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# Prove that f(x)=ln(x + (x^2+1)^0.5 ) is an odd function...solved it by rationalizing but need another way if possible pls :/

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Prove that f(x)=ln(x + (x^2+1)^0.5 ) is an odd function... :/

Oct 28, 2015
edited by Guest  Oct 28, 2015
edited by Guest  Oct 28, 2015
edited by Guest  Oct 28, 2015
edited by Guest  Oct 28, 2015

#5
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Here's an alternative proof: Oct 29, 2015

#1
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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. ://

Oct 29, 2015
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$$-\ln(x+\sqrt{x^2+1}) = \ln\left(\dfrac{1}{x+\sqrt{x^2+1}}\right) =$$

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

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

$$\mbox{so }f(-x)=-f(x) \mbox{ and f is odd.}$$

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Oct 29, 2015
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Thanks Rom,

I should of thought of that I guess. I suppose things often look easy when someone shows you how to do it.

Oct 29, 2015
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I think the questioner knew how to solve it Rom's way (see the question title).  He/she was looking for a possible alternative method!

Oct 29, 2015
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Here's an alternative proof: Alan Oct 29, 2015
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Thank you guys alot...

Thank you Alan for your alternative answer, it helped me alot! I'm new to this site so excuse my late reply

Nov 2, 2015