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What is the range of this combined function?


Finding the Domain and Range of a function composition$x = sec 2y$, Find $dfrac dydx$ in terms of $x$. What about $pm$?Domain and range of an inverse functionWhy does the domain and range of $sqrt x$ contain only positive real numbers?What might this function be?when to use restrictions (domain and range) on trig functionsFinding the Range and the Domain of $f(x)=frac x^21-x$Finding the domain of $(f circ g)(x)$Confusion About Domain and Range of Linear Composite FunctionsRange of a function, with contradictory restriction













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 $f(x) = dfrac1x - 3$ and $g(x) = sqrtx$, 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 $(f circ g)(x) = f(g(x)) = dfrac1sqrtx - 3$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrtx geq 0$, which in turn implies that $y geq - dfrac13$.



Combining these two restrictions, my solution for the range is



$$y in mathbbR mid y geq - dfrac 13 wedge y neq 0 $$



The given solution, however, is:




$$y in mathbbR mid y neq > 0 $$




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 $y geq -dfrac13$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?










share|cite|improve this question











$endgroup$
















    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 $f(x) = dfrac1x - 3$ and $g(x) = sqrtx$, 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 $(f circ g)(x) = f(g(x)) = dfrac1sqrtx - 3$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrtx geq 0$, which in turn implies that $y geq - dfrac13$.



    Combining these two restrictions, my solution for the range is



    $$y in mathbbR mid y geq - dfrac 13 wedge y neq 0 $$



    The given solution, however, is:




    $$y in mathbbR mid y neq > 0 $$




    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 $y geq -dfrac13$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?










    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 $f(x) = dfrac1x - 3$ and $g(x) = sqrtx$, 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 $(f circ g)(x) = f(g(x)) = dfrac1sqrtx - 3$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrtx geq 0$, which in turn implies that $y geq - dfrac13$.



      Combining these two restrictions, my solution for the range is



      $$y in mathbbR mid y geq - dfrac 13 wedge y neq 0 $$



      The given solution, however, is:




      $$y in mathbbR mid y neq > 0 $$




      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 $y geq -dfrac13$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?










      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 $f(x) = dfrac1x - 3$ and $g(x) = sqrtx$, 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 $(f circ g)(x) = f(g(x)) = dfrac1sqrtx - 3$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrtx geq 0$, which in turn implies that $y geq - dfrac13$.



      Combining these two restrictions, my solution for the range is



      $$y in mathbbR mid y geq - dfrac 13 wedge y neq 0 $$



      The given solution, however, is:




      $$y in mathbbR mid y neq > 0 $$




      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 $y geq -dfrac13$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?







      algebra-precalculus functions






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      edited 2 hours ago







      Calculemus

















      asked 3 hours ago









      CalculemusCalculemus

      432317




      432317




















          1 Answer
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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









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            1 Answer
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            oldest

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            active

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            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












              $begingroup$

              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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