229 2.3 Functions ƒ132 = 2 indicates that the ordered pair 13, 22 belongs to ƒ. SOLUTION (a) ƒ1x2 = 3x - 7 ƒ132 = 3132 - 7 Replace x with 3. ƒ132 = 2 Simplify. (b) For ƒ = 51-3, 52, 10, 32, 13, 12, 16, -126, we want ƒ132, the y-value of the ordered pair where x = 3. As indicated by the ordered pair 13, 12, when x = 3, y = 1, so ƒ132 = 1. (c) In the mapping, repeated in Figure 25, the domain element 3 is paired with 5 in the range, so ƒ132 = 5. (d) To evaluate ƒ132 using the graph, find 3 on the x-axis. See Figure 26. Then move up until the graph of ƒ is reached. Moving horizontally to the y-axis gives 4 for the corresponding y-value. Thus, ƒ132 = 4. S Now Try Exercises 67, 69, and 71. –2 3 10 6 5 12 f Domain Range Figure 25 x y 0 2 4 2 4 3 y = f(x) Figure 26 Finding an Expression for ƒ1x2 Consider an equation involving x and y. Assume that y can be expressed as a function ƒ of x. To find an expression for ƒ1x2, use the following steps. Step 1 Solve the equation for y if it is not given in that form. Step 2 Replace y with ƒ1x2. EXAMPLE 8 Writing Equations Using Function Notation Assume that y is a function ƒ of x. Rewrite each equation using function notation. Then find ƒ1-22 and ƒ1p2. (a) y = x2 + 1 (b) x - 4y = 5 SOLUTION (a) Step 1 y = x2 + 1 This equation is already solved for y. Step 2 ƒ1x2 = x2 + 1 Let y = ƒ1x2. Now find ƒ1-22 and ƒ1p . ƒ1-22 = 1-222 + 1 Let x = -2. ƒ1-22 = 4 + 1 ƒ1-22 = 5 ƒ1 p2 = p2 + 1 Let x = p. (b) Step 1 x - 4y = 5 Given equation. -4y = -x + 5 Add -x. y = x - 5 4 Multiply by -1. Divide by 4. Step 2 ƒ1x2 = 1 4 x - 5 4 Let y = ƒ1x2; a - b c = a c - b c If a function f is defined by an equation with x and y (and not with function notation), use the following steps to find ƒ1x2.
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