Question Number 121278 by Ar Brandon last updated on 06/Nov/20
$$\mathrm{The}\:\mathrm{domaine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left(\sqrt{\mathrm{log}_{\mathrm{0}.\mathrm{5}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{7x}+\mathrm{13}\right)}\right)^{−\mathrm{1}} \:\mathrm{is}; \\ $$
Answered by TANMAY PANACEA last updated on 06/Nov/20
$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}×{x}×\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{49}}{\mathrm{4}}+\mathrm{13}−\frac{\mathrm{49}}{\mathrm{4}}\right) \\ $$$$\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}>\mathrm{0} \\ $$$${so}\:\:{g}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13}>\mathrm{0} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{log}}_{\mathrm{0}.\mathrm{5}} \left(\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{7}\boldsymbol{{x}}+\mathrm{13}\right)}} \\ $$$$\boldsymbol{{note}}\:\left(\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{7}\boldsymbol{{x}}+\mathrm{13}\right)\:{can}\:{not}\:{be}\:{equsls}\:{to}\:\mathrm{1} \\ $$$$\:\:{if}\:{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13}=\mathrm{1} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{log}_{\mathrm{0}.\mathrm{5}} \mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{0}}={unddfined} \\ $$$${so}\:\mathrm{1}>\left({x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13}\right)>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13}<\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}<\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)<\mathrm{0} \\ $$$${when}\:{x}>\mathrm{4}\:\:\:\:\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)>\mathrm{0}\:{not}\:{feasibld} \\ $$$${when}\:{x}<\mathrm{3}\:\:\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)>\mathrm{0}\:{not}\:{feasible} \\ $$$${when}\:\mathrm{4}>{x}>\mathrm{3}\:\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)<\mathrm{0}\:{feasible} \\ $$$${so}\:\:\:{domain}\:…\mathrm{4}>{x}>\mathrm{3} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by TANMAY PANACEA last updated on 06/Nov/20
Commented by Ar Brandon last updated on 06/Nov/20
Thanks Sir
Commented by TANMAY PANACEA last updated on 06/Nov/20
$${most}\:{selcome}\:{sir} \\ $$