Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

RolyPoly

  • 2 years ago

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

  • This Question is Closed
  1. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    make sense?

  3. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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!

  7. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah im finished here :)

  8. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  13. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    emm...sorry what do u mean?

  15. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  17. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  18. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes :)

  19. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  20. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  21. mukushla
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    why letting it equal to zero?

  22. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Errr, usual practice.

  23. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  24. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.