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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 = 1 > 0 graph opens up Minimum y = 0. Domain and Range. Domain
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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 = 1 > 0 graph opens up Minimum y = 0
Domain and Range Domain The domain of the Absolute value function is the set of real numbers Range • If the function concave up, the range is greater than or equal to the minimum • If the function concave down, the range is less than or equal to the maximum
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
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
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
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 = 3 2x = - 1 x = 3/2 x = -1/2
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.
Solve the problem and graph Concave up a = 2 Minimum y = 0 Domain: R Range: y 0 Points of intersection (-0.5, 2) (1.5, 2)
Graph of y = |2x – 1| and y = 2 Points of intersections are (-.5, 2) and (1.5, 2)
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