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

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mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1start 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)

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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1see 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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1other 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

RolyPoly
 3 years ago
Best 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!

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1yeah im finished here :)

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

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

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

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0Two 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.

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.11. right and showing f'(x)>0 2. we should just consider the given interval [1,1+t]

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

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

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry! 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.

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

RolyPoly
 3 years ago
Best 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?

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

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

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

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0Oh 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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