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RolyPoly
Group Title
How to get started?
Consider \(\ln x \) in [1, 1+t] (t>0), show that \( \frac{x}{x+1} < \ln (1+x) < x\) .
 2 years ago
 2 years ago
RolyPoly Group Title
How to get started? Consider \(\ln x \) in [1, 1+t] (t>0), show that \( \frac{x}{x+1} < \ln (1+x) < x\) .
 2 years ago
 2 years ago

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mukushla Group TitleBest ResponseYou've already chosen the best response.1
start with defining a function for each part for example\[f(x)=\ln(1+x)\frac{x}{x+1}\]and show that f(x) is strictly increasing (for the other part decreasing)
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
make sense?
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
I'm sorry. I don't quite understand. Why are you defining that function? Or, I should ask what you are doing. (Sorry again!)
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
see u have 2 works to do here one is showing\[\ln(1+x)>\frac{x}{1+x}\]or\[\ln(1+x)\frac{x}{1+x}>0\]so defining\[f(x)=\ln(1+x)\frac{x}{x+1}\]and showing that f(x) is increasing is what u need here
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
other part is showing\[\ln(1+x)<x\]or\[x\ln(1+x)>0\]for this one let\[g(x)=x\ln(1+x)\]and show that g(x) is strictly increasing
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
(Shhh~~~) Would you mind letting me try the second half without giving me the hint? Thanks for your help. I start to understand it!
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
yeah im finished here :)
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
But wait. If I'm doing this way, why do I have to consider lnx in the interval [1, 1+t] for t>0?
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
@mukushla Sorry! I'm calling you back :P
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
no problem rolypoly :) yeah thats the domain for which we want prove that inequality
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Two questions: 1. To show f(x) is increasing, we do differentiation, right? (First derivative) 2. If we just take derivative, we're not really consider the domain, and also lnx.
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
1. right and showing f'(x)>0 2. we should just consider the given interval [1,1+t]
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
How can we ''just consider the given interval''?
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
emm...sorry what do u mean?
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Sorry! You said that we should just consider the given interval. But in the calculation, how can we (show that) we just consider the given interval? Since we're doing it in a ''general'' way.
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
ahh yeah .. when u want to show that f'(x)>0 u must use that
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
\[f'(x) = \frac{1}{1+x}  \frac{(x+1) (x)}{(x+1)^2}=\frac{1}{1+x}\frac{1}{(x+1)^2} = \frac{x}{(x+1)^2}\] Hmm, so far so good?
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Then, it's bad. Once I put f'(x) = 0, I get x=0
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
u see f'(x)>0 for given interval
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
why letting it equal to zero?
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Errr, usual practice.
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Oh wait. For the given interval, f'(x)>0 since 1 is positive and 1+t is also positive for t>0. So, f'(x) >0
 2 years ago
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