Prove that f(x)=ln(x + (x^2+1)^0.5 ) is an odd function... :/
Prove that f(x)=ln(x + (x^2+1)^0.5 ) is an odd function... :/
If this is an odd function then
f(-x)=-f(x)
and since x^2= (-x)^2
that means you are suppose to prove that
\(ln(-x+\sqrt{x^2+1})=-ln(x+\sqrt{x^2+1})\)
Now
\(ln(x+\sqrt{x^2+1})\) \(=sinh^{-1}(x)\)
is the inverse hyperbolic sine function so it is definitely odd. But I do not know how to prove this.
I shall flag other mathematicians to take a look for you. You should get an answer but it meight take a day or so. ://
\(-\ln(x+\sqrt{x^2+1}) = \ln\left(\dfrac{1}{x+\sqrt{x^2+1}}\right) =\)
\(\ln\left(\dfrac{1}{x+\sqrt{x^2+1}} \dfrac{x-\sqrt{x^2+1}}{x-\sqrt{x^2+1}}\right) = \ln\left( \dfrac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}\right)=\)
\(\ln\left(-(x-\sqrt{x^2+1})\right) =\ln\left(-x + \sqrt{x^2+1}\right)\)
\(\mbox{so }f(-x)=-f(x) \mbox{ and f is odd.}\)
.Thanks Rom,
I should of thought of that I guess.
I suppose things often look easy when someone shows you how to do it.