1 / 13

The modulus function

y. x. 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. Sketch the curve y = f(x), using a dashed line for points below the x-axis.

casper
Download Presentation

The modulus function

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. y x 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 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.

  2. y x 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 = 2x - 1 y = -2x +1

  3. Sketch y = (x-1)(x-2) y x

  4. Sketch y = cos x y = cos x x

  5. 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

  6. Examples Solve the equation 2x - 1 = x - 2 Critical values x < ½ ½  x  2 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.

  7. Examples Solve x = 2x + 1 Solve x-3 = 3x + 1 Square both sides Square both sides (x – 3)2 = (3x + 1)2 x2 = (2x + 1)2 x2 - 6x+ 9 = 9x2 + 6x + 1 x2 = 4x2 + 4x + 1 8x2 + 12x – 8 = 0 3x2 + 4x + 1= 0 2x2 + 3x – 2 = 0 (3x + 1)(x + 1) = 0 (2x -1)(x + 2) = 0  x = -1/3 or x = - 1 x = - 2 or x = ½

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

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

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

  11. y x Examples Solve the inequality x + 2 <3x + 1 Square both sides: (x + 2)2< (3x + 1)2  x2 + 4x + 4 < 9x2 + 6x + 1  8x2 + 2x - 3 > 0  (4x + 3)(2x – 1) > 0  x < - ¾ or x > ½

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

  13. 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

More Related