Modulus function

Lets say you have y = -|3x - 1|

when working out where it cuts the axis, particularly the x-coordinate you can do the following

when y = 0, 3x - 1 = 0 therefore x = 1/3

but when you have y = 2|x| - 1 and you are trying to work out the x-coordinate

when y = 0, 2|x| - 1 = 0 therefore |x| = -0.5

Why is it that the modulus sign essentially 'disappears' for the first part above and not the second part?

..

If |function| = 0, then, clearly function = 0

That is why it "disappears" in the first one

This is elementary, but I'm not thinking right. So it's essentially the same as -3x = 0, therefore x = 0 (essentially doing the same above, dividing by -1 and the modulus - if that even makes sense?)

Lets say you have y = -|3x - 1|

when working out where it cuts the axis, particularly the x-coordinate you can do the following

when y = 0, 3x - 1 = 0 therefore x = 1/3

but when you have y = 2|x| - 1 and you are trying to work out the x-coordinate

when y = 0, 2|x| - 1 = 0 therefore |x| = -0.5 hence no x value (but why?)

Why is it that the modulus sign essentially 'disappears' for the first part above and not the second part?

This line is not correct

2|x| -1 = 0
2|x|=1
|x|=0.5

Edited, it's meant to be 2|x| + 1 = 0.

Edited, it's meant to be 2|x| + 1 = 0.

Lets say you have y = -|3x - 1|

when working out where it cuts the axis, particularly the x-coordinate you can do the following

when y = 0, 3x - 1 = 0 therefore x = 1/3

but when you have y = 2|x| + 1 and you are trying to work out the x-coordinate

when y = 0, 2|x| + 1 = 0 therefore |x| = -0.5 hence no x value (but why?)

Why is it that the modulus sign essentially 'disappears' for the first part above and not the second part?

So, let's work through the two examples.

Example 1:

y = -|3x-1| To find where y = 0.

Now, with all questions that use the modulus function, it's best if we first sketch the graph.

Okay, in this case, there is only a single solution.
So:


Example 2:

y = 2|x| +1 To find where y = 0.

Again, the best place to start is by sketching the graph. Here, we easily see that 2|x| +1 never equals 0, by why not?

Well, let's work through it:


Both examples you provided were a little dull. So, to make things interesting:

Example 3:
Looking back at example 1, let's use y = -|3x-1|, but this time to find where y = -5.

So, let's sketch the graph. We can now see there are two solutions! Let's do the maths:

This is incorrect. Take the similar function y = 2|x|-5

x = 2 : y = 2|2| - 5 = -1
x = -2 : y = 2|-2| - 5 = -1

You can't just disregard the negative x-axis with f(|x|).

This is incorrect. Take the similar function y = 2|x|-5

x = 2 : y = 2|2| - 5 = -1
x = -2 : y = 2|-2| - 5 = -1

You can't just disregard the negative x-axis with f(|x|).

And besides, if you read what i had written correctly rather than taking a single quote into isolation, you will find that i said it is replaced with a mirror of the positive x axis.