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Section 1.2 Functions and Graphs (pp. 9–19) Exploration 1 Composing Functions 1. y 3 5 g + f , y 4 5 f + g 2. Domain of y 3 :[ 2 2, 2] Range of y 3 : [0, 2] y 1 : [ 2 4.7, 4.7] by [ 2 2, 4.2] y 2 : [ 2 4.7, 4.7] by [ 2 2, 4.2] y 3 : [ 2 4.7, 4.7] by [ 2 2, 4.2] 3. Domain of y 4 : [0, ); Range of y 4 :( 2‘ ,4] y 4 : [ 2 2, 6] by [ 2 2, 6] 4. y 3 5 y 2 ( y 1 ( x )) 5 ˇ y 1 w ( x w ) w 5 ˇ 4 w 2 w x w 2 w y 4 5 y 1 ( y 2 ( x )) 5 4 2 ( y 2 ( x )) 2 5 4 2 ( ˇ x w ) 2 5 4 2 x , x \$ 0 Quick Review 1.2 1. 3 x 2 1 # 5 x 1 3 2 2 x # 4 x \$2 2 Solution: [ 2 2, ) 2. x ( x 2 2) . 0 Solutions to x ( x 2 2) 5 0: x 5 0, x 5 2 Test x 52 1: 2 1( 2 1 2 2) 5 3 . 0 x ( x 2 2) . 0 is true when x , 0. Test x 5 1: 1(1 2 2) 1 , 0 x ( x 2 2) . 0 is false when 0 , x , 2. Test x 5 3: 3(3 2 2) 5 3 . 0 x ( x 2 2) . 0 is true when x . 2. Solution set: ( 2‘ , 0) < (2, ) 3. ) x 2 3 ) # 4 2 4 # x 2 3 # 4 2 1 # x # 7 Solution set: [ 2 1, 7] 4. ) x 2 2 ) \$ 5 x 2 2 #2 5 or x 2 2 \$ 5 x 3 or x \$ 7 Solution set: ( 2‘ , 2 3] < [7, ) 5. x 2 , 16 Solutions to x 2 5 16: x 4, x 5 4 Test x 6( 2 6) 2 5 36 . 16 x 2 , 16 is false when x ,2 4 Test x 5 0: 0 2 5 0 , 16 x 2 , 16 is true when 2 4 , x , 4 Test x 5 6: 6 2 5 36 . 16 x 2 , 16 is false when x . 4. Solution set: ( 2 4, 4) 6. 9 2 x 2 \$ 0 Solutions to 9 2 x 2 5 0: x 3, x 5 3 Test x 4: 9 2 ( 2 4) 2 5 9 2 16 7 , 0 9 2 x 2 \$ 0 is false when x 3. Test x 5 0: 9 2 0 2 5 9 . 0 9 2 x 2 \$ 0 is true when 2 3 , x , 3. Test x 5 4: 9 2 4 2 5 9 2 16 7 , 0 9 2 x 2 \$ 0 is false when x . 3. Solution set: [ 2 3, 3] 7. Translate the graph of f 2 units left and 3 units downward. 8. Translate the graph of f 5 units right and 2 units upward. 9. (a) f ( x ) 5 4 x 2 2 5 5 4 x 2 2 9 5 0 ( x 1 3)( x 2 3) 5 0 x 3 or x 5 3 (b) f ( x ) 6 x 2 2 5 6 x 2 1 No real solution 10. (a) f ( x ) 5 } 1 x } 5 x 52} 1 5 } (b) f ( x )= 0 } 1 x } 5 0 No solution 11. (a) f ( x 4 ˇ x w 1 w 7 w 5 4 x 1 7 5 16 x 5 9 Check: ˇ 9 w 1 w 7 w 5 ˇ 1 w 6 w 5 4; it checks. (b) f ( x ) 5 1 ˇ x w 1 w 7 w 5 1 x 1 7 5 1 x 6 Check: ˇ 2 w 6 w 1 w 7 w 5 1; it checks. 12. (a) f ( x ) 2 ˇ 3 x w 2 w 1 w 2 x 2 1 8 x 7 (b) f ( x ) 5 3 ˇ 3 x w 2 w 1 w 5 3 x 2 1 5 27 x 5 28 6 Section 1.2

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Section 1.2 Exercises 1. Since A 5 p r 2 5 p 1 } d 2 } 2 2 , the formula is A 5 } p 4 d 2 } , where A represents area and d represents diameter. 2. Let h represent height and let s represent side length. h 2 1 1 } 2 s } 2 2 5 s 2 h 2 5 s 2 2 } 1 4 } s 2 h 2 5 } 3 4 } s 2 h 56} ˇ 2 3 w } s Since side length and height must be positive, the formula is h 5 } ˇ 2 3 w } s .
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