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Unformatted text preview: Section 2.1 1. f (3) = 5(3) + 6 = 15 + 6 = 21 f ( 3) = 5( 3) + 6 = 15 + 6 = 9 f ( a ) = 5( a ) + 6 = 5 a + 6 f ( a ) = 5( a ) + 6 = 5 a + 6 f ( a + 3) = 5( a + 3) + 6 = 5 a + 15 + 6 = 5 a + 21 3. g (0) = 3(0) 2 6(0) 3 = 0 + 0 3 = 3 g ( 1) = 3( 1) 2 6( 1) 3 = 3 + 6 3 = 6 g ( a ) = 3( a ) 2 6( a ) 3 = 3 a 2 6 a 3 g ( a ) = 3( a ) 2 6( a ) 3 = 3 a 2 + 6 a 3 g ( x +1) = 3( x +1) 2 6( x +1) 3 = 3( x 2 +2 x +1) 6 x 6 3 = 3 x 2 +6 x +3 6 x 6 3 = 3 x 2 6 5. f ( a + h ) = 2( a + h ) + 5 = 2 a + 2 h + 5 f ( a ) = 2( a ) + 5 = 2 a + 5 f ( a 2 ) = 2( a 2 ) + 5 = 2 a 2 + 5 f ( a 2 h ) = 2( a 2 h ) + 5 = 2 a 4 h + 5 f (2 a h ) = 2(2 a h ) + 5 = 4 a 2 h + 5 7. s (4) = 2(4) (4) 2 1 = 8 16 1 = 8 15 s (0) = 2(0) (0) 2 1 = 1 = 1 = 0 s ( a ) = 2( a ) ( a ) 2 1 = 2 a a 2 1 s (2 + a ) = 2(2+ a ) (2+ a ) 2 1 = 4+2 a (4+4 a + a 2 ) 1 = 4+2 a a 2 +4 a +3 s ( t + 1) = 2( t +1) ( t +1) 2 1 = 2 t +2 ( t 2 +2 t +1) 1 = 2 t +2 t 2 +2 t 9. f (2) = 2(2) 2 √ (2) 1 = 2(4) √ (2) 1 = 8 1 = 8 f ( a ) = 2( a ) 2 √ ( a ) 1 = 2( a 2 ) √ a 1 = 2 a 2 √ a 1 f ( x + 1) = 2( x +1) 2 √ ( x +1) 1 = 2( x 2 +2 x +1) √ x +1 1 = 2 x 2 +4 x +2 √ x f ( x 1) = 2( x 1) 2 √ ( x 1) 1 = 2( x 2 2 x +1) √ x 1 1 = 2 x 2 4 x +2 √ x 2 11. f ( 2) = ( 2) 2 + 1 = 4 + 1 = 5 f (0) = (0) 2 + 1 = 0 + 1 = 1 f (1) = q (1) = √ 1 = 1 13. f ( 1) = (1 / 2)( 1) 2 + 3 = (1 / 2)(1) + 3 = (1 / 2) + 3 = (5 / 2) f (0) = (1 / 2)(0) 2 + 3 = 0 + 3 = 3 f (1) = 2(1) 2 + 1 = 2(1) + 1 = 2 + 1 = 3 f (2) = 2(2) 2 + 1 = 2(4) + 1 = 8 + 1 = 9 15. f (0) = 2 . f ( x ) = 3 when x = 2 and f ( x ) = 0 when x = 1 . The domain of f ( x ) is given in interval notation by [0 , 6] . The range of f ( x ) is given in interval notation by [ 2 , 6] . 17. The point (2 , √ 3) lies on the graph if g (2) = √ 3. g (2) = q (2) 2 1 = √ 4 1 = √ 3; so that the point is on the graph. 19. The point ( 2 , 3) lies on the graph if f ( 2) = 3. f ( 2) =  ( 2) 1  ( 2)+1 =  3  1 = 3 1 = 3; so that the point is on the graph. 21. The point (1 , 5) lies on the graph if f (1) = 5. f (1) = 2(1) 2 4(1) + c = 2 4 + c = 2 + c ; setting this equal to 5 we find 2 + c = 5 → c = 7 . 23. The function f ( x ) = x 2 + 3 is a polynomial and hence defined for all x so its domain in interval notation is given by (∞ , ∞ ) . 25. The function f ( x ) = 3 x +1 x 2 is a rational function and hence defined for all x for which the denominator is nonzero; the denominator is zero at x = 0 so its domain in interval notation is given by (∞ , 0) ∪ (0 , ∞ ) . 27. The function f ( x ) = √ x 2 + 1 is a radical and hence is defined for all x values for which x 2 + 1 ≥ 0; since this inequality is true for all x, the function’s domain in interval notation is given by (∞ , ∞ ) ....
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 Spring '08
 Smith
 Calculus, lim, Continuous function

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