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Question Number 161590 by geron last updated on 20/Dec/21

f(x)=(1/(1−(√x))) + (1/(1−(√(1−x)))) f(x)_(min) = ?

$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−\sqrt{{x}}}\:+\:\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{1}−\boldsymbol{{x}}}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)_{\boldsymbol{{min}}} =\:? \\ $$

Answered by cortano last updated on 20/Dec/21

Answered by mr W last updated on 20/Dec/21

0<x<1 let x=sin^2 (45°+θ) with −45°<θ<45° f(x)=(1/(1−sin (45°+θ)))+(1/(1−cos (45°+θ))) =(1/(1−sin (45°+θ)))+(1/(1−sin (45°−θ))) f(x)_(min) =2×(1/(1−(1/( (√2)))))=2(2+(√2)) at θ=0° or x=(1/2)

$$\mathrm{0}<{x}<\mathrm{1} \\ $$$${let}\:{x}=\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{45}°+\theta\right)\:{with}\:−\mathrm{45}°<\theta<\mathrm{45}° \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:\left(\mathrm{45}°+\theta\right)}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{45}°+\theta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:\left(\mathrm{45}°+\theta\right)}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:\left(\mathrm{45}°−\theta\right)} \\ $$$${f}\left({x}\right)_{{min}} =\mathrm{2}×\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}=\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right) \\ $$$${at}\:\theta=\mathrm{0}°\:{or}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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