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HELP!!! C3 modulus function graphs I'm very confused:(( Watch

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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
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    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?
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    (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 😢😢
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    (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.
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    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.
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    (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) ?
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    (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
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    - |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.
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    (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? :-)
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    (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
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    (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.
 
 
 
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