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Calle87 Group TitleBest ResponseYou've already chosen the best response.0
should i find f(x) first?
 one year ago

quantum77 Group TitleBest ResponseYou've already chosen the best response.0
This inequality isn't true, are you sure it's not sqrt(1+x)<1+(x/2)? For example if you try x=1 sqrt(1+1)>1+1/2 => sqrt(2)>1.5 is false.
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
yes i wrote it wrong, the sign is the wrong way
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354858780727:dw
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354858978743:dw
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
im just not sure how to apply the MVT
 one year ago

tamtoan Group TitleBest ResponseYou've already chosen the best response.0
square both sides and see what you get .
 one year ago

quantum77 Group TitleBest ResponseYou've already chosen the best response.0
You can square both sides but it says mean value theorem? Does the question give you some lower bound like for all x>0 or x>=1? As the mean value theorem says: for a differentiable function f(x) on the closed interval (a,b) and is continuous on [a,b] then there exists a c where a<=c<=b where: \[f \prime(c)=\frac{ f(b)f(a) }{ ba }\] and you are correct that in this case f(x)=1+(x/2)sqrt(1+x).
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
ya i cant square both sides
 one year ago

tamtoan Group TitleBest ResponseYou've already chosen the best response.0
ty quantum, didn't even know what MVT were :), sorry calle87 :)
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
not sure how to do it with just {a,b} use to having an interval
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354859852306:dw
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
f'(c) is strange
 one year ago

quantum77 Group TitleBest ResponseYou've already chosen the best response.0
Okay so that means that there exists a c where 0<c<=x: That: \[f \prime(c)=\frac{ f(x)f(0) }{ x0 }\] Then, \[1+c/21/(2*\sqrt{1+c})=\frac{ (1+x/2\sqrt(1+x))(11)}{ x }=\frac{ 1+x/2\sqrt(1+x) }{ x }\] Notice that, if c>0: 1/sqrt(1+c)<1 therefore should be able to work out f'(c)>0 for all c>0. Then, \[\frac{ 1+x/2\sqrt(1+x) }\] Since, x>0 the numerator is also greater than 0. Which is the original inequality :)
 one year ago

quantum77 Group TitleBest ResponseYou've already chosen the best response.0
Sorry in the last equation it should have said:\[\frac{ 1+x/2\sqrt(1+x) }{ x }>0\]
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
hmmm, f'(c) why is the one not zero?
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354860607643:dw
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
this is were i get lost
 one year ago

quantum77 Group TitleBest ResponseYou've already chosen the best response.0
Sorry (shoun't be a one). Let me work it out :p
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354860920108:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
dw:1354861141850:dw
 one year ago

quantum77 Group TitleBest ResponseYou've already chosen the best response.0
Haha, we both messed up the derivative... it should be f(c)=1/2*(11/sqrt(1+c)). Which is not hard to show it is greater than 0
 one year ago

quantum77 Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt(1+c)>1 => 1/\sqrt(1+c)<1 => 0.5*(11/\sqrt(1+c))>=0\]
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
so c is a constant
 one year ago

quantum77 Group TitleBest ResponseYou've already chosen the best response.0
c is some constant in the range (0,x)
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
dw:1354861535246:dw
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
mahmit how do you get that?
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354861444641:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
what are you doing? it is solved but with different inequality!!
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
so i can just take sqrt(1+x) as f(x) and apply the MVT to it directly
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
dw:1354861640565:dw
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
wow omg, thank you so much!
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
but instead of > should be<
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
unless for x<0
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
well what do i do with this?
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354861684176:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
dw:1354861815578:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
dw:1354861953258:dw
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
wait how do you just jump from here ?
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354861909373:dw
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
@mahmit2012 do you know what happens to this?
 one year ago

Calle87 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354862035437:dw
 one year ago
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