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Let $f(x)=\ln(x+\sqrt{x^2+1})$. Find a function $g(x)$ such that $g(f(x))=x$ for every $x$. Find $g(2)$.

I don't have even the slightest idea how to solve such question.I tried to transform the rhs of equation $g(\ln(x+\sqrt{x^2+1})) =x$ in form of $f(x)$ but got struck..

Robert Z
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user473333
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3 Answers3

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Since $g(f(x))=x$, $x=g(y)$ is the inverse function of $y=f(x)$. Hence, in order to find $g$, solve the equation $$\ln(x+\sqrt{x^2+1})=y$$ with respect to $x$. Now $$\ln(x+\sqrt{x^2+1})=y\Leftrightarrow \sqrt{x^2+1}=e^y-x,$$ which implies $$x^2+1=(e^y-x)^2=e^{2y}-2e^{y}x+x^2\Leftrightarrow g(y)=x=\frac{e^{2y}-1}{2e^y}=\frac{e^{y}-e^{-y}}{2}.$$ Now you may find $g(2)$.

P.S. Take a look HERE.

Robert Z
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$\cosh^2 z=1+\sinh^2 z$, hence $$ f(\sinh(z))=\log(\sinh z+\cosh z) = \log e^z = z $$ i.e. $f(z)=\text{arcsinh}(z)$ and the problem boils down to computing $\sinh(2)=\frac{e^4-1}{2e^2}\approx \frac{243}{67}$.

Jack D'Aurizio
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Note that for $y=f(x)$: $$f(f^{-1}(x))=f^{-1}(f(x))=x.$$

Also note: $$g(f(x))=x \iff g(x)=f^{-1}(x).$$ To find $f^{-1}(x)$ of the given $y=f(x)=\ln{(x+\sqrt{x^2+1})}$: $$e^y-x=\sqrt{x^2+1} \Rightarrow x=\frac{e^{2y}-1}{2e^y} \Rightarrow f^{-1}(x)=\frac{e^{2x}-1}{2e^x}.$$ Thus: $$g(2)=f^{-1}(2)=\frac{e^{2\cdot 2}-1}{2e^2}=\frac{e^4-1}{2e^2}.$$

farruhota
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