Diyadiya
  • Diyadiya
The given function \[f(x)= \log (x+ \sqrt{x^2+1})\] is an even function ? odd function ? or a periodic function ?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
do you know how to test if function is even odd?
Diyadiya
  • Diyadiya
Yeah f(x)= f(-x) Even f(x)=f(x) Odd
anonymous
  • anonymous
if you just put x=1 and x=-1 and check with calculator results are same just one positive and other negative so it's odd, but idk how to show with x :/

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More answers

anonymous
  • anonymous
\[f(-x)=\log(-x+\sqrt{x^2+1})\]
anonymous
  • anonymous
@ishaan94 knows :D
Diyadiya
  • Diyadiya
Ishaan's not here !
anonymous
  • anonymous
you need to show that \[\log(-x+\sqrt{x^2+1})=-\log(x+\sqrt{x^2+1})\] but i don't know how :/
Diyadiya
  • Diyadiya
okay :)
anonymous
  • anonymous
None :/
apoorvk
  • apoorvk
any even function is that, where f(x)=f(-x), that is, basically, the function takes same values for any particular 'x' as well as '-x' so, the function's graph is symmetrical about the Y-axis. something like this... (considering any arbitrary function) |dw:1332938306166:dw| so you may have guessed what you have to do in here.. tell me if i need to explain more.
anonymous
  • anonymous
it's odd because http://www.wolframalpha.com/input/?i=-1*log+%28x%2B+%5Csqrt%7Bx%5E2%2B1%7D%29%3Dlog+%28-x%2B+%5Csqrt%7B%28-x%29%5E2%2B1%7D%29 but I don't know how to solve it myself
anonymous
  • anonymous
wait is \[\log(x - \sqrt{x^2+1}) =\log(-x + \sqrt{x^2+1})\] If it's then it must be odd
anonymous
  • anonymous
\[-\log (x + \sqrt{x^2 + 1}) \implies\log \frac{1}{(x + \sqrt{x^2+1})} \implies \log \frac{x - \sqrt{x^2 + 1}}{-1}\]
anonymous
  • anonymous
wait it's odd
anonymous
  • anonymous
something is wrong with my head :/
anonymous
  • anonymous
or was
anonymous
  • anonymous
oh..a mistake it should be \(\log \frac{(x - \sqrt{x^2+1})}{1}\)
anonymous
  • anonymous
it's odd, ODD, OOOOODDDDDD
apoorvk
  • apoorvk
for any odd function the graph of the same function would be symmetrical about the line y=x. this means that the functional value of at any 'x' is the negative of that at '-x'. so for the same arbitrary function as in prev. example, the graph for an odd function would be something like this: |dw:1332938696774:dw|
apoorvk
  • apoorvk
and in a periodic function, the function has a period in 'x' after which the graph repeats itself. the best example would be sinx. |dw:1332938943651:dw| now your case, in the question, logarithmic functions are genrally NOT periodic, so it would be safe to start testing it for even or odd..
anonymous
  • anonymous
you worked hard apoorv, too bad you won't get a kiss :troll: +1, for a nice and brief explanation :-)
apoorvk
  • apoorvk
LOL... "kiss". who wanted that in the first place? :p and "brief"?? err... i sense heavy sarcasm. :P
apoorvk
  • apoorvk
ODD.. it is yeah... (i didn't see that i had forgotten to post the actual answer).
anonymous
  • anonymous
I do, but I never got one, it's a virtual platform :/ No no, no sarcasm, you did it nicely, with beautiful graphs... it's so hard to draw graphs in here :/
apoorvk
  • apoorvk
oh-kay thanks. and i ll post in the feedback forums to facilitate a "kiss" option instead of medals for those who want that! :D
Diyadiya
  • Diyadiya
first of all thanks for the explaination @apoorvk but i didn't understand how you solved it @Ishaan94
anonymous
  • anonymous
RAAAAAATIONAALIZAAAAATION...
apoorvk
  • apoorvk
PS- in the first two graphs, the right hand portion has been eclipsed, courtesy faulty diagram posting. so a NOTE that the graph's in both case are asymptotics towards zero, as x approaches infinity.
Diyadiya
  • Diyadiya
\[\log \frac{1}{(x + \sqrt{x^2+1})}\] how did u get this?
anonymous
  • anonymous
tch...tch...\[-\log(x + \sqrt{x^2 + 1}) =-1*\log(x + \sqrt{x^2 + 1})= \log(x + \sqrt{x^2 + 1})^{-1}\]
apoorvk
  • apoorvk
this is how!! |dw:1332939905757:dw|
apoorvk
  • apoorvk
LOL.. "tch tch."
Diyadiya
  • Diyadiya
Okay !! got it Thank youu :D @apoorvk @Ishaan94

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