# HW2 - Section 2.1 1 f(3 = 5(3 6 = 15 6 = 21 f 3 = 5 3 6 =...

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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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HW2 - Section 2.1 1 f(3 = 5(3 6 = 15 6 = 21 f 3 = 5 3 6 =...

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