SECTION 3.5 Inequalities Involving Quadratic Functions 181 Figure 38 y 2 4 22 24 2 4 6 8 10 12 ( , ) (22, 8) f(x) 5 2x2 g(x) 5 x 1 10 x 5 – 2 25 –– 2 Figure 37 f x x x 2 10 2 ( ) = − − y 2 4 2 4 (22, 0) 24 22 26 210 ( , 210.125) x ( , 0) 5 – 2 1 – 4 Figure 39 22 24 0 2 4 Now Work PROBLEMS 5 AND 13 Finding the Domain of a Function Find the domain of the function g x x x 1. 2 ( ) = + + EXAMPLE 3 Solution Since only nonnegative numbers have real square roots, the expression under the square root symbol (the radicand) must be nonnegative. Therefore, finding the domain of g x( ) requires solving the quadratic inequality x x 1 0. 2 + + ≥ In solving the inequality x x 1 0, 2 + + ≥ we set f x x x 1 2 ( ) = + + and graph the function f x( ) to find where f is above or touching the x-axis, which makes f x 0. ( ) ≥ The solution to the inequality is the domain of the function g x . ( ) The graph of the function f x x x 1 2 ( ) = + + has y-intercept 1. There are no x-intercepts. (Do you know why? Check the discriminant.) The vertex is 1 2 , 3 4 . ( ) − Since a 0, > the parabola is concave up and lies above the x-axis for all real numbers x. So, x x 1 0 2 + + ≥ for all real numbers. See Figure 40. We conclude that the domain of the function g x( ) is all real numbers; in interval notation, we write , . ( ) −∞ ∞ Next graph the function f x x x 2 10 2 ( ) = − − to find where f x 0. ( ) < • y-intercept: f 0 10 ( ) = − Evaluate f at 0. • x-intercepts (if any): x x x x x x x x 2 10 0 2 5 2 0 2 5 0 or 2 0 5 2 or 2 2 ( )( ) − − = − + = − = + = = =− The y-intercept is 10; − the x-intercepts are 2− and 5 2 . The vertex is at x b a2 1 4 1 4 . = − = − − = Because f 1 4 10.125, ( ) = − the vertex is 1 4 , 10.125. ( ) − See Figure 37 for the graph. The graph is below the x-axis f x 0 ( ) ( ) < between x 2 = − and x 5 2 . = Because the inequality is strict, the solution set is { } − < < x x 2 5 2 or, using interval notation, 2, 5 2 . ( ) − Option 2 If f x x2 2 ( ) = and g x x 10, ( ) = + then the inequality to be solved is f x g x . ( ) ( ) < Graph the functions f x x2 2 ( ) = and g x x 10. ( ) = + See Figure 38. The graphs intersect where f x g x . ( ) ( ) = Then x x x x x x x x x x 2 10 2 10 0 2 5 2 0 2 5 0 or 2 0 5 2 or 2 2 2 ( )( ) = + − − = − + = − = + = = = − The graphs intersect at the points 2, 8 ( ) − and 5 2 , 25 2 . ( ) Then f x g x ( ) ( ) < on the interval 2, 5 2 , ( ) − where the graph of f is below the graph of g. See Figure 39 for the graph of the solution set. f x g x ( ) ( ) = Factor. Use the Zero-Product Property. Solve f x 0. ( ) = Factor. Use the Zero-Product Property. Now Work PROBLEM 31 Figure 40 f x x x 1 2 ( ) = + + x y (–1, 1) (0, 1) (1, 3) 1 2 3 2 –2 –1 (2 , ) 3 – 4 1 – 2
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