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Question Number 39702 by math khazana by abdo last updated on 10/Jul/18

$$letf\left(x\right)=2\sqrt{x-\sqrt{x-3}+2}$$$$1)find{D}_f$$$$2)calculate{f}^{{}^{\prime }}\left(x\right)$$$$3)determine{f}^{-1}\left(x\right)$$$$4)calculate{\left(f^{-1}\right)}^{{}^{\prime }}\left(x\right)$$$$5)letu\left(x\right)=x^2+4$$$$determinev\left(x\right)=fou\left(x\right)andcalculate$$$$v^{{}^{\prime }}\left(x\right)$$$$6)calculate{\int }_3^{5}f\left(x\right)dx.$$

Commented by math khazana by abdo last updated on 13/Jul/18

$$1)x\in D_f\iff x⩾3andx+2-\sqrt{x-3}⩾0\Rightarrow$$$$x⩾3and{(x+2)}^2⩾x-3\Rightarrow x^2+4x+4-x+3⩾0\Rightarrow$$$$x^2+3x+7⩾0thisinequalityistruebecause$$$$\mathrm{\Delta }=9-28<0\Rightarrow D_f=[3,+\mathrm{\infty }[$$$$2)wehave{f}^2\left(x\right)=4(x-\sqrt{x-3}+2)\Rightarrow$$$$2f^{{}^{\prime }}\left(x\right)f\left(x\right)=4(1-\frac{1}{2\sqrt{x-3}})=4\frac{2\sqrt{x-3}-1}{2\sqrt{x-3}}$$$$=2\frac{2\sqrt{x-3}-1}{\sqrt{x-3}}\Rightarrow f^{{}^{\prime }}\left(x\right)f\left(x\right)=\frac{2\sqrt{x-3}-1}{\sqrt{x-3}}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{2\sqrt{x-3}-1}{2\sqrt{x+2-\sqrt{x-3}}\sqrt{x-3}}.$$

Commented by math khazana by abdo last updated on 13/Jul/18

$$3)f^{-1}\left(x\right)=y\iff x=f\left(y\right)\Rightarrow x=2\sqrt{y+2-\sqrt{y-3}}$$$$\Rightarrow x^2=4(y+2-\sqrt{y-3})\Rightarrow$$$$x^2=4y+8-4\sqrt{y-3}\Rightarrow 4\sqrt{y-3}=4y+8-x^2\Rightarrow$$$$16(y-3)={(4y+8-x^2)}^2\Rightarrow$$$$16y-48={(4y+8)}^2-2x^2(4y+8)+x^4$$$$16y-48=16y^2+64y+64-8{yx}^2-16x^2+x^4\Rightarrow$$$$x^4-8(y+2)x^2+16y^2+64-16y+48=0\Rightarrow$$$$x^4-8(y+2)x^2+16y^2-16y+112=0$$$${\mathrm{\Delta }}^{{}^{\prime }}={(-2(y+2))}^2-16y^2+16y-112$$$$=4(y^2+4y+4)-16y^2+16y-112$$$$=-8y^2+32y+16-112...becontinued...$$

Commented by maxmathsup by imad last updated on 14/Jul/18

$$5)wehavev\left(x\right)=f\left(u\left(x\right)\right)=2\sqrt{u\left(x\right)-\sqrt{u\left(x\right)-3}+2}$$$$=2\sqrt{x^2+4-\sqrt{x^2+1}+2}$$$$=2\sqrt{x^2-\sqrt{x^2+1}+6}$$$$wehave{v}^2\left(x\right)=4\{x^2-\sqrt{x^2+1}+2\}\Rightarrow$$$$2v\left(x\right)v^{{}^{\prime }}\left(x\right)=4{\{x^2+2-\sqrt{x^2+1}\}}^{{}^{\prime }}=4\{2x-\frac{x}{\sqrt{x^2+1}}\}\Rightarrow$$$$v\left(x\right)v^{{}^{\prime }}\left(x\right)=2\left\{\frac{2x\sqrt{x^2+1}-x}{\sqrt{x^2+1}}\right\}\Rightarrow v^{{}^{\prime }}\left(x\right)=\frac{2}{v\left(x\right)}\frac{2x\sqrt{x^2+1}-x}{\sqrt{x^2+1}}$$$$=\frac{2x\sqrt{x^2+1}-x}{(\sqrt{x^2+6-\sqrt{x^2+1})(\sqrt{x^2+1)}}}.$$

Commented by maxmathsup by imad last updated on 14/Jul/18

$$6){\int }_3^{5}f\left(x\right)dx=2{\int }_3^5\sqrt{x+2-\sqrt{x-3}}dxchangement\sqrt{x-3}=tgivex=3+t^2$$$${\int }_3^{5}f\left(x\right)dx=2{\int }_0^{\sqrt{2}}\sqrt{5+t^2-t}2tdt=4{\int }_0^{\sqrt{2}}t\sqrt{t^2-t+5}dt$$$$=4{\int }_0^{\sqrt{2}}t\sqrt{{(t-\frac{1}{2})}^2+5-\frac{1}{4}}dt=4{\int }_0^{\sqrt{2}}t\sqrt{{(t-\frac{1}{2})}^2+\frac{19}{4}}dt$$$${=}_{t-\frac{1}{2}=\frac{\sqrt{19}}{2}sh\left(u\right)}4{\int }_{-arsh\left(\frac{1}{\sqrt{19}}\right)}^{argsh\left(\frac{2\sqrt{2}-1}{\sqrt{19}}\right)}(\frac{1}{2}+\frac{\sqrt{19}}{2}sh\left(u\right))ch\left(u\right)\frac{\sqrt{19}}{2}ch\left(u\right)du$$$$=\sqrt{19}{\int }_{-argsh\left(\frac{1}{\sqrt{19}}\right)}^{argsh\left(\frac{2\sqrt{2}-1}{\sqrt{19}}\right)}(1+\sqrt{19}sh\left(u\right))ch\left(u\right)du$$$$=\sqrt{19}{\int }_{-argsh\left(\frac{1}{\sqrt{19}}\right)}^{argsh\left(\frac{2\sqrt{2}-1}{\sqrt{19}}\right)}ch\left(u\right)du+\sqrt{19}{\int }_{-argsh\left(\frac{1}{\sqrt{19}}\right)}^{argsh\left(\frac{2\sqrt{2}-1}{\sqrt{19}}\right)}sh\left(u\right)ch\left(u\right)du$$$$=\sqrt{19\{}\frac{2\sqrt{2}-1}{\sqrt{19}}+\frac{1}{\sqrt{19}}\}+\frac{\sqrt{19}}{2}\{{\left(\frac{2\sqrt{2}-1}{\sqrt{19}}\right)}^2-\left(\frac{1}{19}\right)\}.$$

Commented by ajfour last updated on 14/Jul/18

$$Greatworksir!$$

Commented by math khazana by abdo last updated on 15/Jul/18

$$thankyousir.$$

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