{−4, −1 > 1 x2 Solution set: 0} . − 2 > 0 x2 ( ) d. > 0 f x ( ) ( ) c. = f x g x − + 1 > x2 0 − + 1 = 4 + 1 x2 x − 1 > 0 x2 0 = + 4x x2 (1 ) (1 ) ( − + > 0 x x 0 = + 4) x x = 1; = −1 = −4; = 0 x x x x ( = and = 1 . –1 x − x The ) 2 > )( are x + 2 0 x The are = 2 and = 2 . x x – ( ) = − 2 p x x2 { The solution set is x } −1 < < 1 or, x (−1, interval notation, 1) .
x2 −1 > 1 x2 − 2 > 0 c. f ( x ) = g ( x ) − x2 + 1 = 4 x + 1 0 = x2 + 4x 0 = x ( x + 4) x = −4; x = 0 Solution set: {−4, 0} . d. f ( x ) > 0 − x2 + 1 > 0 1 − x2 > 0 (1 − x ) (1 + x ) > 0 x = 1; x = −1 ( x − 2 )( x + 2 ) > 0 The are x = –1 and x = 1 . The are x = – 2 and x = 2 . p ( x ) = x2 − 2 The solution set is { x −1 < x < 1 } or, interval notation, (−1, 1) .
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx