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Question Number 121278 by Ar Brandon last updated on 06/Nov/20

The domaine of the function f(x)=((√(log_(0.5) (x^2 −7x+13))))^(−1) is;

$$\mathrm{The}\:\mathrm{domaine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left(\sqrt{\mathrm{log}_{\mathrm{0}.\mathrm{5}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{7x}+\mathrm{13}\right)}\right)^{−\mathrm{1}} \:\mathrm{is}; \\ $$

Answered by TANMAY PANACEA last updated on 06/Nov/20

x^2 −7x+13 (x^2 −2×x×(7/2)+((49)/4)+13−((49)/4)) (x−(7/2))^2 +(3/4)>0 so g(x)=x^2 −7x+13>0 f(x)=(1/( (√(log_(0.5) (x^2 −7x+13))))) note (x^2 −7x+13) can not be equsls to 1 if x^2 −7x+13=1 f(x)=(1/( (√(log_(0.5) 1))))=(1/0)=unddfined so 1>(x^2 −7x+13)>0 x^2 −7x+13<1 x^2 −7x+12<0 (x−3)(x−4)<0 when x>4 (x−3)(x−4)>0 not feasibld when x<3 (x−3)(x−4)>0 not feasible when 4>x>3 (x−3)(x−4)<0 feasible so domain ...4>x>3

$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}×{x}×\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{49}}{\mathrm{4}}+\mathrm{13}−\frac{\mathrm{49}}{\mathrm{4}}\right) \\ $$$$\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}>\mathrm{0} \\ $$$${so}\:\:{g}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13}>\mathrm{0} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{log}}_{\mathrm{0}.\mathrm{5}} \left(\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{7}\boldsymbol{{x}}+\mathrm{13}\right)}} \\ $$$$\boldsymbol{{note}}\:\left(\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{7}\boldsymbol{{x}}+\mathrm{13}\right)\:{can}\:{not}\:{be}\:{equsls}\:{to}\:\mathrm{1} \\ $$$$\:\:{if}\:{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13}=\mathrm{1} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{log}_{\mathrm{0}.\mathrm{5}} \mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{0}}={unddfined} \\ $$$${so}\:\mathrm{1}>\left({x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13}\right)>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{13}<\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}<\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)<\mathrm{0} \\ $$$${when}\:{x}>\mathrm{4}\:\:\:\:\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)>\mathrm{0}\:{not}\:{feasibld} \\ $$$${when}\:{x}<\mathrm{3}\:\:\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)>\mathrm{0}\:{not}\:{feasible} \\ $$$${when}\:\mathrm{4}>{x}>\mathrm{3}\:\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)<\mathrm{0}\:{feasible} \\ $$$${so}\:\:\:{domain}\:...\mathrm{4}>{x}>\mathrm{3} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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

$${most}\:{selcome}\:{sir} \\ $$

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