To solve this question we will solve the given equation for y and we will analyze the possible values of y (that will be the range of the given function).
Assuming that 2x+1≠0, and multiplying the given equation by 2x+1 we get:
[tex]\begin{gathered} y\cdot(2x+1)=\frac{2}{2x+1}\cdot(2x+1), \\ 2yx+y=2. \end{gathered}[/tex]Subtracting y from the above equation we get:
[tex]\begin{gathered} 2yx+y-y=2-y, \\ 2yx=2-y\text{.} \end{gathered}[/tex]Dividing the above equation by 2y we get:
[tex]\begin{gathered} \frac{2yx}{2y}=\frac{2-y}{2y}, \\ x=\frac{2-y}{2y} \end{gathered}[/tex]Therefore, y can be all real numbers except zero.
Answer:
[tex](-\infty,0)\cup(0,\infty)\text{.}[/tex]