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

Mathematics
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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)
make sense?
I'm sorry. I don't quite understand. Why are you defining that function? Or, I should ask what you are doing. (Sorry again!)

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Other answers:

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
other part is showing\[\ln(1+x)0\]for this one let\[g(x)=x-\ln(1+x)\]and show that g(x) is strictly increasing
(Shhh~~~) Would you mind letting me try the second half without giving me the hint? Thanks for your help. I start to understand it!
yeah im finished here :)
But wait. If I'm doing this way, why do I have to consider lnx in the interval [1, 1+t] for t>0?
@mukushla Sorry! I'm calling you back :P
no problem rolypoly :) yeah thats the domain for which we want prove that inequality
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.
1. right and showing f'(x)>0 2. we should just consider the given interval [1,1+t]
How can we ''just consider the given interval''?
emm...sorry what do u mean?
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.
ahh yeah .. when u want to show that f'(x)>0 u must use that
\[f'(x) = \frac{1}{1+x} - \frac{(x+1) -(x)}{(x+1)^2}=\frac{1}{1+x}-\frac{1}{(x+1)^2} = \frac{x}{(x+1)^2}\] Hmm, so far so good?
yes :)
Then, it's bad. Once I put f'(x) = 0, I get x=0
u see f'(x)>0 for given interval
why letting it equal to zero?
Errr, usual practice.
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

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