The modulus function The modulus of x, written  x  and is defined by  x  = x if x  0  x  = -x if x < 0. Sketch of y =  x  y x y =  x  y = x.

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Presentation transcript:

The modulus function The modulus of x, written  x  and is defined by  x  = x if x  0  x  = -x if x < 0. Sketch of y =  x  y x y =  x  y = x y =  x  y = - x Sketch the curve y = f(x), using a dashed line for points below the x-axis Reflect any part of the curve below the x-axis in the x-axis.

Sketch y =  2x - 1  The graph of y =  2x - 1  has the right-hand branch with equation y = 2x – 1. The left-hand branch is the negative of 2x – 1, i.e. y = -2x + 1. y x y = 2x - 1 y = -2x +1

Sketch y =  (x-1)(x-2)  x y

Sketch y =  cos x  x y =  cos x 

Equations involving modulus Solve the equation  2x - 1  = 3 y = 3 Left-hand branch: -2x + 1 = 3  x = - 1 Right-hand branch: 2x - 1 = 3  x = 2 Solution: x = -1 or x = 2

Examples Solve the equation  2x - 1  =  x - 2  Critical values x 2  2x - 1  -(2x – 1) 2x – 1 2x - 1  x - 2  -(x – 2) -(x – 2) x - 2 x < ½ : -2x + 1 = -x + 2  x = - 1 ½  x  2 : 2x - 1 = -x + 2  x = 1 x > 2 : 2x - 1 = x - 2  x = -1 Not consistent. Solution: x = -1 or x = 1 Other methods: Square both sides or graphical method.

Examples Solve  x  =  2x + 1  Square both sides x 2 = (2x + 1) 2 x 2 = 4x 2 + 4x + 1 3x 2 + 4x + 1= 0 (3x + 1)(x + 1) = 0  x = -1/3 or x = - 1 Solve  x-3  =  3x + 1  Square both sides (x – 3) 2 = (3x + 1) 2 x 2 - 6x+ 9 = 9x 2 + 6x + 1 8x x – 8 = 0 2x 2 + 3x – 2 = 0 (2x -1)(x + 2) = 0 x = - 2 or x = ½

Inequalities involving modulus  x  < a  -a < x < a  x  > a  x a Example  x  < 2  -2 < x <  x  > 2  x

Examples Solve the inequality  2x - 1  < 3 y = 3  2x - 1  < 3  - 3 < 2x – 1 < 3  -2 < 2x < 4  -1 < x < 2

Examples Solve the inequality  2x - 3   5 y = 5  2x - 5   5  2x – 3  -5 or 2x – 3  5  x  -1 or x  4

Examples Solve the inequality  x + 2  <  3x + 1  Square both sides: (x + 2) 2 < (3x + 1) 2  x 2 + 4x + 4 < 9x 2 + 6x + 1  8x 2 + 2x - 3 > 0  (4x + 3)(2x – 1) > 0  x ½ y x

Examples Solve the inequality  x + 2  > 2x + 1 Critical values x < -2 x  - 2  x + 2  -x - 2 x + 2 2x + 1x + 1 2x +1 when x 2x + 1  x < - 1 when x  - 2: x + 2 > 2x + 1  x < 1 x y Solution x < 1

Example A graph has equation y = 2x +  x + 2 . Express y as a linear functions of x ( y = mx + c). When x < -2 : y = 2x – x – 2 = x – 2  y = x - 2 When x  -2 : y = 2x + x + 2 = 3x + 2  y = 3x + 2