106 CHAPTER 2 Functions and Their Graphs Figure 51 (–4, 8) (–2, 4) (1, 1) (4, 2) (0, 0) y x 25 5 8 23 (a) (c) The y-intercept of the graph of the function is f 0 . ( ) Because the equation for f when x 0 = is f x x, ( ) = the y-intercept is f 0 0 0. ( ) = = The x-intercepts of the graph of a function are the real solutions to the equation f x 0. ( ) = To find the x-intercepts of f, solve f x 0 ( ) = for each equation of the function, and then determine which values of x, if any, satisfy the domain of the equation. f x x x 0 2 0 0 ( ) = − = = x 0 < f x x x 0 0 0 ( ) = = = x 0 ≥ The result on the left x 0 ( ) = is discarded because it does not satisfy the condition x 0. < The result on the right x 0 ( ) = is obtained under the condition x 0, ≥ so 0 is an x-intercept. The only intercept is 0, 0 . ( ) (d) To graph f, graph each equation. First graph the line y x2 = − and keep only the part for which x 0. < Then graph the square root function y x = for x 0. ≥ See Figure 51(a). Figure 51(b) show the inputs used to graph in Desmos.. (e) From the graph, we conclude that the range of f is the set of nonnegative real numbers, or the interval 0, . [ )∞ Analyzing a Piecewise-defined Function A piecewise-defined function f is defined as ( ) = − + − ≤ < = > ⎧ ⎨ ⎪⎪ ⎪ ⎩ ⎪⎪ ⎪⎪ f x x x x x x 2 1 if 3 1 2 if 1 if 1 2 (a) Find f 2 , ( ) − f 1 , ( ) and f 2 . ( ) (b) Find the domain of f. (c) Locate any intercepts. (d) Graph f. (e) Use the graph to find the range of f. Solution EXAMPLE 4 (a) To find f 2 , ( ) − observe that 3 2 1, − ≤− < so when x 2, = − the equation for f is given by f x x2 1. ( ) = − + Then f 2 2 2 1 5 ( ) ( ) − = − − + = f x x x 2 1 if 3 1 ( ) =− + − ≤ < To find f 1 , ( ) observe that when x 1, = the equation for f is f x 2. ( ) = So, f 1 2 ( ) = f x x 2 if 1 ( ) = = When x 2, = the equation for f is f x x .2 ( ) = Then f 2 2 4 2 ( ) = = f x x x if 1 2 ( ) = > (b) The domain of f is the union of the domains of each equation in the piecewisedefined function. So the domain of f is 3,1 1 1, 3, . [ ) { } ( ) [ ) − ∪ ∪ ∞ = − ∞ The domain of f is 3, [ ) − ∞ in interval notation, or x x 3 { } ≥− in set notation. (c) The y-intercept of the graph of the function is f 0 . ( ) Because the equation for f when x 0 = is f x x2 1, ( ) = − + the y-intercept is f 0 2 0 1 1. ( ) = − ⋅ + = The x-intercepts of the graph of a function f are the real solutions of the equation f x 0. ( ) = To find the x-intercepts of f, solve f x 0 ( ) = for each equation of the function, and then determine what values of x, if any, are in the domain of the equation. ( ) = − + = − = − = f x x x x 0 2 1 0 2 1 1 2 x 3 1 − ≤ < ( ) = = f x 0 2 0 No solution x 1 = ( ) = = = f x x x 0 0 0 2 x 1> (b)
RkJQdWJsaXNoZXIy NjM5ODQ=