Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    13
    ReputationRep:
    Name:  image.jpeg
Views: 34
Size:  61.4 KB Okay guys, I'm really confused to the answer of q4b, why is it a reflection in the y axis?? And why is the shape symmetrical?? I would've thought the left side of the axis (for negative values of X) would be more of a shape similar to the original but a reflection in the X axis as we are considering the modulus of the function . And why are the coordinates of X 1 and -1 shouldn't it be 1 and -3/2 (if you multiply eveything by sf 1/2) ??

    Edit: the question is asking for f(|2X|) just to clarify as it looks a bit blurred
    Online

    10
    ReputationRep:
    Note that:

     \lvert x \rvert = \begin{cases} x, &  \text{if} \,  x \geq 0,\\ 

-x & \text{if} \, x < 0 \end{cases}.

    Therefore for:
    x = -1, f(|x|) = f(|-1|) = f(1),

    x= -2, f(|-2|) = f(2),

    etc.

    Can you see why we have a symmetric graph?
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by simon0)
    Note that:

     \lvert x \rvert = \begin{cases} x, &  \text{if} \,  x \geq 0,\\ 

-x & \text{if} \, x < 0 \end{cases}.

    Therefore for:
    x = -1, f(|x|) = f(|-1|) = f(1),

    x= -2, f(|-2|) = f(2),

    etc.

    Can you see why we have a symmetric graph?

    Yes thank you! It's because all the X values in the negatives will have the same y values as to the ones on the positives so it's a reflection in the Y axis? but now I'm confused on why y=|f(x)| is reflected in the X axis ? What is the difference between them two ? I'm getting mixed up 😢😢
    Online

    10
    ReputationRep:
    (Original post by nadiakms)
    ...It's because all the X values in the negatives will have the y value as to the ones on the positives so it's a reflection in the Y axis?
    Yes.


    (Original post by nadiakms)
    ...but now I'm confused on why y=|f(x)| is reflected in the X axis ? What is the difference between them two ?
    There is a difference between f(|x|) and |f(x)|.

    Note |a| is the "absolute value" of a, so we are only interested in the value of a rather than the sign of a.

    For f(|x|), f takes the absolute value of x.
    (So f(|-2|) = f(2), f(|5|) = f(5) ).

    For |f(x)|, this is the absolute value of y.
    Online

    10
    ReputationRep:
    Compare the following for f(x) = sin(x).

    - sin(|x|)
    Note we take the sin of the absolute value of x, f can still be negative:

     \sin( \lvert - \pi / 2 \rvert ) \, = \, \sin( \pi /2 ) = 1,

     \sin( \lvert - \pi \rvert ) = \sin( \pi ),

     \sin( \lvert -3 \pi / 2 \rvert ) = \sin( 3 \pi / 2 ) = -1.
    Attached Images
     
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by simon0)
    Yes.




    There is a difference between f(|x|) and |f(x)|.

    Note |a| is the "absolute value" of a, so we are only interested in the value of a rather than the sign of a.

    For f(|x|), f takes the absolute value of x.
    (So f(|-2|) = f(2), f(|5|) = f(5).

    For |f(x)|, this is the absolute value of y.
    Thank youuuuu😭💕 So are you saying with |f(x)| you work out the y value as its dependent on x then you take the modulus, which will tell you the absolute value of y but with f(|x|) we want the absolute value of x so we take the modulus of that only, modulus measures the magnitude so all X will be positive which is essentially why the graph is a reflection as mentioned before, Is my understanding correct? Also is this anyway related to graph transformations like the y=-f(X) and y=f(-X) ?
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by simon0)
    Compare the following for f(x) = sin(x).

    - sin(|x|)
    Note we take the sin of the absolute value of x, f can still be negative:

     \sin( \lvert - \pi / 2 \rvert ) \, = \, \sin( \pi /2 ) = 1,

     \sin( \lvert - \pi \rvert ) = \sin( \pi ),

     \sin( \lvert -3 \pi / 2 \rvert ) = \sin( 3 \pi / 2 ) = -1.
    I'm confused on the last one, why is it still a negative even after we've taken the modulus
    Online

    10
    ReputationRep:
    - |f(x)| = |sin(x) |
    Note here we take the absolute value of sin(x).

    So here negative values of sin(x) are taken to be positive.

    So:

     \lvert \sin( \pi / 2 ) \rvert  = \lvert 1 \rvert = 1,

     \lvert \sin ( 3\pi / 2) \rvert = \lvert -1 \rvert = 1.
    Attached Images
     
    Online

    10
    ReputationRep:
    (Original post by nadiakms)
    I'm confused on the last one, why is it still a negative even after we've taken the modulus
    The function is: sin(|x|).

    We are told here to take the absolute value of x, that is all.

    -----------------------------------------------------------

    In this case:   \lvert x_{0} \rvert = \lvert -3 \pi / 2 \rvert = 3 \pi / 2.

    Therefore  \sin( \lvert x_{0} \rvert ) = \sin( 3 \pi / 2) = -1.

    (Note, different from:  | \sin( x_{0} ) | , as  \sin(x_{0}) = \sin( - 3 \pi / 2 ) = 1.
    Therefore  | \sin( x_{0} ) | = | 1 | = 1 ).

    Can you see the difference now? :-)
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by simon0)
    The function is: sin(|x|).

    We are told here to take the absolute value of x, that is all.

    -----------------------------------------------------------

    In this case:   \lvert x_{0} \rvert = \lvert - 3 \pi / 2 \rvert = 3 \pi / 2. .

    Therefore  \sin( \lvert x_{0} \rvert ) = \sin( \lvert 3 \pi / 2 \rvert )  = \sin( 3 \pi / 2) = -1.

    Can you see the difference now? :-)
    Thank you ever so much! I can finally sleep in peace
    Online

    10
    ReputationRep:
    (Original post by nadiakms)
    Thank youuuuu😭💕 So are you saying with |f(x)| you work out the y value as its dependent on x then you take the modulus, which will tell you the absolute value of y but with f(|x|) we want the absolute value of x so we take the modulus of that only, modulus measures the magnitude so all X will be positive which is essentially why the graph is a reflection as mentioned before, Is my understanding correct?
    Yes.

    (Original post by nadiakms)
    Also is this anyway related to graph transformations like the y=-f(X) and y=f(-X) ?
    Similar concept in regards that "f(-x)" and "-f(x)" are different.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
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

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.