chapter 1 41 - 28 SECTION 1.6 INVEHSE FUNCTIONS AND...

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Unformatted text preview: 28. SECTION 1.6 INVEHSE FUNCTIONS AND LOGARITHMS 41 An exponential model is y 2 abt. where 400 (millions) a 2 19976760197589 X 10—9 and b 2 10129334321697. This model gives y(1925) z 111 million. y(2010) m 330 million. and y(2020) % 375 million. 18900 2030 1.6 Inverse Functions and Logarithms 1. 10. 11. 12. 13. 14. (a) See Definition 1. (b) It must pass the Horizontal Line Test. (a) f‘1(y) 2 a: (2) flag) 2 y for any y in B. The domain of J"1 is B and the range of f—1 is A. (b) See the steps in (5). (c) Reflect the graph of f about the line y 2 I}. f is not one-to—one because 2 75 6. but f(2) 2 2.0 2 f(6). f is one—[done since for any two different domain values. there are different range values. No horizontal line intersects the graph of f more than once. Thus. by the Horizontal Line Test. f is one-to—one. The horizontal line 3,! 2 0 (the m—axis) intersects the graph of f in more than one point. Thus. by the Horizontal Line Test. f is not one—to—one. The horizontal line y 2 0 (the :c—axis) intersects the graph of f in more than one point. Thus. by the Horizontal Line Test, f is not one-to—one. No horizontal line intersects the graph of f more than once. Thus. by the Horizontal Line Test. f is one—to—one. . The graph of f(:v) 2 1(a: + 5) is a line with slope %. It passes the Horizontal Line Test. so f is one—to-one. 2 Algebraic solution: Ile 75 232, then :01 + 5 2 $2 + 5 2> 2(351 + 5) 2 %(332 + 5) 2> f(m1) 2 f(x2). so f is one—t0~one. The graph of f(:c) 2 1 + 4x — 1:2 is a parabola with axis of symmetry an 2 —% 2 ~fi 2 2. Pick any x—values equidistant from 2 to find two equal function values. For example. f(1) 2 4 and f(3) 2 4. so f is not 1-1. g(m) 2 III 2 g(—1) 2 1 2 9(1). 309 is not one—to—one. down. Thus. even iftl does not equal t2, f(t1) may equal f(t2). so f is not 1—1. f is not 1~1 because eventually we all stop growing and therefore. there are two times at which we have the same height. ...
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