Turn on thread page Beta

What is the domain and range of f(x) = 1/ln(x^2)? watch

    • Thread Starter
    Offline

    15
    ReputationRep:
    any help is appreciated
    Offline

    19
    ReputationRep:
    (Original post by zezno)
    any help is appreciated
    Well typically domain should be given...but I assume it just wants you to write which values the function can be defined on. Well, remember that we can't divide by 0, and that lnx is defined for x > 0. As for the range, for what values of k can you solve the equation 1/ln(x^2) = k?
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by 13 1 20 8 42)
    Well typically domain should be given...but I assume it just wants you to write which values the function can be defined on. Well, remember that we can't divide by 0, and that lnx is defined for x > 0. As for the range, for what values of k can you solve the equation 1/ln(x^2) = k?
    the Question comes with the domain x is equal to or greater than 0.

    My thought process:

    The denominator can't be 0 as you said.

    so x^2 > 0

    so would the domain be (-infinity, 0), (0, infinity)?

    Would the range be (0, infinity)? no idea lol
    Offline

    2
    ReputationRep:
    I believe the domain would be x≠1, as that would give ln(1), which equals 0 and would leave f(x) undefined.
    I think the range would be f(x) > 0.
    Offline

    19
    ReputationRep:
    (Original post by zezno)
    the Question comes with the domain x is equal to or greater than 0.

    My thought process:

    The denominator can't be 0 as you said.

    so x^2 > 0

    so would the domain be (-infinity, 0), (0, infinity)?

    Would the range be (0, infinity)? no idea lol
    Well, for one, if the question gives that x >/= 0, then you can't include anything from (-infinity, 0). But otherwise, we have ln(1) = 0, not ln(0) = 0 (though of course ln0 is undefined anyhow so 0 can't be included for that reason..)

    Well, the range should contain that interval, but remember ln can be negative as well.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by 13 1 20 8 42)
    Well, for one, if the question gives that x >/= 0, then you can't include anything from (-infinity, 0). But otherwise, we have ln(1) = 0, not ln(0) = 0 (though of course ln0 is undefined anyhow so 0 can't be included for that reason..)

    Well, the range should contain that interval, but remember ln can be negative as well.
    so what would be the domain and range? My head hurts lol
    Offline

    19
    ReputationRep:
    (Original post by zezno)
    so what would be the domain and range? My head hurts lol
    Okay, whatever. So if the question indeed gives x >/=0, you can't have x = 0, or x = 1. Those are your only restrictions.
    As for 1/ln(x^2); well, suppose 1/ln(x^2) = k where k is a real number. Then we can solve; we get ln(x^2) = 1/k, so x^2 = e^(1/k), and then x = e^(1/2k). Of course, this doesn't work when k = 0. But otherwise it works, right, be k negative or positive. So that should tell you the range.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by 13 1 20 8 42)
    Okay, whatever. So if the question indeed gives x >/=0, you can't have x = 0, or x = 1. Those are your only restrictions.
    As for 1/ln(x^2); well, suppose 1/ln(x^2) = k where k is a real number. Then we can solve; we get ln(x^2) = 1/k, so x^2 = e^(1/k), and then x = e^(1/2k). Of course, this doesn't work when k = 0. But otherwise it works, right, be k negative or positive. So that should tell you the range.
    ln(x^2) cannot be equal to 0

    therefore x^2 =e^0
    therefore x^2 = 1
    so x = +- root 1
    so x cannot be equal to +-1



    So I think the domain will be (0, infinity), but x cannot be equal to +-1 ?

    please could you tell me if this is correct and what the range is, it's my first time learning this so I'm getting confused a lot
    Offline

    19
    ReputationRep:
    (Original post by zezno)
    ln(x^2) cannot be equal to 0

    therefore x^2 =e^0
    therefore x^2 = 1
    so x = +- root 1
    so x cannot be equal to +-1



    So I think the domain will be (0, infinity), but x cannot be equal to +-1 ?

    please could you tell me if this is correct and what the range is, it's my first time learning this so I'm getting confused a lot
    Yes, if x has to be greater than or equal to zero, then indeed the domain is (0, infinity) but missing out 1 (note that you don't need to say that x =/= -1, because -1 is not in this interval)

    As for the range, I mean, you can always solve the equation 1/ln(x^2) = k for k =/= 0 right? So the range is just all real numbers apart from zero.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by 13 1 20 8 42)
    Yes, if x has to be greater than or equal to zero, then indeed the domain is (0, infinity) but missing out 1 (note that you don't need to say that x =/= -1, because -1 is not in this interval)

    As for the range, I mean, you can always solve the equation 1/ln(x^2) = k for k =/= 0 right? So the range is just all real numbers apart from zero.
    thanks for all your help!


    I can just write the range as (-infinity, infinity) except for 0 right? as this is writing all real numbers
    Offline

    19
    ReputationRep:
    (Original post by zezno)
    thanks for all your help!


    I can just write the range as (-infinity, infinity) except for 0 right? as this is writing all real numbers
    Yeah. The classy way is R \ {0} or (-infinity, infinity) \ {0}
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by 13 1 20 8 42)
    Yeah. The classy way is R \ {0} or (-infinity, infinity) \ {0}
    Now for this question:

    x^-1>0
    (x-1)(x+1)>0

    So would the domain be:

    (-infin, -1), (1, infin)

    should I also include x=0 in the domain because of the numerator?

    also, what would the range be?
    Attached Images
     
    Offline

    15
    ReputationRep:
    (Original post by 13 1 20 8 42)
    Yeah. The classy way is R \ {0} or (-infinity, infinity) \ {0}
     \mathbb{R}\setminus \{0\} .:cool::cool::cool:
 
 
 
The home of Results and Clearing

3,037

people online now

1,567,000

students helped last year

University open days

  1. Sheffield Hallam University
    City Campus Undergraduate
    Tue, 21 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Wed, 22 Aug '18
  3. University of Buckingham
    Postgraduate Open Evening Postgraduate
    Thu, 23 Aug '18
Poll
How are you feeling about GCSE results day?
Help with your A-levels

All the essentials

The adventure begins mug

Student life: what to expect

What it's really like going to uni

Rosette

Essay expert

Learn to write like a pro with our ultimate essay guide.

Uni match

Uni match

Our tool will help you find the perfect course for you

Study planner

Create a study plan

Get your head around what you need to do and when with the study planner tool.

Study planner

Resources by subject

Everything from mind maps to class notes.

Hands typing

Degrees without fees

Discover more about degree-level apprenticeships.

A student doing homework

Study tips from A* students

Students who got top grades in their A-levels share their secrets

Study help links and info

Can you help? Study help unanswered threadsRules and posting guidelines

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.