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

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mukushlaBest 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)
 one year ago

RolyPolyBest 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!)
 one year ago

mukushlaBest 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
 one year ago

mukushlaBest 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
 one year ago

RolyPolyBest 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!
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
yeah im finished here :)
 one year ago

RolyPolyBest 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?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
@mukushla Sorry! I'm calling you back :P
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
no problem rolypoly :) yeah thats the domain for which we want prove that inequality
 one year ago

RolyPolyBest 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.
 one year ago

mukushlaBest 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]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
How can we ''just consider the given interval''?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
emm...sorry what do u mean?
 one year ago

RolyPolyBest 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.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
ahh yeah .. when u want to show that f'(x)>0 u must use that
 one year ago

RolyPolyBest 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?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Then, it's bad. Once I put f'(x) = 0, I get x=0
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
u see f'(x)>0 for given interval
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
why letting it equal to zero?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Errr, usual practice.
 one year ago

RolyPolyBest 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
 one year ago
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