Question

How to use MVT to show for any n>0\(\dfrac{n}{2n+1}<\sqrt{n^2+n}-n <\dfrac{1}{2}\)
Please, help

Answer 1

n >0, 2n+1>0, hence the far left >0, so that we have interval (0,1/2)

Answer 2

To apply MVT, we just show that \(\sqrt{n^2+n}-n >\dfrac{n}{2n+1}\) in (0,1/2)

Answer 3

i'm very short in time let's see

Answer 4

so did you show that part?

Answer 5

try to define a function where x>0
\[f(x)=\sqrt{x^2+x}-x-1/2\]

Answer 6

you need to show \[f(x)=f(0)+f'(c)x\leq0\]

Answer 7

I would like to know the middle part

Answer 8

that's for max
min do another one

Answer 9

oh, got you.

Answer 10

that would show min<blah<max
which is exactly what the MVT is about

Answer 11

there may be couple of other ways to do this
but go for the simplest hehe

Answer 12

in your calc level
they would only say few stuff and bam you got the result
but that's advanced for me hehe

Answer 13

ok, thanks for help. Let me try.

Answer 14

ok!
no problem
i'm still here for a few mins
let's see if that works

Answer 15

You can go, I need time to observe it and to arrange the stuff neatly. If I miss a small mistake, my credit is out. That is my prof's policy. hehehe

Answer 16

*make, not miss

Answer 17

haha that's a tough prof :)
i would hate him for that, unfair haha

Answer 18

Nope, I love him, he trained me to be a good student who knows how to put the stuff flawless.

Answer 19

eh okay! may be he works for you after all :)

Answer 20

I got a C last semester with him. hehehe... but still love him.

Answer 21

haha i can't judge until i'm part of his lectures you know better than me :)

Answer 22

I like the kind of profs like MIT people
you see the prof for calc he is awesome that is if you watch some mit videos

Answer 23

i let you finish this friend :)
good luck

Answer 24

ty

Answer 25

I don't think it works on this way :(
My attempt:
Let \(f(x) = \sqrt{x^2+x}-x -\dfrac{1}{2}\)
we need show \(f(x) = f(0) +f'(c) x < 0\)
\(f(0) = -\dfrac{1}{2}\) and \(f'(x) =\dfrac{2x+1}{2\sqrt{x^2+x}}-1\)
hence \(f(x) =-\dfrac{3}{2}+\dfrac{2c+1}{2\sqrt{c^2+c}}x\)
This is an equation of a line with the slope >0 , it can't be <0 if x >0

Answer 26

You made an error I guess

Answer 27

point out, please.

Answer 28

You didn't use 0<c<x for the interval
Also how did you get - 3/2 check that again

Answer 29

Let me close and post it again. I messed it up. :)

Answer 30

yea, you don't really want my help here loser, my googleing would be the only saving grace