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