Show that f(x) = log(x+√x^2 +1 ) is an odd function .
Show \(f(-x)=-f(x)\)
Do you know about the Rationalizing of Denominator??
just put -x in place of x
Guys we have to solve it further it is not easy as you are thinking.. We have to prove by rationalization process..
yeah i know ..
so what is the final answer ? .. can any one explain it more clearly .. please ?
Find -f(x) first : \(-f(x) = -\log(x + \sqrt{x^2 + 1})\) --------1
Now find \(f(-x)\)..
okay
\[f(-x) = \log(-x + \sqrt{x^2 + 1})\] Now we have to prove they are equal..
.. okay
\[-f(x) = -\log(x + \sqrt{x^2 + 1}) = \log(x + \sqrt{x^2 + 1})^{-1} = \log(\frac{1}{x + \sqrt{x^2 + 1}})\] \[\implies \log(\frac{1}{x + \sqrt{x^2 + 1}} \times \frac{x - \sqrt{x^2 +1}}{x - \sqrt{x^2 +1}})\] \[\implies \log(\frac{x - \sqrt{x^2 + 1}}{x^2 - x^2 - 1}) \implies \log(-x + \sqrt{x^2 + 1}) \implies f(-x)\] So, \[f(-x) = -f(x)\]
thank you very much :)
Welcome dear..
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
thank you sami for the help
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