anonymous
  • anonymous
Prove that sqrt(1+x)>1+(x/2) using MVT
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
should i find f(x) first?
anonymous
  • anonymous
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.
anonymous
  • anonymous
yes i wrote it wrong, the sign is the wrong way

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
|dw:1354858780727:dw|
anonymous
  • anonymous
|dw:1354858978743:dw|
anonymous
  • anonymous
im just not sure how to apply the MVT
tamtoan
  • tamtoan
square both sides and see what you get .
anonymous
  • anonymous
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: \[f \prime(c)=\frac{ f(b)-f(a) }{ b-a }\] and you are correct that in this case f(x)=1+(x/2)-sqrt(1+x).
anonymous
  • anonymous
x>0
anonymous
  • anonymous
ya i cant square both sides
tamtoan
  • tamtoan
ty quantum, didn't even know what MVT were :), sorry calle87 :)
anonymous
  • anonymous
not sure how to do it with just {a,b} use to having an interval
anonymous
  • anonymous
s'ok :}
anonymous
  • anonymous
|dw:1354859852306:dw|
anonymous
  • anonymous
f'(c) is strange
anonymous
  • anonymous
Okay so that means that there exists a c where 00: 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
  • anonymous
Sorry in the last equation it should have said:\[\frac{ 1+x/2-\sqrt(1+x) }{ x }>0\]
anonymous
  • anonymous
hmmm, f'(c) why is the one not zero?
anonymous
  • anonymous
|dw:1354860607643:dw|
anonymous
  • anonymous
this is were i get lost
anonymous
  • anonymous
Sorry (shoun't be a one). Let me work it out :p
anonymous
  • anonymous
|dw:1354860920108:dw|
anonymous
  • anonymous
|dw:1354861141850:dw|
anonymous
  • anonymous
Haha, we both messed up the derivative... it should be f(c)=1/2*(1-1/sqrt(1+c)). Which is not hard to show it is greater than 0
anonymous
  • anonymous
\[\sqrt(1+c)>1 => 1/\sqrt(1+c)<1 => 0.5*(1-1/\sqrt(1+c))>=0\]
anonymous
  • anonymous
so c is a constant
anonymous
  • anonymous
c is some constant in the range (0,x)
anonymous
  • anonymous
|dw:1354861535246:dw|
anonymous
  • anonymous
mahmit how do you get that?
anonymous
  • anonymous
|dw:1354861444641:dw|
anonymous
  • anonymous
what are you doing? it is solved but with different inequality!!
anonymous
  • anonymous
MVT
anonymous
  • anonymous
so i can just take sqrt(1+x) as f(x) and apply the MVT to it directly
anonymous
  • anonymous
|dw:1354861640565:dw|
anonymous
  • anonymous
yes.
anonymous
  • anonymous
wow omg, thank you so much!
anonymous
  • anonymous
but instead of > should be<
anonymous
  • anonymous
unless for x<0
anonymous
  • anonymous
well what do i do with this?
anonymous
  • anonymous
|dw:1354861684176:dw|
anonymous
  • anonymous
|dw:1354861815578:dw|
anonymous
  • anonymous
x>0
anonymous
  • anonymous
|dw:1354861953258:dw|
anonymous
  • anonymous
wait how do you just jump from here ?
anonymous
  • anonymous
|dw:1354861909373:dw|
anonymous
  • anonymous
@mahmit2012 do you know what happens to this?
anonymous
  • anonymous
|dw:1354862035437:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.