Question Number 25279 by NECx last updated on 07/Dec/17
$${Find}\:{the}\:{domain}\:{and}\:{range}\:{of} \\ $$$${f}\left({x}\right)=\sqrt{\left(\frac{{x}−\mathrm{5}}{{x}−\mathrm{3}}\right)} \\ $$
Answered by SAGARSTARK last updated on 07/Dec/17
$$\left(\frac{{x}−\mathrm{5}}{{x}−\mathrm{3}}\right)\geqslant\mathrm{0}\:\boldsymbol{\mathrm{x}}\in\left(−\infty,\mathrm{3}\right)\cup\left[\mathrm{5},\infty\right)\:{using}\: \\ $$$${wavy}\:{curve}\:{method} \\ $$
Commented by Rasheed.Sindhi last updated on 08/Dec/17
$$\mathrm{And}\:\mathrm{what}\:\mathrm{about}\:\mathrm{range}? \\ $$
Commented by NECx last updated on 08/Dec/17
$${thanks}.{How}\:{about}\:{the}\:{range}? \\ $$$$ \\ $$
Answered by jota+ last updated on 08/Dec/17
$${Domf}\left({x}\right)=\left(−\infty\:\:\mathrm{3}\right)\cup\left[\mathrm{5}\:\:+\infty\right) \\ $$$${if}\:\:{y}={f}\left({x}\right)\:{then} \\ $$$${x}=\:\frac{\mathrm{3}{y}^{\mathrm{2}} −\mathrm{5}}{{y}^{\mathrm{2}} −\mathrm{1}}\:\:\:{and}\:\:{y}\:\geqslant\:\mathrm{0} \\ $$$${Ranf}\left({x}\right)=\left\{\mathrm{0}\right\}\cup{R}^{+} −\left\{\mathrm{1}\right\} \\ $$