( ( ) 0 ≥ − g. ≥ 1 x f x − + 4 > 1 x2 ≥ − 0 x2 3 )( ) 3 + 3 x The are = 3 and = 3 . x x – ( } ( ) −2 or > 2 or, x = p x (−∞, (2, ) The using interval notation, −2) ∪ ∞ − x . ( ) set e. ≤ 0 g x )( ) { 3 + 3 x is x − + 4 ≤ 0 x2 − ≤ 4 0 x2 < x (2 ) (2 ) − + ≤ 0 x x { The solution set is x = = The zeros are and 2 . x x ≤ ≤ –2 3 x − 3} or, using interval notation, ⎤⎦ 3, 3 . 29. = − 4; x2 ( ) = − + 4 g x x2 ( ) f x ( ) a. = 0 f x { } The solution set is ≤ −2 or ≥ 2 or, x x x − 4 = 0 x2 ( ( [2, (−∞, ) − + = 2) 2) using interval notation, −2] ∪ ∞ x x 0 . = 2; = −2 x x {−2, ( ) ( Solution set: 2} . f. > f x g ) ( ) x b. = 0 g x − 4 > − + 4 x2 x2 2 ( − = − + = 2 4) 0 4 0 x2 x2 ( ( 4 − = 0 x2 2 − 2) + 2) = 0 x x (2 ) (2 ) − + = = 2; = −2 0 x x x x = = −2 2; x x {−2, Solution set: 2} . {−2, Solution set: 2} . ( ) > d. 0 f x ( ) ( c. = f x g − > 4 0 x2 ( ( ) − 2) + 2) > 0 x x x = = The zeros are and 2 . x x − 4 = − + 4 –2 x2 x2 2x2 − 8 = 0 2x − 8 > 0
g. f ( x ) ≥ 1 − x2 + 4 > 1 0 ≥ x2 − 3 0 ≥ ( x − 3 )( x + 3 ) The are x = – 3 and x = 3 . p ( x ) = ( x − 3 )( x + 3 ) The set is { x x < −2 or x > 2 } or, using interval notation, (−∞, −2) ∪ (2, ∞ ) . e. g ( x ) ≤ 0 − x2 + 4 ≤ 0 4 − x2 ≤ 0 (2 − x ) (2 + x ) ≤ 0 The solution set is { x − 3 ≤ x ≤ 3} or, The zeros are x = –2 and x = 2 . using interval notation, 3, 3 ⎤⎦ . 29. f ( x ) = x2 − 4; g ( x ) = − x2 + 4 a. f ( x ) = 0 x2 − 4 = 0 ( x − 2) ( x + 2) = 0 x = 2; x = −2 Solution set: {−2, 2} . b. g ( x ) = 0 The solution set is { x x ≤ −2 or x ≥ 2 } or, using interval notation, (−∞, −2] ∪ [2, ∞ ) . f. f ( x ) > g ( x ) x2 − 4 > − x2 + 4 2 − x2 + 4 = 0 4 − x2 = 0 (2 − x ) (2 + x ) = 0 x = 2; x = −2 Solution set: {−2, 2} . c. f ( x ) = g ( x ) x2 − 4 = − x2 + 4 2x2 − 8 = 0 2 ( x2 − 4) = 0 2 ( x − 2) ( x + 2) = 0 x = 2; x = −2 Solution set: {−2, 2} . d. f ( x ) > 0 x2 − 4 > 0 ( x − 2) ( x + 2) > 0 The zeros are x = –2 and x = 2 . 2x − 8 > 0
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