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Question Number 125230 by mohammad17 last updated on 09/Dec/20

Answered by mathmax by abdo last updated on 09/Dec/20

$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+..}}}\Rightarrow {\mathrm{f}}^2\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{f}\left(\mathrm{x}\right)\Rightarrow {\mathrm{f}}^2\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)-\mathrm{x}=0$$$$\mathrm{withf}⩾0\to \mathrm{\Delta }=1-4(-\mathrm{x})=4\mathrm{x}+1\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1+\sqrt{4\mathrm{x}+1}}{2}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1-\sqrt{4\mathrm{x}+1}}{2}\mathrm{but}\mathrm{f}⩾0\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1+\sqrt{4\mathrm{x}+1}}{2}$$$$\mathrm{f}\mathrm{defined}\mathrm{on}[-\frac{1}{4},+\mathrm{\infty }[\mathrm{and}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{2}\times \frac{4}{2\sqrt{4\mathrm{x}+1}}=\frac{1}{\sqrt{4\mathrm{x}+1}}$$$$\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(6\right)=\frac{1}{\sqrt{25}}=\frac{1}{5}$$

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