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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

### Best Answer

#5
+28188
+10

Here's an alternative proof:

Oct 29, 2015

### 6+0 Answers

#1
+105634
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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
#2
+6045
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$$-\ln(x+\sqrt{x^2+1}) = \ln\left(\dfrac{1}{x+\sqrt{x^2+1}}\right) =$$

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

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

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

.
Oct 29, 2015
#3
+105634
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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
#4
+28188
+5

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
#5
+28188
+10
Best Answer

Here's an alternative proof:

Alan Oct 29, 2015
#6
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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