Show that $f(x) = \log(x + \sqrt {x^2+1})$ is an odd function

$$f(x)+f(-x)=\log\left[\left(x+\sqrt{x^2+1}\right)\cdot\left(-x+\sqrt{x^2+1}\right)\right]=\log(x^2+1-x^2)=0.$$


$$f(-x) = \ln\left(-x + \sqrt{x^2 + 1}\right) = \ln\left( \left(-x + \sqrt{x^2 + 1}\right) \frac{x +\sqrt{x^2 + 1}}{x +\sqrt{x^2 + 1}} \right) \\ = \ln\left(\frac{1}{x +\sqrt{x^2 + 1}}\right) = -f(x)$$


Hint:

use $$(a-b)*(a+b) = a^2 - b^2$$ and $$\log\frac{1}{a} = - \log(a) $$ properties.

Tags:

Functions

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