Question Number 207043 by necx122 last updated on 04/May/24
$${Solve}\:{for}\:{the}\:{domain}\:{and}\:{range}\:{of}\:{f}\left({x}\right)\:{if} \\ $$$${f}\left({x}\right)\:=\:\sqrt{\frac{\mathrm{3}{x}−\mathrm{5}}{{x}+\mathrm{3}}} \\ $$
Answered by Frix last updated on 04/May/24
![1. x+3≠0 x≠−3 2. ((3x−5)/(x+3))≥0 (3x−5<0∧x+3<0)∨(3x−5≥0∧x+3>0) x<−3∨x≥(5/3) [=domain] f(x)=y=(√((3x−5)/(x+3))) ⇔ x=−((3y^2 +5)/(y^2 −3)) ⇒ f(x)≥0∧f(x)≠(√3) [=range]](https://www.tinkutara.com/question/Q207052.png)
$$\mathrm{1}.\:{x}+\mathrm{3}\neq\mathrm{0} \\ $$$${x}\neq−\mathrm{3} \\ $$$$\mathrm{2}.\:\frac{\mathrm{3}{x}−\mathrm{5}}{{x}+\mathrm{3}}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{3}{x}−\mathrm{5}<\mathrm{0}\wedge{x}+\mathrm{3}<\mathrm{0}\right)\vee\left(\mathrm{3}{x}−\mathrm{5}\geqslant\mathrm{0}\wedge{x}+\mathrm{3}>\mathrm{0}\right) \\ $$$${x}<−\mathrm{3}\vee{x}\geqslant\frac{\mathrm{5}}{\mathrm{3}}\:\left[=\mathrm{domain}\right] \\ $$$${f}\left({x}\right)={y}=\sqrt{\frac{\mathrm{3}{x}−\mathrm{5}}{{x}+\mathrm{3}}}\:\Leftrightarrow\:{x}=−\frac{\mathrm{3}{y}^{\mathrm{2}} +\mathrm{5}}{{y}^{\mathrm{2}} −\mathrm{3}} \\ $$$$\Rightarrow\:{f}\left({x}\right)\geqslant\mathrm{0}\wedge{f}\left({x}\right)\neq\sqrt{\mathrm{3}}\:\left[=\mathrm{range}\right] \\ $$
Commented by necx122 last updated on 05/May/24
$${Thank}\:{you}\:{so}\:{much}. \\ $$$$\:{Now},\:{for}\:{getting}\:{the}\:{values}\:{of}\:{the} \\ $$$$\:{inequality}\:\frac{\mathrm{3}{x}−\mathrm{5}}{{x}+\mathrm{3}}\:\geqslant\mathrm{0} \\ $$$${when}\:{do}\:{you}\:{decide}\:{to}\:{use}\:\left(\mathrm{3}{x}−\mathrm{5}\geqslant\mathrm{0}\:\wedge\:{x}+\mathrm{3}>\mathrm{0}\right) \\ $$$${and}\:\left(\mathrm{3}{x}−\mathrm{5}\:<\mathrm{0}\:\wedge\:{x}+\mathrm{3}\:<\mathrm{0}\right). \\ $$$$\left.\mathrm{1}\right){Can}\:{these}\:\mathrm{2}\:{inequalities}\:{satisfy}\:{the} \\ $$$${same}\:{range}\:{of}\:{values}? \\ $$$$\left.\mathrm{2}\right){How}\:{do}\:{we}\:{choose}\:{which}\:{to}\:{go}\:{with}? \\ $$$$\left.\mathrm{3}\right){what}\:{specific}\:{rules}\:{are}\:{there}\:{for} \\ $$$${solving}\:{rational}\:{inequalities}? \\ $$$${Thank}\:{you}\:{so}\:{much}.\:{I}\:{anticipate}\:{your} \\ $$$${response}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 05/May/24
$$\frac{\mathrm{3}{x}−\mathrm{5}}{{x}+\mathrm{3}}\geqslant\mathrm{0}\:\Leftrightarrow\:\left(\mathrm{3}{x}−\mathrm{5}\right)\left({x}+\mathrm{3}\right)\geqslant\mathrm{0}\:\wedge\:{x}+\mathrm{3}\neq\mathrm{0} \\ $$$${it}'{s}\:{easier}\:{to}\:{treat}\:\left(\mathrm{3}{x}−\mathrm{5}\right)\left({x}+\mathrm{3}\right)\geqslant\mathrm{0}. \\ $$$${we}\:{get}\:{directly}\:{x}\leqslant−\mathrm{3}\:\vee\:{x}\geqslant\frac{\mathrm{5}}{\mathrm{3}}. \\ $$$${see}\:{diagram}\:{below}. \\ $$$${for}\:{y}\geqslant\mathrm{0}\:\Rightarrow{x}\leqslant{x}_{\mathrm{1}} \:\vee\:{x}\geqslant{x}_{\mathrm{2}} \\ $$$${for}\:{y}\leqslant\mathrm{0}\:\Rightarrow{x}_{\mathrm{1}} \leqslant\:{x}\leqslant{x}_{\mathrm{2}} \\ $$
Commented by mr W last updated on 05/May/24
Commented by Frix last updated on 05/May/24
![((g(x))/(h(x)))≥0 means that either both functions are >0 or both functions are <0. [(7/4)>0 and ((−7)/(−4)) is also >0] h(x)=0 is allowed but h(x)=0 leads to the undefined (((x))/0). I wrote (g(x)≥0∧h(x)>0)∨(g(x)<0∧h(x)<0) where the ∨ means “or” so we get 2 regions](https://www.tinkutara.com/question/Q207062.png)
$$\frac{{g}\left({x}\right)}{{h}\left({x}\right)}\geqslant\mathrm{0}\:\mathrm{means}\:\mathrm{that}\:\mathrm{either}\:\mathrm{both}\:\mathrm{functions} \\ $$$$\mathrm{are}\:>\mathrm{0}\:\mathrm{or}\:\mathrm{both}\:\mathrm{functions}\:\mathrm{are}\:<\mathrm{0}. \\ $$$$\left[\frac{\mathrm{7}}{\mathrm{4}}>\mathrm{0}\:\mathrm{and}\:\frac{−\mathrm{7}}{−\mathrm{4}}\:\mathrm{is}\:\mathrm{also}\:>\mathrm{0}\right] \\ $$$${h}\left({x}\right)=\mathrm{0}\:\mathrm{is}\:\mathrm{allowed}\:\mathrm{but}\:{h}\left({x}\right)=\mathrm{0}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{undefined}\:\frac{\left({x}\right)}{\mathrm{0}}. \\ $$$$\mathrm{I}\:\mathrm{wrote}\:\left({g}\left({x}\right)\geqslant\mathrm{0}\wedge{h}\left({x}\right)>\mathrm{0}\right)\vee\left({g}\left({x}\right)<\mathrm{0}\wedge{h}\left({x}\right)<\mathrm{0}\right) \\ $$$$\mathrm{where}\:\mathrm{the}\:\vee\:\mathrm{means}\:“\mathrm{or}''\:\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\mathrm{2}\:\mathrm{regions} \\ $$