.
The critical points are 6-2x = 0, i.e. x = 3; and x-1=0, i.e. x = 1.
If x < 1, then 6 - 2x > 0 and hence |6-2x} = (6-2x);
x - 1 < 0 and hence |x-1| = (1-x);
thus the entire function is f(x) = (6-2x) + (1-x) - 2x = 7 - 5x.
If 1 < x < 3, then 6 - 2x > 0 and hence |6-2x} = (6-2x);
x - 1 > 0 and hence |x-1| = (x-1);
thus the entire function is f(x) = (6-2x) + (x-1) - 2x = 5 - 3x.
Lastly, if x > 3 then 6 - 2x < 0 and hence |6-2x} = (-6+2x);
x - 1 > 0 and hence |x-1| = (x-1);
thus the entire function is f(x) = (-6+2x) + (x-1) - 2x = -7 + x.
So you have the expressions for f(x) piecewise linear and can draw it like this
Plot y = |6-2x|+|X-1|-2X
On plotting absolute value functions see the lessons
How to plot functions containing Linear Terms under the Absolute Value sign. Lesson 1
How to plot functions containing Linear Terms under the Absolute Value sign. Lesson 2
How to plot functions containing Linear Terms under the Absolute Value sign. Lesson 3
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"Plotting Absolute values functions ".
The strategy is to break up the entire set of real numbers into sub-domains (ranges) where the absolute value of linear term
is a linear function, and then to plot all these piece-wise linear functions.