Modulus: y=|f(x)| & y=f|(x)|

Ricochet J · 16 Dec 2006 at 13:40

I understand the stuff, I just need to know why! It's not a h/w thread! It's a "Why is it?" thread!

Right, i'm at home revising for Core 3 Maths A2 Level! Just going through some modulus stuff as I always loose marks there.

I understand that when

y=|f(x)|
x>0 drawing y=|f(x)| is the same as drawing y=f(x)
x<0 drawing y=|f(x)| is the same as drawing y=-f(x), hence a relection in the x-axis

y=f|(x)|
x>0 drawing y=f|(x)| is the same as drawing y=f(x)
x<0 drawing y=f|(x)| is the same as drawing y=f(-x), hence a reflection in the y-axis

Now, that's all jolly good. I understand that, however, when i'm drawing the function it always helps me to draw in two parts. Let me give you an example.

Example
f(x)=3x-2
If I draw y=|f(x)| the drawing will look like y=3x-2 and y=-(3x-2) combined. Like so:

HOWEVER, If I draw y=f|(x)| the drawing will look like y=3x-2 and ??WHAT?? combined.
Like so:

All I need to know is what the hell is my missing function!

I can still do it without knowing it, but it will make my life A HELL of a lot more easier if I knew how to figure it out! And it's not just that example. I want to know how to work it out for any modulus question

I'm sure it's something easy!

Help

rdizzle · 16 Dec 2006 at 13:45

surely -3x-2

mayb not tho, not done that in a while.

Ricochet J · 16 Dec 2006 at 13:47

rdizzle said:
surely -3x-2

mayb not tho, not done that in a while.

Actually, I think you're right!

rdizzle · 16 Dec 2006 at 13:47

glad i could help

touch · 16 Dec 2006 at 13:49

-3x+2 i believe

Ricochet J · 16 Dec 2006 at 13:52

touch said:
-3x+2 i believe

-3x+2 would be y=|f(x)|

I'm sure that's not it. However, I reckon it's -3x-2

touch · 16 Dec 2006 at 13:55

eXSBass said:
If I draw y=|f(x)| the drawing will look like y=3x-2 and y=-(3x-2) combined.

y=-(3x-2) != -3x -2

y=-(3x-2) = -3x+2 (if you say here that y=|f(x)| = -3x+2, surely y=f|(x)| cannot be the same?)

y = |f(x)| = -3x+2 (!= f|(x)|)

or am i completely wrong?

Ricochet J · 16 Dec 2006 at 14:00

touch said:
yeh, think im wrong

Nope, I said

"If I draw y=|f(x)| the drawing will look like y=3x-2 and y=-(3x-2) combined"

Edit: I'm going to revise transformations. Maybe that'll help.

Thanks for replies.

But, i'll be back in an hour to see if I get anymore replies

touch · 16 Dec 2006 at 14:01

lol, dont mind me, i'v read it over a few times and ninja-edited my comment about 10 times and have completely confused myself

Psiko · 16 Dec 2006 at 15:03

eXSBass said:
I understand that when

y=|f(x)|
x>0 drawing y=|f(x)| is the same as drawing y=f(x)
x<0 drawing y=|f(x)| is the same as drawing y=-f(x), hence a relection in the x-axis

This is not true, although you have done the sketch correctly.

f(x) > 0, then y = |f(x)| is y = f(x)
f(x) < 0, then y = |f(x)| is y = -f(x).

So, to obtain this graph from the graph of f(x), keep the top half (y>0) the same, and reflect the bottom half (y<0) in the x-axis, which is what you have done.

For y = f|(x)|, just draw y = f(x) for x>0 and then reflect this in the y-axis, to obtain the negative side. I'm not sure how you came to your sketch. For this example, y = f|(x)| = 3|x| - 2. This is just a |x| graph which has been stretched in the y direction and translated in the y-direction, so it will look exactly like a normal modulus graph, that has been stretched a bit and moved round.