Understanding Modulus Equations and Their Applications: A Clarification Guide

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SUMMARY

The discussion clarifies the evaluation of the function f(x) = 1 - |1 - x| on the interval [-2, 2]. It establishes that the absolute value expression |1 - x| changes based on the value of x, specifically at x = 1. For x ≤ 1, |1 - x| simplifies to 1 - x, leading to f(x) = x. Conversely, for x ≥ 1, |1 - x| becomes x - 1, resulting in f(x) = 2 - x. This bifurcation allows for a piecewise definition of the function.

PREREQUISITES
  • Understanding of absolute value functions
  • Familiarity with piecewise functions
  • Basic knowledge of inequalities
  • Graphing functions on a Cartesian plane
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  • Study piecewise function definitions and their applications
  • Learn about absolute value equations and inequalities
  • Explore graphing techniques for piecewise functions
  • Investigate the properties of modulus functions in calculus
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Mathematics students, educators, and anyone seeking to deepen their understanding of modulus equations and their applications in piecewise function analysis.

ElDavidas
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hi

I've been reading through some notes and I can't see where a step comes from.

I understand that

[tex]|x| = \left\{\begin{array}{cc} x,& \mbox{ if } x\geq 0\\-x, & \mbox{ if } x \leq0 \end{array}\right[/tex]

The equation I'm stuck on reads as

[tex]f(x) = 1 - | 1- x | \mbox{on} [-2,2][/tex]

[tex]= \left\{\begin{array}{cc} 2-x,& \mbox{ if } x\geq 1\\x, & \mbox{ if } x \leq 1 \end{array}\right[/tex]

Can someone explain this step?
 
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ElDavidas said:
hi

I've been reading through some notes and I can't see where a step comes from.

I understand that

[tex]|x| = \left\{\begin{array}{cc} x,& \mbox{ if } x\geq 0\\-x, & \mbox{ if } x \leq0 \end{array}\right[/tex]

The equation I'm stuck on reads as

[tex]f(x) = 1 - | 1- x | \mbox{on} [-2,2][/tex]

[tex]= \left\{\begin{array}{cc} 2-x,& \mbox{ if } x\geq 1\\x, & \mbox{ if } x \leq 1 \end{array}\right[/tex]

Can someone explain this step?
Well, when 1-x>=0, then |1-x|=1-x
Thus, when x<=1, we have f(x)=1-|1-x|=1-(1-x)=x
 
ElDavidas said:
hi

I've been reading through some notes and I can't see where a step comes from.

I understand that

[tex]|x| = \left\{\begin{array}{cc} x,& \mbox{ if } x\geq 0\\-x, & \mbox{ if } x \leq0 \end{array}\right[/tex]

The equation I'm stuck on reads as

[tex]f(x) = 1 - | 1- x | \mbox{on} [-2,2][/tex]

[tex]= \left\{\begin{array}{cc} 2-x,& \mbox{ if } x\geq 1\\x, & \mbox{ if } x \leq 1 \end{array}\right[/tex]

Can someone explain this step?
The crucial point about an absolute value (modulus) is to determine when the quantity inside the modulus changes from negative to positive and vice-versa: and that occurs, of course, where it is equal to 0.

Since the quantity inside the absolute value is 1- x, that will be 0 when
1- x= 0 or when x= 1. That means we can write this as two separate functions for x< 1 and x> 1.

If x< 1, 1- x is positive and |1- x|= 1- x.
If x< 1, 1- |1- x|= 1- (1- x)= x.

If x> 1, 1- x is negative and |1- x|= -(1- x)= x- 1. If x> 1, 1- |1- x|=
1-(-(1-x))= 1+ 1- x= 2- x.

If x= 1, of course 1-x= 0 so 1- |1-x|= 1. which is correct for either of the formulas so it doesn't matter where you put the "equals".
 

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