441 4.1 Inverse Functions Another approach is to select points on the graph of ƒ and use symmetry to find corresponding points on the graph of ƒ -1. For example, suppose the point 1a, b2 shown in Figure 8 is on the graph of a one-to-one function ƒ. Then the point 1b, a2 is on the graph of ƒ -1. The line segment connecting 1a, b2 and 1b, a2 is perpendicular to, and cut in half by, the line y = x. The points 1a, b2 and 1b, a2 are “mirror images” of each other with respect to y = x. Thus, we can find the graph of ƒ −1 from the graph of ƒ by locating the mirror image of each point in ƒ with respect to the line y =x. (a, b) (b, a) y = x y a 0 a b b x Figure 8 EXAMPLE 7 Graphing ƒ−1 Given the Graph of ƒ In each set of axes in Figure 9, the graph of a one-to-one function ƒ is shown in blue. Graph ƒ -1 in red. SOLUTION In Figure 9, the graphs of two functions ƒ shown in blue are given with their inverses shown in red. In each case, the graph of ƒ -1 is a reflection of the graph of ƒ with respect to the line y = x. x y (3, 1) (0, –4) (–4, 0) f –1 f y = x (1, 3) 0 –2 5 –2 5 Figure 9 x y (4, 2) f y = x (2, 4) 0 –2 5 –2 5 f –1 (1, 1) S Now Try Exercises 77 and 81. EXAMPLE 8 Finding the Inverse of a Function (Restricted Domain) Let ƒ1x2 = 2x + 5, x Ú -5. Find ƒ -11x2. SOLUTION The domain of ƒ is restricted to the interval 3-5, ∞2. Function ƒ is one-to-one because it is an increasing function and thus has an inverse function. Now we find the equation of the inverse. ƒ1x2 = 2x + 5, x Ú -5 Given function y = 2x + 5, x Ú -5 Replace ƒ1x2 with y. Step 1 x = 2y + 5, y Ú -5 Interchange x and y. Step 2 x2 = A 2y + 5 B 2 Square each side. (1+11)111 * x2 = y + 5 A 2a B 2 = a for a Ú 0 Solve for y. y = x2 - 5 Subtract 5. Rewrite. However, we cannot define ƒ -11x2 as x2 - 5. The domain of ƒ is 3-5, ∞2, and its range is 30, ∞2. The range of ƒ is the domain of ƒ -1, so ƒ -1 must be defined as follows. Step 3 ƒ -11x2 = x2 - 5, x Ú 0 As a check, the range of ƒ -1, 3-5, ∞2, is the domain of ƒ.
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