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Question Number 46419 by maxmathsup by imad last updated on 25/Oct/18

$$letf\left(x\right)=\sqrt{x+1-\sqrt{x-1}}$$$$1)find{D}_f$$$$2)determine{f}^{-1}\left(x\right)$$$$3)find\int f\left(x\right)dx$$$$4)dtetrmine\int f^{-1}\left(x\right)$$$$5)letg\left(x\right)={\left(ch\left(x\right)\right)}^{2}calculatefog\left(x\right)and{\left(fog\right)}^{{}^{\prime }}\left(x\right).$$

Commented by maxmathsup by imad last updated on 28/Oct/18

$$1)x\in D_f\iff x+1-\sqrt{x-1}⩾0andx⩾1\Rightarrow x+1⩾\sqrt{x-1}andx⩾1$$$$\Rightarrow {(x+1)}^2⩾x-1andx⩾1\Rightarrow x^2+2x+1-x+1⩾0andx⩾1$$$$\Rightarrow x(x+1)⩾0andx⩾1\Rightarrow D_f=[1,+\mathrm{\infty }[$$$$2)letf\left(x\right)=y\Rightarrow x=f^{-1}\left(y\right)$$$$f\left(x\right)=y\Rightarrow x+1-\sqrt{x-1}=y^2\Rightarrow (x+1)-y^2=\sqrt{x-1}\Rightarrow$$$${(x+1-y^2)}^2=x-1\Rightarrow {(x+1)}^2-2y^2(x+1)+y^4-x+1=0\Rightarrow$$$$x^2+2x+1-2y^{2}x-2y^2+y^4-x+1=0\Rightarrow$$$$x^2+(1-2y^2)x+y^4-2y^2+2=0thisequationhave{s}^t\iff \mathrm{\Delta }⩾0\Rightarrow$$$${(1-2y^2)}^2-4(y^4-2y^2+2)⩾0\Rightarrow 4y^4-4y^2+1-4y^4+8y^2-8⩾0\Rightarrow$$$$4y^2-7⩾0so{x}_1=\frac{2y^2-1+\sqrt{4y^2-7}}{2}and{x}_2=\frac{2y^2-1-\sqrt{4y^2-7}}{2}$$$$f^{-1}\left(x\right)=\frac{2x^2-1\stackrel{-}{+}\sqrt{4x^2-7}}{2}$$

Commented by maxmathsup by imad last updated on 28/Oct/18

$$3)\int f\left(x\right)dx=\int \sqrt{x-1+2-\sqrt{x-1}}dxchangement\sqrt{x-1}=tgivex-1=t^2\Rightarrow$$$$\int f\left(x\right)dx=\int \sqrt{t^2+2-t}2tdt=2\int t\sqrt{t^2-t+2}dt$$$$=2\int t\sqrt{t^2-2\frac{1}{2}t+\frac{1}{4}+2-\frac{1}{4}}dt=2\int t\sqrt{{(t-\frac{1}{2})}^2+\frac{7}{4}}dt$$$${=}_{t-\frac{1}{2}=\frac{\sqrt{7}}{2}sh\left(t\right)}2\int (\frac{\sqrt{7}}{2}sh\left(t\right)+\frac{1}{2})\frac{\sqrt{7}}{2}ch\left(t\right)\frac{\sqrt{7}}{2}ch\left(t\right)dt$$$$=\frac{7}{4}\int (1+\sqrt{7}sh\left(t\right){ch}^2\left(t\right)dt$$$$=\frac{7}{4}\int {ch}^{2}tdt+\frac{7\sqrt{7}}{4}\int sh\left(t\right){ch}^2\left(t\right)dtbut\int {ch}^{2}tdt=\int \frac{1+ch\left(2t\right)}{2}dt$$$$=\frac{t}{2}+\frac{1}{4}sh\left(2t\right)=\frac{t}{2}+\frac{1}{2}sh\left(t\right)ch\left(t\right)and\int sh\left(t\right){ch}^{2}tdt=\frac{1}{3}{ch}^{3}t\Rightarrow$$$$\int f\left(x\right)dx=\frac{7}{8}t+\frac{7}{8}sht)ch\left(t\right)+\frac{7\sqrt{7}}{12}{ch}^3\left(t\right)+c$$$$=\frac{7}{8}\sqrt{x-1}+\frac{7}{8}sh(\sqrt{x-1)}ch\left(\sqrt{x-1}\right)+\frac{7\sqrt{7}}{12}{ch}^3\left(\sqrt{x-1}\right)+c.$$$$$$

Commented by maxmathsup by imad last updated on 28/Oct/18

$$4)wehave2\int f^{-1}\left(x\right)dx=\int (2x^2-1)dx\stackrel{-}{+}\int \sqrt{4x^2-7}dxbut$$$$\int (2x^2-1)dx=\frac{2}{3}{x}^3-x+c_1$$$$\int \sqrt{4x^2-7}dx=\int 2\sqrt{x^2-\frac{7}{4}}dx{=}_{x=\frac{\sqrt{7}}{2}ch\left(t\right)}2\int \frac{\sqrt{7}}{2}sh\left(t\right)\frac{\sqrt{7}}{2}sh\left(t\right)dt$$$$=\frac{7}{2}\int {sh}^2\left(t\right)dt=\frac{7}{4}\int (ch\left(2t\right)-1)dt$$$$=\frac{7}{8}sh\left(2t\right)-\frac{7t}{4}+c_2$$$$=\frac{7}{4}sh\left(t\right)ch\left(t\right)-\frac{7}{4}t+c_2=\frac{7}{4}\sqrt{{\left(\frac{2x}{\sqrt{7}}\right)}^2-1}\frac{2x}{\sqrt{7}}-\frac{7}{4}argch\left(\frac{2x}{\sqrt{7}}\right)+c_2$$$$=\frac{\sqrt{7}}{2}\sqrt{\frac{4x^2}{7}-1}-\frac{7}{4}ln(\frac{2x}{\sqrt{7}}+\sqrt{\frac{4x^2}{7}-1})+c_2\Rightarrow$$$$\int f^{-1}\left(x\right)dx=\frac{x^3}{3}-\frac{x}{2}\stackrel{-}{+}\{\frac{\sqrt{7}}{4}\sqrt{\frac{4x^2}{7}-1}-\frac{7}{8}ln(\frac{2x}{\sqrt{7}}+\sqrt{\frac{4x^2}{7}-1})+C.$$

Commented by maxmathsup by imad last updated on 28/Oct/18

$$erroroftypo$$$$\int f^{-1}\left(x\right)dx=\frac{x^3}{3}-\frac{x}{2}\stackrel{-}{+}\{\frac{\sqrt{7}}{4}x\sqrt{\frac{4x^2}{7}-1}-\frac{7}{8}ln(\frac{2x}{\sqrt{7}}+\sqrt{\frac{4x^2}{7}-1})+C$$

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