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Modulus Graph

Modulus Graph. 2 different types of Modulus Graphs covered. y=f(|x|) y=|f(x)|. y=|f(x)|. First sketch the graph of y = x . To get the graph of y =| x |, reflect in the x -axis the portion of the line below it. What results is a V-shaped graph with the vertex at (0, 0) as shown. y=|cos x|.

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Modulus Graph

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  1. Modulus Graph

  2. 2 different types of Modulus Graphs covered • y=f(|x|) • y=|f(x)|

  3. y=|f(x)| • First sketch the graph of y=x. To get the graph of y=|x|, reflect in the x-axis the portion of the line below it. • What results is a V-shaped graph with the vertex at (0, 0) as shown.

  4. y=|cos x| Example: Draw graph of y =|cos x| Solution : We first draw the graph of y = cos x . Then, we reflect the negative part of the graph in x-axis to get y =|cos x|

  5. y=|lg x| Example: Draw graph of y =|lg x| . Solution : We first draw graph y = lg x . Then, we reflect the negative part of the graph in x-axis to get y=|lg x| .

  6. What about y= −|f(x)|? • First sketch the graph of y=|x|. To get the graph of y= −|x|, reflect in the x-axis the portion of the line below it. • What results is a inverted V-shaped graph with the vertex at (0, 0) as shown.

  7. y=f(|x|) • y=f(x) ⇒ y=f(|x|) • remove left half of the graph • take the mirror image of right half of the graph in y-axis

  8. y=f(|x|) Example: Draw graph of y=sin|x| Solution : First we draw graph of sin x. In order to obtain the graph of y=sin|x|, we remove left half of the graph and take the mirror image of right half of the graph of in y-axis.

  9. Analysis: Whether it is the positive or negative x-value, they will have the same y-value.

  10. y=f(|x|) Example: Draw graph of y=e|x+1| Solution :We first draw graph of y = ex . Then, we shift the graph left by 1 unit to obtain the graph of ex+1. At x = 0, y=e0+1=e . In order to obtain the graph of y=e|x+1| , we remove left part of the graph and take the mirror image of right half of the graph of y=ex+1 in y-axis.

  11. Difference between y=|x| and y=f|x| y = sin |x| y = |sin x|

  12. Ex 1(a)

  13. Ex 1(b)

  14. Ex 1(c)

  15. Ex 1(d) (3, 5) (−1, 3)

  16. Ex 1(e)

  17. Ex 1(f)

  18. Ex 1(g)

  19. Ex 1(h)

  20. Ex 1(i)

  21. Ex 1(j)

  22. Ex 1(k)

  23. Ex 1(l)

  24. y 3 1 x 0.5 -1 (a) For y ≤ 3, -1 ≤ x ≤ 2 (b) Point of intersection: (0, 1) and (2, 3) Example -1 2 (c) |2x – 1| > x + 1  x < 0 and x > 2

  25. Definition The absolute value of modulus of a real number a is denoted by |a| and can be defined as: As can be seen, |a| is the value of a without regarding its sign and it is always positive or zero.

  26. Introduction • From an analytic geometry point of view, the absolute value of a real number is that number's distance from zero along the real number line. • To be more specific, the absolute value of the difference of two real numbers is the distance between them.

  27. Example 1 Given that f(x) = |x2 - 6x|, find the value of f(1)

  28. Solution 1 f(1) = |12 – 6(1)| = |-5| = 5

  29. Properties of absolute values • For real values of a and b (b ≠0) • |a| ≥ 0 • |-a| = |a| • |ab| = |a||b| • |a/b|= |a|/|b| • |an| = |a|n • |a±b| ≠ |a| ± |b|

  30. Example 2 Simplify |π - 4|, giving your answers in exact form.

  31. Solution 2 |π – 4| = -(π – 4) = 4 - π

  32. Example 3 Solve |x – 3| = 2 |x – 3| = 2 x – 3 = 2 x = 5 √ |x – 3| = 2 3 – x = 2 x = 1 √

  33. Example 4 Solve |2x – 4| = 1 |2x – 4| = 1 2x – 4 = 1 x = 2.5 √ |2x – 4| = 1 4 – 2x = 1 x = 1.5 √

  34. Example 3 Solve x = |x2 – 2|

  35. Solution 3 x2 – 2 = x x2 – x – 2 = 0 (x – 2)(x + 1) = 0 x = 2 OR -1 x2 – 2 = -x X2 + x – 2 = 0 (x + 2)(x – 1) = 0 x = -2 OR 1 But x = |x2 – 2| ≥ 0. So x = 1 OR 2

  36. Simple equations involving modulus expressions • In general, equations of the form |a| = |b| can be solved as follows • |a| = |b|  a = b or a = -b

  37. Example 4 Solve |2x + 3| = |x|

  38. Solution 4 2x + 3 = x X = -3 2x + 3 = -x x = -1

  39. Simple equations involving modulus expressions • Alternatively, we can square both sides and we get our answers nevertheless. • |a| = |b| a2 = b2

  40. Exercise 2 Solve |x + 3| = 1 |x + 3| = 1 x + 3 = 1 x = −2 √ |x + 3| = 1 –x – 3 = 1 x = − 4 √

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