What is the range of this combined function?












4












$begingroup$


I am attempting a Functions and Inverses self-test offered by the University of Toronto, and I'm trying to understand why my answer for question (4) differs from the given one.



Given $f(x) = dfrac{1}{x - 3}$ and $g(x) = sqrt{x}$, we are asked to find the domain and range of the combined function
$$(f circ g)(x)$$



My solution for the domain matches the given one, and I won't bother reproducing it here, but my solution for the range does not. This is how I determined the range:



Since $(f circ g)(x) = f(g(x)) = dfrac{1}{sqrt{x} - 3}$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrt{x} geq 0$, which in turn implies that $y geq - dfrac{1}{3}$.



Combining these two restrictions, my solution for the range is



$${y in mathbb{R} mid y geq - dfrac {1}{3} wedge y neq 0 }$$



The given solution, however, is:




$${y in mathbb{R} mid y neq > 0 }$$




I'm not sure what the $neq >$ notation means. I'm assuming it's a typo, and it's actually supposed to be just a $neq$ sign. But even so, why isn't the $y geq -dfrac{1}{3}$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?






algebra-precalculus functions







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$endgroup$

















    4












    $begingroup$


    I am attempting a Functions and Inverses self-test offered by the University of Toronto, and I'm trying to understand why my answer for question (4) differs from the given one.



    Given $f(x) = dfrac{1}{x - 3}$ and $g(x) = sqrt{x}$, we are asked to find the domain and range of the combined function
    $$(f circ g)(x)$$



    My solution for the domain matches the given one, and I won't bother reproducing it here, but my solution for the range does not. This is how I determined the range:



    Since $(f circ g)(x) = f(g(x)) = dfrac{1}{sqrt{x} - 3}$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrt{x} geq 0$, which in turn implies that $y geq - dfrac{1}{3}$.



    Combining these two restrictions, my solution for the range is



    $${y in mathbb{R} mid y geq - dfrac {1}{3} wedge y neq 0 }$$



    The given solution, however, is:




    $${y in mathbb{R} mid y neq > 0 }$$




    I'm not sure what the $neq >$ notation means. I'm assuming it's a typo, and it's actually supposed to be just a $neq$ sign. But even so, why isn't the $y geq -dfrac{1}{3}$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?






    algebra-precalculus functions







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I am attempting a Functions and Inverses self-test offered by the University of Toronto, and I'm trying to understand why my answer for question (4) differs from the given one.



      Given $f(x) = dfrac{1}{x - 3}$ and $g(x) = sqrt{x}$, we are asked to find the domain and range of the combined function
      $$(f circ g)(x)$$



      My solution for the domain matches the given one, and I won't bother reproducing it here, but my solution for the range does not. This is how I determined the range:



      Since $(f circ g)(x) = f(g(x)) = dfrac{1}{sqrt{x} - 3}$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrt{x} geq 0$, which in turn implies that $y geq - dfrac{1}{3}$.



      Combining these two restrictions, my solution for the range is



      $${y in mathbb{R} mid y geq - dfrac {1}{3} wedge y neq 0 }$$



      The given solution, however, is:




      $${y in mathbb{R} mid y neq > 0 }$$




      I'm not sure what the $neq >$ notation means. I'm assuming it's a typo, and it's actually supposed to be just a $neq$ sign. But even so, why isn't the $y geq -dfrac{1}{3}$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?






      algebra-precalculus functions







      share|cite|improve this question











      $endgroup$




      I am attempting a Functions and Inverses self-test offered by the University of Toronto, and I'm trying to understand why my answer for question (4) differs from the given one.



      Given $f(x) = dfrac{1}{x - 3}$ and $g(x) = sqrt{x}$, we are asked to find the domain and range of the combined function
      $$(f circ g)(x)$$



      My solution for the domain matches the given one, and I won't bother reproducing it here, but my solution for the range does not. This is how I determined the range:



      Since $(f circ g)(x) = f(g(x)) = dfrac{1}{sqrt{x} - 3}$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrt{x} geq 0$, which in turn implies that $y geq - dfrac{1}{3}$.



      Combining these two restrictions, my solution for the range is



      $${y in mathbb{R} mid y geq - dfrac {1}{3} wedge y neq 0 }$$



      The given solution, however, is:




      $${y in mathbb{R} mid y neq > 0 }$$




      I'm not sure what the $neq >$ notation means. I'm assuming it's a typo, and it's actually supposed to be just a $neq$ sign. But even so, why isn't the $y geq -dfrac{1}{3}$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?






      algebra-precalculus functions




      algebra-precalculus functions






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      share|cite|improve this question













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      edited 2 hours ago







      Calculemus

















      asked 3 hours ago









      CalculemusCalculemus

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          $begingroup$

          The range is $(0,infty) cup (-infty, -frac 1 3]$. To see this write the range as ${frac 1 {t-3}: t geq 0, t neq 3}$. Find ${frac 1 {t-3}:0 leq t < 3}$ and ${frac 1 {t-3}: 3 < t <infty)}$ separately. These can be written as ${frac 1 s:-3 leq s < 0}$ and ${frac 1 s: 0 < s <infty)}$. Can you compute the range now?






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            $begingroup$

            The range is $(0,infty) cup (-infty, -frac 1 3]$. To see this write the range as ${frac 1 {t-3}: t geq 0, t neq 3}$. Find ${frac 1 {t-3}:0 leq t < 3}$ and ${frac 1 {t-3}: 3 < t <infty)}$ separately. These can be written as ${frac 1 s:-3 leq s < 0}$ and ${frac 1 s: 0 < s <infty)}$. Can you compute the range now?






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              The range is $(0,infty) cup (-infty, -frac 1 3]$. To see this write the range as ${frac 1 {t-3}: t geq 0, t neq 3}$. Find ${frac 1 {t-3}:0 leq t < 3}$ and ${frac 1 {t-3}: 3 < t <infty)}$ separately. These can be written as ${frac 1 s:-3 leq s < 0}$ and ${frac 1 s: 0 < s <infty)}$. Can you compute the range now?






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                The range is $(0,infty) cup (-infty, -frac 1 3]$. To see this write the range as ${frac 1 {t-3}: t geq 0, t neq 3}$. Find ${frac 1 {t-3}:0 leq t < 3}$ and ${frac 1 {t-3}: 3 < t <infty)}$ separately. These can be written as ${frac 1 s:-3 leq s < 0}$ and ${frac 1 s: 0 < s <infty)}$. Can you compute the range now?






                share|cite|improve this answer









                $endgroup$



                The range is $(0,infty) cup (-infty, -frac 1 3]$. To see this write the range as ${frac 1 {t-3}: t geq 0, t neq 3}$. Find ${frac 1 {t-3}:0 leq t < 3}$ and ${frac 1 {t-3}: 3 < t <infty)}$ separately. These can be written as ${frac 1 s:-3 leq s < 0}$ and ${frac 1 s: 0 < s <infty)}$. Can you compute the range now?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Kavi Rama MurthyKavi Rama Murthy

                78.5k53572




                78.5k53572






























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