$$f(x) = \log(\sqrt{x^2+1}+x)$$ I can't figure out, why this function is odd. I mean, of course, its graph shows, it's odd, but when I investigated $f(-x)$, I couldn't find way to $-\log(\sqrt{x^2+1}+x)$.
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If$$f(x) = \log(\sqrt{x^2+1}+x)$$ then $$f(-x) = \log \left(\sqrt{(-x)^2+1}-x\right)=$$ $$= \log \left((\sqrt{x^2+1}-x)\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)=$$ $$= \log \left(\frac{1}{\sqrt{x^2+1}+x}\right)=- \log({\sqrt{x^2+1}+x})=-f(x)$$ |
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Hint: $(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x) = 1$. |
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We have $f(-x)=\log \left(\frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{(\sqrt{x^2+1}+x)}\right)=\log\left(\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right)=-\log(\sqrt{x^2+1}+x)=-f(x)$. |
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Another thing to add is that the Taylor series (of odd functions, if it exists) has only odd powers $$ x-{\frac {1}{6}}{x}^{3}+{\frac {3}{40}}{x}^{5}+\dots.$$ |
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This function is another command for showing $f(x)=\text{arcsinh}(x)$ being increasing continous in $[0,\infty]$. We know that $f(x)=\sinh(x)$ is an odd one-one function. $$f(x)=\text{arcsinh(x)}\to \sinh(f(x))=x$$ so if $x\to -x$ then $$\sinh(f(-x))=-x\longrightarrow -\sinh(f(-x))=x\longrightarrow\sinh(-f(-x))=x=\sinh(f(x))$$ so $f(-x)=-f(x)$. This means $f(x)$ is an odd function. There is another different approach for this. See this link http://ddmf.msr-inria.inria.fr/1.8/ddmf
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To add another hint to Dan's answer, consider that $-\log a = \log \frac1a$ and then simplify the radical out of the denominator for your function. |
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Note by differentiating that $$f(x)=\int_0^x \frac{dt}{\sqrt{t^2+1}}.$$ The result now follows from the fact that $\dfrac{1}{\sqrt{t^2+1}}$ is even. |
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