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THE ABSOLUTE VALUE FUNCTION. Properties of The Absolute Value Function Vertex (2, 0) f (x)=|x -2| +0 vertex (x,y) = (-(-2), 0) Maximum or Minimum? a =

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Presentation on theme: "THE ABSOLUTE VALUE FUNCTION. Properties of The Absolute Value Function Vertex (2, 0) f (x)=|x -2| +0 vertex (x,y) = (-(-2), 0) Maximum or Minimum? a ="— Presentation transcript:

1 THE ABSOLUTE VALUE FUNCTION

2 Properties of The Absolute Value Function Vertex (2, 0) f (x)=|x -2| +0 vertex (x,y) = (-(-2), 0) Maximum or Minimum? a = 1 > 0 graph opens up Minimum y = 0

3 Domain and Range Domain The domain of the Absolute value function is the set of real numbers Range 1. If the function concave up, the range is greater than or equal to the minimum 2. If the function concave down, the range is less than or equal to the maximum

4 Equations Involving Absolute Value Functions If |x|= a where a is a positive real number Then either x = a or x = -a Examples If |x| = 3 then either x = 3 or x = -3 If|x+1| = 5 then either x+1 = 5 or x+1 = 5 If |2x-3| = 4 then either 2x - 3 = 4 or 2x - 3 = -4

5 Inequalities Involving Absolute Value Functions If |x| < a where a is a positive real number Then –a < x < a If |x| a then –a x a Examples If |x| < 3 then r -3 < x < 3 If |x+1| 5 then -5 x + 1 5 -6 x 4 If |2x-3| < 4 then either -4 < 2x - 3 < 4

6 Solution of Equations Involving Absolute Value Functions The solution of two functions f(x) and g (x) is their point of intersection If f (x) = |x-1| and g (x) = 3 Then the solution of f (x) and g (x) is the point (s) |x – 1| = 3

7 Example Solve the problem and graph f (x) = |2x - 1| g (x) = 2 Let f (x) = g (x) If |2x – 1| = 2 Then 2x – 1 = 2 or 2x -1 = -2 + 1 1 +1 +1 2x = 32x = - 1 x = 3/2 x = - 1/2

8 Solve the problem and graph (cont.) Vertex ( 1/2,0) Y- intercept x = 0 |2(0)-1| = 1 X – intercept y = 0 |2x-1| = 0 2x – 1 =0 x = ½ Note the x intercept is the x coordinate of the vertex.

9 Solve the problem and graph Concave up a = 2 Minimum y = 0 Domain: RRange: y 0 Points of intersection (-0.5, 2) (1.5, 2)

10 Graph of y = |2x – 1| and y = 2 Points of intersections are (-.5, 2) and (1.5, 2)

11 Solve the problem and graph (cont.) f (x) = |2x – 1| < g (x) = 2 |2x -1| < 2 From the graph we can see that |2x -1| < 2 f (x) is below the line y =2 When -.5 < x < 1.5


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