## anonymous 3 years ago Prove that sqrt(1+x)>1+(x/2) using MVT

1. anonymous

should i find f(x) first?

2. 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.

3. anonymous

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

4. anonymous

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5. anonymous

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6. anonymous

im just not sure how to apply the MVT

7. tamtoan

square both sides and see what you get .

8. 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).

9. anonymous

x>0

10. anonymous

ya i cant square both sides

11. tamtoan

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

12. anonymous

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

13. anonymous

s'ok :}

14. anonymous

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15. anonymous

f'(c) is strange

16. anonymous

Okay so that means that there exists a c where 0<c<=x: That: $f \prime(c)=\frac{ f(x)-f(0) }{ x-0 }$ Then, $1+c/2-1/(2*\sqrt{1+c})=\frac{ (1+x/2-\sqrt(1+x))-(1-1)}{ x }=\frac{ 1+x/2-\sqrt(1+x) }{ x }$ 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 :)

17. anonymous

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

18. anonymous

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

19. anonymous

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20. anonymous

this is were i get lost

21. anonymous

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

22. anonymous

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23. anonymous

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24. 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

25. anonymous

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

26. anonymous

so c is a constant

27. anonymous

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

28. anonymous

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29. anonymous

mahmit how do you get that?

30. anonymous

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31. anonymous

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

32. anonymous

MVT

33. anonymous

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

34. anonymous

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35. anonymous

yes.

36. anonymous

wow omg, thank you so much!

37. anonymous

but instead of > should be<

38. anonymous

unless for x<0

39. anonymous

well what do i do with this?

40. anonymous

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41. anonymous

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42. anonymous

x>0

43. anonymous

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44. anonymous

wait how do you just jump from here ?

45. anonymous

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46. anonymous

@mahmit2012 do you know what happens to this?

47. anonymous

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