Question

Show that f(x) = log(x+√x^2 +1 ) is an odd function .

Answer 1

Show \(f(-x)=-f(x)\)

Answer 2

Do you know about the Rationalizing of Denominator??

Answer 3

just put -x in place of x

Answer 4

Guys we have to solve it further it is not easy as you are thinking..

We have to prove by rationalization process..

Answer 5

yeah i know ..

Answer 6

so what is the final answer ? .. can any one explain it more clearly .. please ?

Answer 7

Find -f(x) first :

\(-f(x) = -\log(x + \sqrt{x^2 + 1})\) --------1

Answer 8

Now find \(f(-x)\)..

Answer 9

okay

Answer 10

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

Now we have to prove they are equal..

Answer 11

.. okay

Answer 12

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

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

\[\implies \log(\frac{x - \sqrt{x^2 + 1}}{x^2 - x^2 - 1}) \implies \log(-x + \sqrt{x^2 + 1}) \implies f(-x)\]

So,

\[f(-x) = -f(x)\]

Answer 13

thank you very much :)

Answer 14

Welcome dear..

Answer 15

the function is niether odd nor even if it is odd then it should fill the following criteria
f(-x)=-f(x)
but it does not fulfill this criteria so i would say it it is niether odd nor even

Answer 16

thank you sami for the help