Hey there! Sign in to join this conversationNew here? Join for free

Modulus Function Question Watch

Announcements
    • Thread Starter
    Offline

    2
    ReputationRep:
    Name:  image.jpeg
Views: 73
Size:  434.7 KB
    Don't understand the question ☹️
    (Ignore bottom question).
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (Original post by TSRforum)
    Name:  image.jpeg
Views: 73
Size:  434.7 KB
    Don't understand the question ☹️
    (Ignore bottom question).
    A different example which will give you a hint:

    x^2 is always positive so | x^2 | is identical to x^2 since taking the absolute value has no effect for a positive number.

    Now what can you say about 2x - 1 when x > 0.5 ?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by notnek)
    A different example which will give you a hint:

    x^2 is always positive so | x^2 | is identical to x^2 since taking the absolute value has no effect for a positive number.

    Now what can you say about 2x - 1 when x > 0.5 ?
    Y is greater than zero?

    Still don't understand ..


    Posted from TSR Mobile
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (Original post by TSRforum)
    Y is greater than zero?

    Still don't understand ..


    Posted from TSR Mobile
    When x > 0.5, 2x - 1 is always positive (check this).

    So | 2x - 1 | is the same as 2x - 1.

    S y = x + | 2x - 1 | is the same as _____________
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by notnek)
    When x > 0.5, 2x - 1 is always positive (check this).

    So | 2x - 1 | is the same as 2x - 1.

    S y = x + | 2x - 1 | is the same as _____________
    3x-1?


    Posted from TSR Mobile
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    Correct.
    Offline

    15
    ReputationRep:
    When x>1/2, |2x-1|=2x-1
    When x<1/2, |2x-1|=1-2x
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by B_9710)
    When x>1/2, |2x-1|=2x-1
    When x<1/2, |2x-1|=1-2x
    What about this question:

    Find the solution of eq. |x-3| + |x+3| = 6

    (Requires sketching y=|x|, y=|x-3| and y= |x-3| + |x+3|)


    Posted from TSR Mobile
    Offline

    15
    ReputationRep:
    (Original post by TSRforum)
    What about this question:

    Find the solution of eq. |x-3| + |x+3| = 6

    (Requires sketching y=|x|, y=|x-3| and y= |x-3| + |x+3|)


    Posted from TSR Mobile
    Split it up. Consider what happens to the modulus functions when x>3, when x<-3, when -3<x<3.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by B_9710)
    Split it up. Consider what happens to the modulus functions when x>3, when x<-3, when -3<x<3.
    Don't understand what you mean...


    Posted from TSR Mobile
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by TSRforum)
    Don't understand what you mean...


    Posted from TSR Mobile
    Also how would I go about sketching y= |x-3| + |x+3|?


    Posted from TSR Mobile
    Offline

    17
    ReputationRep:
    (Original post by TSRforum)
    Name:  image.jpeg
Views: 73
Size:  434.7 KB
    Don't understand the question ☹️
    (Ignore bottom question).
    |x|= x for x>=0 pr -x for x<0
    When is 2x-1>=0 or <0?


    Posted from TSR Mobile
    Offline

    17
    ReputationRep:
    (Original post by TSRforum)
    Also how would I go about sketching y= |x-3| + |x+3|?


    Posted from TSR Mobile
    Split between regions x>3, x<-3 and -3<x<3


    Posted from TSR Mobile
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by physicsmaths)
    |x|= x for x>=0 pr -x for x<0
    When is 2x-1>=0 or <0?


    Posted from TSR Mobile
    Already done the question 🙂
    Need help with the second question I posted


    Posted from TSR Mobile
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by physicsmaths)
    Split between regions x>3, x<-3 and -3<x<3


    Posted from TSR Mobile
    Don't understand...


    Posted from TSR Mobile
    Offline

    15
    ReputationRep:
    (Original post by TSRforum)
    Don't understand...


    Posted from TSR Mobile
    When x>3, |x-3|=x-3 and |x+3|=x+3.
    Use this logic to show what happens to each of the modulus functions in the 3 regions I said earlier.
    Make sense or still not sure?
    Offline

    1
    ReputationRep:
    (Original post by B_9710)
    When x>3, |x-3|=x-3 and |x+3|=x+3.
    Use this logic to show what happens to each of the modulus functions in the 3 regions I said earlier.
    Make sense or still not sure?
    You see you. I like you. I give you rep and follow. Continue being pure hearted
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Should Spain allow Catalonia to declare independence?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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