Question Number 184819 by cortano1 last updated on 12/Jan/23
$$\:\:{For}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:,\:{maximum}\:{value} \\ $$$$\:\:{of}\:{f}\left({x}\right)={x}\sqrt{\mathrm{1}−{x}+\sqrt{\mathrm{1}−{x}}}\:{is}\:\_\_ \\ $$
Answered by Frix last updated on 12/Jan/23
![f′(x)=0 −((5x−4+2(3x−2)(√(1−x)))/(4(√(1−x))(√(1−x+(√(1−x))))))=0 2(3x−2)(√(1−x))=4−5x Squaring and transforming x(x^2 −((59)/(36))x+(2/3))=0 x=0 [obviously wrong as f′(0)=(√2)] x=(8/9) [also wrong as f′((8/9))=−1] x=(3/4) [f′((3/4))=0] f((3/4))=((3(√3))/8)](https://www.tinkutara.com/question/Q184850.png)
$${f}'\left({x}\right)=\mathrm{0} \\ $$$$−\frac{\mathrm{5}{x}−\mathrm{4}+\mathrm{2}\left(\mathrm{3}{x}−\mathrm{2}\right)\sqrt{\mathrm{1}−{x}}}{\mathrm{4}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{1}−{x}+\sqrt{\mathrm{1}−{x}}}}=\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{3}{x}−\mathrm{2}\right)\sqrt{\mathrm{1}−{x}}=\mathrm{4}−\mathrm{5}{x} \\ $$$$\mathrm{Squaring}\:\mathrm{and}\:\mathrm{transforming} \\ $$$${x}\left({x}^{\mathrm{2}} −\frac{\mathrm{59}}{\mathrm{36}}{x}+\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\left[\mathrm{obviously}\:\mathrm{wrong}\:\mathrm{as}\:{f}'\left(\mathrm{0}\right)=\sqrt{\mathrm{2}}\right] \\ $$$${x}=\frac{\mathrm{8}}{\mathrm{9}}\:\left[\mathrm{also}\:\mathrm{wrong}\:\mathrm{as}\:{f}'\left(\frac{\mathrm{8}}{\mathrm{9}}\right)=−\mathrm{1}\right] \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{4}}\:\left[{f}'\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{0}\right] \\ $$$${f}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$
Answered by Frix last updated on 12/Jan/23
![Let t=(√(1−x))≥0 ⇔ x=1−t^2 f(t)=(1−t^2 )(√(t^2 +t)) f′(t)=0 −((6t^3 +5t^2 −2t−1)/(2(√(t^2 +t))))=0 t=−1 [wrong] t=−(1/3) [wrong] t=(1/2) ⇒ x=(3/4)](https://www.tinkutara.com/question/Q184851.png)
$$\mathrm{Let}\:{t}=\sqrt{\mathrm{1}−{x}}\geqslant\mathrm{0}\:\Leftrightarrow\:{x}=\mathrm{1}−{t}^{\mathrm{2}} \\ $$$${f}\left({t}\right)=\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\sqrt{{t}^{\mathrm{2}} +{t}} \\ $$$${f}'\left({t}\right)=\mathrm{0} \\ $$$$−\frac{\mathrm{6}{t}^{\mathrm{3}} +\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}}{\mathrm{2}\sqrt{{t}^{\mathrm{2}} +{t}}}=\mathrm{0} \\ $$$${t}=−\mathrm{1}\:\left[\mathrm{wrong}\right] \\ $$$${t}=−\frac{\mathrm{1}}{\mathrm{3}}\:\left[\mathrm{wrong}\right] \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{x}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$