calculus1-bb

Thomas' Calculus, Early Transcendentals, Media Upgrade (11th Edition)

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Chapter 1 Additional and Advanced Exercises 51 2. (a) (b) 3. There are (infinitely) many such function pairs. For example, f(x) 3x and g(x) 4x satisfy œœ f(g(x)) f(4x) 3(4x) 12x 4(3x) g(3x) g(f(x)). œœœ 4. Yes, there are many such function pairs. For example, if g(x) (2x 3) and f(x) x , then œ± œ $ "Î$ (f g)(x) f(g(x)) f (2x 3) (2x 3) 2x 3. ‰œ œ ± œ ± œ ± ab $$ "Î$ 5. If f is odd and defined at x, then f( x) f(x). Thus g( x) f( x) 2 f(x) 2 whereas ²œ ² ²œ²²œ ² ² g(x) (f(x) 2) f(x) 2. Then g cannot be odd because g( x) g(x) f(x) 2 f(x) 2 ² œ² ² œ² ± Ê ² ± 4 0, which is a contradiction. Also, g(x) is not even unless f(x) 0 for all x. On the other hand, if f is Êœ œ even, then g(x) f(x) 2 is also even: g( x) f( x) 2 f(x) 2 g(x). ² œ ² ² œ 6. If g is odd and g(0) is defined, then g(0) g( 0) g(0). Therefore, 2g(0) 0 g(0) 0. œ²œ ² œÊ œ 7. For (x y) in the 1st quadrant, x y 1 x ß± œ ± kk kk x y 1 x y 1. For (x y) in the 2nd Í ±œ± Í œ ß q u a d r a n t , x y x1 xyx1 ± œ± Ͳ±œ± y 2x 1. In the 3rd quadrant, x y x 1 Íœ± ± œ± x y x 1 y 2x 1. In the 4th Ͳ²œ± Í œ² ² quadrant, x y x 1 x ( y) x 1 ±œ ± Í ± ² œ ± y 1. The graph is given at the right. Íœ ² 8. We use reasoning similar to Exercise 7. (1) 1st quadrant: y y x x ± kk 2y 2x y x. ÍœÍ œ (2) 2nd quadrant: y y x x ± 2y x ( x) 0 y 0. ± ² œ Í œ (3) 3rd quadrant: y y x x ± y ( y) x ( x) 0 0 ͱ²œ±²Íœ points in the 3rd quadrant Ê all satisfy the equation. (4) 4th quadrant: y y x x ± y ( y) 2x 0 x. Combining ͱ²œ Íœ these results we have the graph given at the right: 9. If f is even and odd, then f( x) f(x) and f( x) f(x) f(x) f(x) for all x in the domain of f. ² ²œ Ê œ ² Thus 2f(x) 0 f(x) 0. œÊ œ 10. (a) As suggested, let E(x) E( x) E(x) E is an ² œ œ œ Ê f(x) f( x) f( x) f( ( x)) f(x) f( x) ±² ² ±²² ## # even function. Define O(x) f(x) E(x) f(x) . Then œ f(x) f( x) f(x) f( x) ²²
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52 Chapter 1 Functions O( x) O(x) O is an odd function ±œ œ œ ± œ ± Ê f( x) f( ( x)) f( x) f(x) f(x) f( x) ±±±± ±± ## # Š‹ f(x) E(x) O(x) is the sum of an even and an odd function. Êœ² (b) Part (a) shows that f(x) E(x) O(x) is the sum of an even and an odd function. If also œ² f(x) E (x) O (x), where E is even and O is odd, then f(x) f(x) 0 E (x) O (x) ± œ œ ² "" " " ab (E(x) O(x)). Thus, E(x) E (x) O (x) O(x) for all x in the domain of f (which is the same as the ±² ± domain of E E and O O ). Now (E E )( x) E( x) E ( x) E(x) E (x) (since E and E are ± ± œ ± ± ± œ ± " " " " even) (E E )(x) E E is even.
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calculus1-bb - Chapter 1 Additional and Advanced Exercises...

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