SECTION 5.4 Logarithmic Functions 317 Figure 39 x x x y 3 3 (1, 0) , 21) ( (e, 1) (e, 21) 3 (a) y 5 In x y (1, 0) 21 21 21 21 21 1 1 1 3 (b) y 5 2In x 1 –e 1 –e ( ) 1 –e , 1 ( ) y 5 3 1 (3, 0) (e 1 2, 21) 3 (c) y 5 2In (x 2 2) 1 2,1 x 5 2 x 5 0 x 5 0 Multiply by 21; reflect about the x-axis. Replace x by x 2 2; shift right 2 units. (b) To obtain the graph of ( ) = − − y x ln 2 , begin with the graph of = y x ln and use transformations. See Figure 39. (c) The range of ( ) ( ) = − − f x x ln 2 is the set of all real numbers. The vertical asymptote is = x 2. [Do you see why? The original asymptote ( ) = x 0 is shifted to the right 2 units.] (d) To find −f ,1 begin with ( ) = − − y x ln 2 . The inverse function is defined implicitly by the equation ( ) = − − x y ln 2 Now solve for y. ( ) − = − = − = + − − x y e y y e ln 2 2 2 x x The inverse of f is ( ) = + − − f x e 2. x 1 (e) The domain of −f 1 equals the range of f, which is the set of all real numbers, from part (c).The range of −f 1 is the domain of f, which is ( )∞ 2, in interval notation. (f) To graph −f ,1 use the graph of f in Figure 39(c) and reflect it about the line = y x. See Figure 40. We could also graph ( ) = + − − f x e 2 x 1 using transformations. Isolate the logarithm. Change to exponential form. Solve for y. Figure 40 x y 5 5 (21, e12) (1, 12) (0, 3) (3, 0) 21 21 1 1 2, 1) 1 –e x 5 2 y 5 2 y 5 x f 21(x) 5 e 2x 1 2 f (x) 5 2ln(x 2 2) (e 1 2, 21) 1 –e ( Now Work PROBLEM 75
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