## anonymous 4 years ago How to get started? Consider $$\ln x$$ in [1, 1+t] (t>0), show that $$\frac{x}{x+1} < \ln (1+x) < x$$ .

1. 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)

2. anonymous

make sense?

3. 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!)

4. 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

5. anonymous

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

6. 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!

7. anonymous

yeah im finished here :)

8. 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?

9. anonymous

@mukushla Sorry! I'm calling you back :P

10. anonymous

no problem rolypoly :) yeah thats the domain for which we want prove that inequality

11. 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.

12. anonymous

1. right and showing f'(x)>0 2. we should just consider the given interval [1,1+t]

13. anonymous

How can we ''just consider the given interval''?

14. anonymous

emm...sorry what do u mean?

15. 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.

16. anonymous

ahh yeah .. when u want to show that f'(x)>0 u must use that

17. 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?

18. anonymous

yes :)

19. anonymous

Then, it's bad. Once I put f'(x) = 0, I get x=0

20. anonymous

u see f'(x)>0 for given interval

21. anonymous

why letting it equal to zero?

22. anonymous

Errr, usual practice.

23. 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