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should i find f(x) first?

yes i wrote it wrong, the sign is the wrong way

|dw:1354858780727:dw|

|dw:1354858978743:dw|

im just not sure how to apply the MVT

square both sides and see what you get .

x>0

ya i cant square both sides

ty quantum, didn't even know what MVT were :), sorry calle87 :)

not sure how to do it with just {a,b} use to having an interval

s'ok :}

|dw:1354859852306:dw|

f'(c) is strange

Sorry in the last equation it should have said:\[\frac{ 1+x/2-\sqrt(1+x) }{ x }>0\]

hmmm, f'(c) why is the one not zero?

|dw:1354860607643:dw|

this is were i get lost

Sorry (shoun't be a one). Let me work it out :p

|dw:1354860920108:dw|

|dw:1354861141850:dw|

\[\sqrt(1+c)>1 => 1/\sqrt(1+c)<1 => 0.5*(1-1/\sqrt(1+c))>=0\]

so c is a constant

c is some constant in the range (0,x)

|dw:1354861535246:dw|

mahmit how do you get that?

|dw:1354861444641:dw|

what are you doing? it is solved but with different inequality!!

MVT

so i can just take sqrt(1+x) as f(x) and apply the MVT to it directly

|dw:1354861640565:dw|

yes.

wow omg, thank you so much!

but instead of > should be<

unless for x<0

well what do i do with this?

|dw:1354861684176:dw|

|dw:1354861815578:dw|

x>0

|dw:1354861953258:dw|

wait how do you just jump from here ?

|dw:1354861909373:dw|

@mahmit2012 do you know what happens to this?

|dw:1354862035437:dw|