Prove that sqrt(1+x)>1+(x/2) using MVT

- anonymous

Prove that sqrt(1+x)>1+(x/2) using MVT

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- anonymous

should i find f(x) first?

- 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

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

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## More answers

- anonymous

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- anonymous

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- anonymous

im just not sure how to apply the MVT

- tamtoan

square both sides and see what you get .

- 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

x>0

- anonymous

ya i cant square both sides

- tamtoan

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

- anonymous

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

- anonymous

s'ok :}

- anonymous

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- anonymous

f'(c) is strange

- 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

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

- anonymous

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

- anonymous

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- anonymous

this is were i get lost

- anonymous

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

- anonymous

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- anonymous

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- 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

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

- anonymous

so c is a constant

- anonymous

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

- anonymous

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- anonymous

mahmit how do you get that?

- anonymous

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- anonymous

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

- anonymous

MVT

- anonymous

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

- anonymous

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- anonymous

yes.

- anonymous

wow omg, thank you so much!

- anonymous

but instead of > should be<

- anonymous

unless for x<0

- anonymous

well what do i do with this?

- anonymous

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- anonymous

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- anonymous

x>0

- anonymous

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- anonymous

wait how do you just jump from here ?

- anonymous

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- anonymous

@mahmit2012 do you know what happens to this?

- anonymous

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