Question #294830

Domain and renge f(x)=1/2x-p

1
Expert's answer
2022-02-08T07:40:44-0500

Let us find the domain and range of f(x)=12xp.f(x)=\frac{1}{2x-p}.

It follows that the domain is equal to the set R{p2}.\R\setminus\{\frac{p}2\}.

If y=0y=0 then the equation y=12xpy=\frac{1}{2x-p} has no solution. If y0y\ne 0 then the equation y=12xpy=\frac{1}{2x-p} is equivalent to 2xp=1y2x-p=\frac{1}{y}, and hence has a solution x=12y+p2x=\frac{1}{2y}+\frac{p}2 . Therefore, f(x)=f(12y+p2)=y.f(x)=f(\frac{1}{2y}+\frac{p}2)=y. We conclude that the range of the function ff is equal to the set R{0}.\R\setminus\{0\}.

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