Combining functions and finding domain











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1
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$f(x) = ln (2 - x)$



$g(x) = frac {1}{√x}$



I am asked to find the domain of $frac gf$



So $frac gf$ would be $frac {1}{√x}$ over $ln (2 - x)
$
which would be $frac{ln (2-x)}{√x}$



so the domain is $0 < x < 2$



Is this correct?



Thank you!






algebra-precalculus







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  • $g(x)/f(x)$ is not the function you wrote. Check it.
    – james watt
    Nov 22 at 18:06










  • Edited. I think my domain is still right?
    – M Do
    Nov 22 at 18:09










  • No, $g/f$ is not correct.
    – james watt
    Nov 22 at 18:10

















up vote
1
down vote

favorite












$f(x) = ln (2 - x)$



$g(x) = frac {1}{√x}$



I am asked to find the domain of $frac gf$



So $frac gf$ would be $frac {1}{√x}$ over $ln (2 - x)
$
which would be $frac{ln (2-x)}{√x}$



so the domain is $0 < x < 2$



Is this correct?



Thank you!






algebra-precalculus







share|cite|improve this question
























  • $g(x)/f(x)$ is not the function you wrote. Check it.
    – james watt
    Nov 22 at 18:06










  • Edited. I think my domain is still right?
    – M Do
    Nov 22 at 18:09










  • No, $g/f$ is not correct.
    – james watt
    Nov 22 at 18:10















up vote
1
down vote

favorite









up vote
1
down vote

favorite











$f(x) = ln (2 - x)$



$g(x) = frac {1}{√x}$



I am asked to find the domain of $frac gf$



So $frac gf$ would be $frac {1}{√x}$ over $ln (2 - x)
$
which would be $frac{ln (2-x)}{√x}$



so the domain is $0 < x < 2$



Is this correct?



Thank you!






algebra-precalculus







share|cite|improve this question















$f(x) = ln (2 - x)$



$g(x) = frac {1}{√x}$



I am asked to find the domain of $frac gf$



So $frac gf$ would be $frac {1}{√x}$ over $ln (2 - x)
$
which would be $frac{ln (2-x)}{√x}$



so the domain is $0 < x < 2$



Is this correct?



Thank you!






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 Nov 22 at 18:32









gimusi

92.7k94495




92.7k94495










asked Nov 22 at 18:04









M Do

124




124












  • $g(x)/f(x)$ is not the function you wrote. Check it.
    – james watt
    Nov 22 at 18:06










  • Edited. I think my domain is still right?
    – M Do
    Nov 22 at 18:09










  • No, $g/f$ is not correct.
    – james watt
    Nov 22 at 18:10




















  • $g(x)/f(x)$ is not the function you wrote. Check it.
    – james watt
    Nov 22 at 18:06










  • Edited. I think my domain is still right?
    – M Do
    Nov 22 at 18:09










  • No, $g/f$ is not correct.
    – james watt
    Nov 22 at 18:10


















$g(x)/f(x)$ is not the function you wrote. Check it.
– james watt
Nov 22 at 18:06




$g(x)/f(x)$ is not the function you wrote. Check it.
– james watt
Nov 22 at 18:06












Edited. I think my domain is still right?
– M Do
Nov 22 at 18:09




Edited. I think my domain is still right?
– M Do
Nov 22 at 18:09












No, $g/f$ is not correct.
– james watt
Nov 22 at 18:10






No, $g/f$ is not correct.
– james watt
Nov 22 at 18:10












1 Answer
1






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0
down vote













HINT



Note that



$$frac g f = frac1{sqrt x log (2-x)}$$



and we need




  • for the definition of $f$: $$2-x>0$$

  • for the definition of $g$: $$xge 0$$

  • for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$


for there we obtain that





  • $x>0$

  • $2-x>0$

  • $2-xneq 1$


Solve that system to find the domain.






share|cite|improve this answer























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    1 Answer
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    up vote
    0
    down vote













    HINT



    Note that



    $$frac g f = frac1{sqrt x log (2-x)}$$



    and we need




    • for the definition of $f$: $$2-x>0$$

    • for the definition of $g$: $$xge 0$$

    • for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$


    for there we obtain that





    • $x>0$

    • $2-x>0$

    • $2-xneq 1$


    Solve that system to find the domain.






    share|cite|improve this answer



























      up vote
      0
      down vote













      HINT



      Note that



      $$frac g f = frac1{sqrt x log (2-x)}$$



      and we need




      • for the definition of $f$: $$2-x>0$$

      • for the definition of $g$: $$xge 0$$

      • for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$


      for there we obtain that





      • $x>0$

      • $2-x>0$

      • $2-xneq 1$


      Solve that system to find the domain.






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        HINT



        Note that



        $$frac g f = frac1{sqrt x log (2-x)}$$



        and we need




        • for the definition of $f$: $$2-x>0$$

        • for the definition of $g$: $$xge 0$$

        • for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$


        for there we obtain that





        • $x>0$

        • $2-x>0$

        • $2-xneq 1$


        Solve that system to find the domain.






        share|cite|improve this answer














        HINT



        Note that



        $$frac g f = frac1{sqrt x log (2-x)}$$



        and we need




        • for the definition of $f$: $$2-x>0$$

        • for the definition of $g$: $$xge 0$$

        • for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$


        for there we obtain that





        • $x>0$

        • $2-x>0$

        • $2-xneq 1$


        Solve that system to find the domain.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 at 18:31

























        answered Nov 22 at 18:10









        gimusi

        92.7k94495




        92.7k94495






























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