f-x-ln-x-x-2-1-find-f-1-x-plzz-help-

Question Number 27608 by chernoaguero@gmail.com last updated on 10/Jan/18

$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{ln}\left(\mathrm{x}\:+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$ \\ $$$$\mathrm{find}\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\mathrm{plzz}\:\mathrm{help} \\ $$

Commented by abdo imad last updated on 11/Jan/18

f(x)=y ⇔ f^(−1) (y)=x but f(x)=y ⇔ ln(x+(√(x^2 +1)) ) =y x+(√(x^2 +1)) =e^y ⇔ (√(x^2 +1 )) = e^y −x and e^y >x ⇔ x^2 +1 =(e^y −x)^2 ⇔ x^2 = (e^y −x)^2 −1 ⇔x^2 = e^(2y) −2e^y x +x^2 −1⇔ 2e^y x = e^(2y) −1 ⇔ x=((e^(2y) −1)/(2e^y )) = ((e^y −e^(−y) )/2) = sinhy f^(−1) (x)= ((e^x −e^(−x) )/2) =shx .

$${f}\left({x}\right)={y}\:\:\:\Leftrightarrow\:{f}^{−\mathrm{1}} \left({y}\right)={x}\:\:\:{but}\:{f}\left({x}\right)={y}\:\Leftrightarrow\:{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\right)\:={y} \\ $$$${x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\:={e}^{{y}} \:\:\Leftrightarrow\:\:\:\:\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}\:}\:=\:{e}^{{y}} \:−{x}\:\:{and}\:{e}^{{y}} \:>{x} \\ $$$$\Leftrightarrow\:\:{x}^{\mathrm{2}} +\mathrm{1}\:=\left({e}^{{y}} −{x}\right)^{\mathrm{2}} \:\:\Leftrightarrow\:{x}^{\mathrm{2}} =\:\left({e}^{{y}} \:−{x}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\Leftrightarrow{x}^{\mathrm{2}} =\:{e}^{\mathrm{2}{y}} \:\:−\mathrm{2}{e}^{{y}} {x}\:+{x}^{\mathrm{2}} −\mathrm{1}\Leftrightarrow\:\mathrm{2}{e}^{{y}} {x}\:=\:{e}^{\mathrm{2}{y}} −\mathrm{1} \\ $$$$\Leftrightarrow\:{x}=\frac{{e}^{\mathrm{2}{y}} −\mathrm{1}}{\mathrm{2}{e}^{{y}} }\:=\:\frac{{e}^{{y}} \:−{e}^{−{y}} }{\mathrm{2}}\:=\:{sinhy} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\:\frac{{e}^{{x}} \:−{e}^{−{x}} }{\mathrm{2}}\:={shx}\:\:. \\ $$$$ \\ $$$$ \\ $$

Commented by chernoaguero@gmail.com last updated on 11/Jan/18

Thank u sir but what do u mean by sinhy an shx

$$\mathrm{Thank}\:\mathrm{u}\:\mathrm{sir}\:\mathrm{but}\:\mathrm{what}\:\mathrm{do}\:\:\mathrm{u}\:\mathrm{mean} \\ $$$$\mathrm{by}\:\mathrm{sinhy}\:\mathrm{an}\:\mathrm{shx} \\ $$

Commented by Joel578 last updated on 11/Jan/18