Question

What is the domain and range of a rational function?

Answer 1

$$\begin{gathered} \mathrm{Rational}\mathrm{function},f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}\mathrm{where}P\left(x\right)\mathrm{and}Q\left(x\right)\mathrm{are}\mathrm{polynomial}\mathrm{and}Q\left(x\right)\ne 0. \\ \mathrm{Therefore},\mathrm{domain}\mathrm{of}f\left(x\right)i\mathrm{s}\mathrm{all}\mathrm{real}\mathrm{numbers}\mathrm{except}\mathrm{those}\mathrm{values}\mathrm{of}x\mathrm{for}\mathrm{which}Q\left(x\right)=0. \\ \mathrm{The}\mathrm{range}\mathrm{of}\mathrm{rational}\mathrm{function}\mathrm{is}\mathrm{different}\mathrm{for}\mathrm{different}\mathrm{functions}. \\ \mathrm{Ex}:f\left(x\right)=\frac{1}{x^2} \\ \mathrm{Domain}: \\ {x}^2\mathrm{is}0\mathrm{for}x=0 \\ \mathrm{Therefore},\mathrm{domain}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}R-\left\{0\right\} \\ \mathrm{Range}: \\ \mathrm{The}\mathrm{value}\mathrm{of}{x}^2\mathrm{will}\mathrm{always}\mathrm{be}\mathrm{positive}\mathrm{and}\mathrm{so}\frac{1}{x^2}\mathrm{will}\mathrm{also}\mathrm{be}\mathrm{always}\mathrm{positive}. \\ \mathrm{Hence},\mathrm{range}\mathrm{of}f\left(x\right)\mathrm{is}\left(0,\infty \right) \end{gathered}$$