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anonymous
 4 years ago
Prove that sqrt(1+x)>1+(x/2) using MVT
anonymous
 4 years ago
Prove that sqrt(1+x)>1+(x/2) using MVT

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0should i find f(x) first?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i wrote it wrong, the sign is the wrong way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354858780727:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354858978743:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im just not sure how to apply the MVT

tamtoan
 4 years ago
Best ResponseYou've already chosen the best response.0square both sides and see what you get .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya i cant square both sides

tamtoan
 4 years ago
Best ResponseYou've already chosen the best response.0ty quantum, didn't even know what MVT were :), sorry calle87 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not sure how to do it with just {a,b} use to having an interval

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354859852306:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay 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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry in the last equation it should have said:\[\frac{ 1+x/2\sqrt(1+x) }{ x }>0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmmm, f'(c) why is the one not zero?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354860607643:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is were i get lost

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry (shoun't be a one). Let me work it out :p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354860920108:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861141850:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt(1+c)>1 => 1/\sqrt(1+c)<1 => 0.5*(11/\sqrt(1+c))>=0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0c is some constant in the range (0,x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861535246:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mahmit how do you get that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861444641:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what are you doing? it is solved but with different inequality!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i can just take sqrt(1+x) as f(x) and apply the MVT to it directly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861640565:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wow omg, thank you so much!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but instead of > should be<

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well what do i do with this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861684176:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861815578:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861953258:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait how do you just jump from here ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354861909373:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012 do you know what happens to this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1354862035437:dw
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