I need to show that $f(x) = \log(x + \sqrt{x^2+1})$ is an odd function and from what I can understand from this question (found while searching): What is an odd function?, I have to show that$f(-x)=-f(x)$. I have struggled to figure it out for hours but I couldn't find a solution.
Thanks in advance!
$$f(x)+f(-x)=\log\left[\left(x+\sqrt{x^2+1}\right)\cdot\left(-x+\sqrt{x^2+1}\right)\right]=\log(x^2+1-x^2)=0.$$
$$f(-x) = \ln\left(-x + \sqrt{x^2 + 1}\right) = \ln\left( \left(-x + \sqrt{x^2 + 1}\right) \frac{x +\sqrt{x^2 + 1}}{x +\sqrt{x^2 + 1}} \right) \\ = \ln\left(\frac{1}{x +\sqrt{x^2 + 1}}\right) = -f(x)$$
Hint:
use $$(a-b)*(a+b) = a^2 - b^2$$ and $$\log\frac{1}{a} = - \log(a) $$ properties.
By definition, $f(-x)$ is odd if and only if $f(-x)=-f(x)$
Now, we have $$f(x)=\log(x+\sqrt{x^2+1})$$ setting $x=-x$, we get $$f(-x)=\log((-x)+\sqrt{(-x)^2+1})$$ $$f(-x)=\log(-x+\sqrt{x^2+1})$$ $$=\log(\sqrt{x^2+1}-x)$$$$=-\log\left(\frac{1}{\sqrt{x^2+1}-x}\right)$$ $$=-\log\left(\frac{\sqrt{x^2+1}+x}{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}\right)$$ $$=-\log\left(\frac{x+\sqrt{x^2+1}}{(\sqrt{x^2+1})^2-(x)^2}\right)$$ $$=-\log\left(x+\sqrt{x^2+1}\right)$$$$=-f(x)$$ Hence the given function $f(x)$ is odd.
Can we Use The fact that If $f'(x)$ is an even function, Then function $f(x)$ is odd.
or If If $f'(x)$ is an odd function, Then function $f(x)$ is even.
So Here $$\displaystyle f(x) = \ln \left(x+\sqrt{x^2+1}\right)\;,$$ Then $$\displaystyle \frac{d}{dx}(f(x)) = \frac{d}{dx}\left[\ln \left(x+\sqrt{x^2+1}\right)\right]$$
So $$\displaystyle f'(x) = \frac{1}{(x+\sqrt{x^2+1})}\cdot \left(1+\frac{2x}{2\sqrt{x^2+1}}\right) = \frac{1}{\sqrt{x^2+1}}$$
So Here $$\displaystyle f'(x) = \frac{1}{\sqrt{x^2+1}}\;,$$ So $$f'(-x) = f'(x)$$ So Here $f(x)$ is an even function.
So Function $f(x) = \ln \left(x+\sqrt{x^2+1}\right)$ is an Odd function.
f(x)= log(x+√x^2+1) f(-x)=log(-x+√x^2+1) f(x)+f(-x)=log[(x+√x^2+x).(x+√x^2+1) f(x)=(x^2+1-x^2) f(1)= 0 Hance,f(x)Isan odd function
