Question

show that if f is odd and find the inverse of f
f(x)= ln(x+sqrt(x^2+1))

Answer 1

f is odd if f(-x)=-f(x)

Answer 2

@ mukushl how?????

Answer 3

\[f(-x)=\ln (-x+\sqrt{1+x^2)}\]
and
\[-f(x)=-\ln (x+\sqrt{1+x^2)}\]

Answer 4

\[
f(x)= \sinh^{-1}(x)
\]
and sinh is odd.
The inverse of f is sinh(x)

Answer 5

See http://en.wikipedia.org/wiki/Hyperbolic_function

Answer 6

eliassaab's way is much better

Answer 7

@mukushla f(-x) not equal -f(x)

Answer 8

so f(x) is not odd

Answer 9

But in question it is written SHOW THAT so f(x) is an odd function..

Answer 10

This is what I want

Answer 11

@waterineyes
is f an odd function?

Answer 12

im wondering because it seems f(-x) not equal -f(x)

Answer 13

Question says f is an odd function we have to just show it and I am unable in doing that..
Do you think that ln(......) is any formula or not??

Answer 14

im sorry
\[\frac{1}{x+\sqrt{1+x^2}}=-x+\sqrt{1+x^2}\]
then f is odd

Answer 15

Guys, f(x) is odd. See my post above and copy and paste in your browser to see that.
http://www.wolframalpha.com/input/?i=Plot [Log[x+%2B+Sqrt[1+%2B+x^2]]%2C+{x%2C+-100%2C+100}]

Answer 16

That is what I was thinking...

Answer 17

I think openstudy should change the way medals are awarded. The only one who got a medal here is one who did not contribute to the solution.

Answer 18

for the inverse:
\[y=\ln(x+\sqrt{x^2+1})\]
inverse is given by:
\[x=\ln(y+\sqrt{y^2+1})\]
\[e^x=e^{\ln(y+\sqrt{y^2+1})}\]
\[e^x=y+\sqrt{y^2+1}\]
\[e^x/y=1+\sqrt{(y^2+1)/y^2}\]
\[e^x/y=1+\sqrt{1+1/y^2}\]
\[e^x/y-1=\sqrt{1+1/y^2}\]
\[e^{2x}/y^2-2e^x/y+1=1/y^2+1\]
\[e^{2x}-2e^xy+y^2=1+y^2\]
\[y=(e^{2x}-1)/2e^x\] <-answer

Answer 19

but icant find perfect answer for f is odd

Answer 20

You want to prove f is odd??

Answer 21

yes

Answer 22

Do you know what is an odd function??

Answer 23

f(-x) = -f(x)

Answer 24

Yes you got it very right..

How we find first f(-x) by replacing x by (-x)

Okay??

Answer 25

\[f(−x)= \ln(-x + \sqrt{1 + x^2})\]

Got or not??

Answer 26

And now we find -f(x):

\[-f(x) = -\ln(x + \sqrt{1 + x^2})\]

Got till here?

Answer 27

What you think @obada these two expressions are equal or not??

Answer 28

not equal

Answer 29

I will prove they are equal just be cool...

See, here we need to solve more so that we can prove they are equal..

You know one of the properties of Log is:

\[a.\ln(x ) = \ln(x)^a\]

This is called Power Rule...

Answer 30

im waiting you waterineyes

Answer 31

\[(-1)\ln(x + \sqrt{1 + x^2}) = \ln(x + \sqrt{1 + x^2})^{-1}\]

\[\ln(\frac{1}{x + \sqrt{1 + x^2}})\]

This is because:
\[a^{-1} = \frac{1}{a}\]

So, Rationalizing the denominator we get:

\[= \ln(\frac{1}{x + \sqrt{1 + x^2}} \times \frac{x - \sqrt{1 + x^2}}{x -\sqrt{1 + x^2}})\]

\[= \ln(\frac{x - \sqrt{1 + x^2}}{x^2 - 1 - x^2}) = \ln(\frac{x - \sqrt{1 + x^2}}{-1})\]

\[ = \ln(-x + \sqrt{1 + x^2)}\] (Just taken the -common from Numerator and this - cancels with denominator..)

Answer 32

Now what you think the result matches with f(-x) or not???

Answer 33

thank you very much @waterineyes just you answer me
yes now iknow how is odd ?
thx again

Answer 34

No need to say thanks..

Welcome dear...

Answer 35

@eliassaab, do you want to say anything else??

Answer 36

I do not know why are we still discussing it. I gave a one line proof that f is odd at the very beginning and what is the inverse of it.

Answer 37

Well, math can be solved in more than one way..
You gave your one..
so,, we gave our one...