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Question Number 184819 by cortano1 last updated on 12/Jan/23

$$For0⩽x⩽1,maximumvalue$$$$off\left(x\right)=x\sqrt{1-x+\sqrt{1-x}}is\mathrm{\_}\mathrm{\_}$$

Answered by Frix last updated on 12/Jan/23

$$f^{\prime }\left(x\right)=0$$$$-\frac{5x-4+2(3x-2)\sqrt{1-x}}{4\sqrt{1-x}\sqrt{1-x+\sqrt{1-x}}}=0$$$$2(3x-2)\sqrt{1-x}=4-5x$$$$\mathrm{Squaring}\mathrm{and}\mathrm{transforming}$$$$x(x^2-\frac{59}{36}x+\frac{2}{3})=0$$$$x=0[\mathrm{obviously}\mathrm{wrong}\mathrm{as}{f}^{\prime }\left(0\right)=\sqrt{2}]$$$$x=\frac{8}{9}[\mathrm{also}\mathrm{wrong}\mathrm{as}{f}^{\prime }\left(\frac{8}{9}\right)=-1]$$$$x=\frac{3}{4}[f^{\prime }\left(\frac{3}{4}\right)=0]$$$$f\left(\frac{3}{4}\right)=\frac{3\sqrt{3}}{8}$$

Answered by Frix last updated on 12/Jan/23

$$\mathrm{Let}t=\sqrt{1-x}⩾0\iff x=1-t^2$$$$f\left(t\right)=(1-t^2)\sqrt{t^2+t}$$$$f^{\prime }\left(t\right)=0$$$$-\frac{6t^3+5t^2-2t-1}{2\sqrt{t^2+t}}=0$$$$t=-1\left[\mathrm{wrong}\right]$$$$t=-\frac{1}{3}\left[\mathrm{wrong}\right]$$$$t=\frac{1}{2}\Rightarrow x=\frac{3}{4}$$

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