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

$$\mathrm{The}\mathrm{domaine}\mathrm{of}\mathrm{the}\mathrm{function}$$$$\mathrm{f}\left(\mathrm{x}\right)={\left(\sqrt{{\log }_{0.5}({\mathrm{x}}^2-7\mathrm{x}+13)}\right)}^{-1}\mathrm{is};$$

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

$$x^2-7x+13$$$$(x^2-2\times x\times \frac{7}{2}+\frac{49}{4}+13-\frac{49}{4})$$$${(x-\frac{7}{2})}^2+\frac{3}{4}>0$$$$sog\left(x\right)=x^2-7x+13>0$$$$\mathit{f}\left(\mathit{x}\right)=\frac{1}{\sqrt{{\mathit{l}\mathit{o}\mathit{g}}_{0.5}({\mathit{x}}^2-7\mathit{x}+13)}}$$$$\mathit{n}\mathit{o}\mathit{t}\mathit{e}({\mathit{x}}^2-7\mathit{x}+13)cannotbeequslsto1$$$$if{x}^2-7x+13=1$$$$f\left(x\right)=\frac{1}{\sqrt{{log}_{0.5}1}}=\frac{1}{0}=unddfined$$$$so1>(x^2-7x+13)>0$$$$x^2-7x+13<1$$$$x^2-7x+12<0$$$$(x-3)(x-4)<0$$$$whenx>4(x-3)(x-4)>0notfeasibld$$$$whenx<3(x-3)(x-4)>0notfeasible$$$$when4>x>3(x-3)(x-4)<0feasible$$$$sodomain...4>x>3$$$$$$$$$$$$$$

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

$$mostselcomesir$$

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