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

Note that:

$\lvert x \rvert = \begin{cases} x, & \text{if} \, x \geq 0,\\ [br]-x & \text{if} \, x < 0 \end{cases}.$

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?

...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?

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

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.

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

The function is: sin(|x|).

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

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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 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) ?