## RolyPoly Group Title 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

1. mukushla Group Title

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. mukushla Group Title

make sense?

3. RolyPoly Group Title

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. mukushla Group Title

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. mukushla Group Title

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. RolyPoly Group Title

(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. mukushla Group Title

yeah im finished here :)

8. RolyPoly Group Title

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. RolyPoly Group Title

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

10. mukushla Group Title

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

11. RolyPoly Group Title

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. mukushla Group Title

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

13. RolyPoly Group Title

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

14. mukushla Group Title

emm...sorry what do u mean?

15. RolyPoly Group Title

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. mukushla Group Title

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

17. RolyPoly Group Title

$f'(x) = \frac{1}{1+x} - \frac{(x+1) -(x)}{(x+1)^2}=\frac{1}{1+x}-\frac{1}{(x+1)^2} = \frac{x}{(x+1)^2}$ Hmm, so far so good?

18. mukushla Group Title

yes :)

19. RolyPoly Group Title

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

20. mukushla Group Title

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

21. mukushla Group Title

why letting it equal to zero?

22. RolyPoly Group Title

Errr, usual practice.

23. RolyPoly Group Title

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