgroup$
      – J. W. Tanner
      Jan 29 at 14:58










    • $begingroup$
      To second @J.W.Tanner, this is at best a comment. Plots are never rigorous proofs of anything. Please consider deleting this and transferring it to the comments section.
      $endgroup$
      – TheSimpliFire
      Feb 13 at 9:35


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Assuming that you are considering this as a function from real numbers to real numbers, you can rearrange to write:$$f(x)=sqrt{(x-2)^2+1}$$
    So the radicand is positive no matter what real number $x$ is, and therefore the domain is all real numbers.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      Assuming that you are considering this as a function from real numbers to real numbers, you can rearrange to write:$$f(x)=sqrt{(x-2)^2+1}$$
      So the radicand is positive no matter what real number $x$ is, and therefore the domain is all real numbers.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        Assuming that you are considering this as a function from real numbers to real numbers, you can rearrange to write:$$f(x)=sqrt{(x-2)^2+1}$$
        So the radicand is positive no matter what real number $x$ is, and therefore the domain is all real numbers.






        share|cite|improve this answer











        $endgroup$



        Assuming that you are considering this as a function from real numbers to real numbers, you can rearrange to write:$$f(x)=sqrt{(x-2)^2+1}$$
        So the radicand is positive no matter what real number $x$ is, and therefore the domain is all real numbers.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 29 at 15:00

























        answered Jan 29 at 7:16









        alex.jordanalex.jordan

        39.6k660122




        39.6k660122























            1












            $begingroup$

            The domain is all real numbers. See this graph.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              a graph is not a proof
              $endgroup$
              – J. W. Tanner
              Jan 29 at 14:58










            • $begingroup$
              To second @J.W.Tanner, this is at best a comment. Plots are never rigorous proofs of anything. Please consider deleting this and transferring it to the comments section.
              $endgroup$
              – TheSimpliFire
              Feb 13 at 9:35
















            1












            $begingroup$

            The domain is all real numbers. See this graph.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              a graph is not a proof
              $endgroup$
              – J. W. Tanner
              Jan 29 at 14:58










            • $begingroup$
              To second @J.W.Tanner, this is at best a comment. Plots are never rigorous proofs of anything. Please consider deleting this and transferring it to the comments section.
              $endgroup$
              – TheSimpliFire
              Feb 13 at 9:35














            1












            1








            1





            $begingroup$

            The domain is all real numbers. See this graph.






            share|cite|improve this answer









            $endgroup$



            The domain is all real numbers. See this graph.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 29 at 7:12









            BadAtGeometryBadAtGeometry

            300215




            300215












            • $begingroup$
              a graph is not a proof
              $endgroup$
              – J. W. Tanner
              Jan 29 at 14:58










            • $begingroup$
              To second @J.W.Tanner, this is at best a comment. Plots are never rigorous proofs of anything. Please consider deleting this and transferring it to the comments section.
              $endgroup$
              – TheSimpliFire
              Feb 13 at 9:35


















            • $begingroup$
              a graph is not a proof
              $endgroup$
              – J. W. Tanner
              Jan 29 at 14:58










            • $begingroup$
              To second @J.W.Tanner, this is at best a comment. Plots are never rigorous proofs of anything. Please consider deleting this and transferring it to the comments section.
              $endgroup$
              – TheSimpliFire
              Feb 13 at 9:35
















            $begingroup$
            a graph is not a proof
            $endgroup$
            – J. W. Tanner
            Jan 29 at 14:58




            $begingroup$
            a graph is not a proof
            $endgroup$
            – J. W. Tanner
            Jan 29 at 14:58












            $begingroup$
            To second @J.W.Tanner, this is at best a comment. Plots are never rigorous proofs of anything. Please consider deleting this and transferring it to the comments section.
            $endgroup$
            – TheSimpliFire
            Feb 13 at 9:35




            $begingroup$
            To second @J.W.Tanner, this is at best a comment. Plots are never rigorous proofs of anything. Please consider deleting this and transferring it to the comments section.
            $endgroup$
            – TheSimpliFire
            Feb 13 at 9:35



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