How to get started? Consider $\ln x $ in [1, 1+t] (t0), show that $ \frac{x}{x+1}

Question

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

anonymous · Answer

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 · Answer

make sense?

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

anonymous · Answer

(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 · Answer

yeah im finished here :)

anonymous · Answer

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 · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

anonymous · Answer

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

anonymous · Answer

emm...sorry what do u mean?

anonymous · Answer

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 · Answer

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

anonymous · Answer

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

anonymous · Answer

yes :)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

why letting it equal to zero?

anonymous · Answer

Errr, usual practice.

anonymous · Answer

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