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Mathematics Dictionary
Dr. K. G. Shih
Absolute Operation


  • Q01 | - Solve abs(x-2) = 3
  • Q02 | - Sketch graph of y = abs(x-1) + abs(x+1)
  • Q03 | - Solve abs(x-2) + abs(x+3) = 6
  • Q04 | - Study graph of y = abs(x+2) + abs(x+3)
  • Q05 | - Solve abs(x - 1) + abs(x + 1) = c
  • Q06 | - Diagram :


Q01. Solve abs(x-2) = 3

Procedures as below :
  • It is true (x-2) = 3
  • Hence x = 5
  • It is also true (x-2) = -3
  • Hence x = -1
Graphic solution
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Q02. Sketch graph of y = abs(x-1) + abs(x+1)

Graph is given in ABH program 07 01
Slope
    1. The slope is +2 and y = +2*x if x > 1
    2. The slope is -2 and y = -2*x if x < -1
Graph of y = abs(x-1) + abs(x+1)
  • When x GT 1, y = 2*x.
  • When x between -1 and 1, y = 2.
  • When x LT -1, y = -2*x.

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Q03 Solve abs(x-2) + abs(x+3) = 6

Grahic solution is given in ABH program 07 03
  • * Start software ABH find graphic solution from program 07 03
  • Solution
    • 1. y = +(x-2) + (x+3) if x is greater than 2
    • 2. y = -(x-2) - (x+3) if x is less than -3
    • 3. y = 5 if x is between -3 and 2
    • 4. The solution are the intesections of y = abs(x-2) + abs(x+3) and y = 6

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Q04 Study graph of y = abs(x+2) + abs(x+3)

Solutions
  • Slope = 0 and y = 1 when -3 < x < -2
  • Slope = 2 and y = 2*x+5 when x greater than -2
  • Slope =-2 and y =-2*x-5 when x less than -3
  • Range = 1 when -3 < x < -2
  • Range > 1 when x greater than -2
  • Range > 1 when x less than -3
Graph
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Q05 Solve abs(x - 1) + abs(x + 1) = c

Graphic solution Example Solve |x-1| + |x+1| = 3.
  • This equations contains 3 equations :
    • When x GT +1, |x-1| + |x+1| GT 2.
    • When x between -1 and 1, |x-1| + |x+1| = 2.
    • When x LT -1, |x-1| + |x+1| GT 2.
  • Hence there are two roots.
    • When x GT +1, y = 2*x. Hence +2*x = 3 and x = +1.5.
    • When x LT -1, y =-2*x. Hence -2*x = 3 and x = -1.5.
Example Solve |x-1| + |x+1| = 1.
  • This equations contains 3 equations :
    • When x GT +1, |x-1| + |x+1| GT 2.
    • When x between -1 and 1, |x-1| + |x+1| = 2.
    • When x LT -1, |x-1| + |x+1| GT 2.
  • Hence there are no solutions.
Example Solve |x-1| + |x+1| = 2.
  • This equations contains 3 equations :
    • When x GT +1, |x-1| + |x+1| GT 2.
    • When x between -1 and 1, |x-1| + |x+1| = 2.
    • When x LT -1, |x-1| + |x+1| GT 2.
  • Hence the solution is x between -1 and 1.

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Q05


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Q06 Diagram of intersection of y = x^2 - 6*x + 8 with it inverse


y = x^2 - 6*x + 8 and its inverse


y = x^2 + 0.25 and its inverse


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