ss_3_8

# ss_3_8 - Section 3.8 Solving Polynomial Inequalities...

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Section 3.8 Solving Polynomial Inequalities Example. Find all x such that x 2 + x - 2 < 0. Solution. x 2 + x - 2=( x +2)( x - 1). The expression ( x x - 1) changes sign at x = - 2and x =1. If x< - 2, then ( x x - 1) > 0. If - 2 <x< 1, then ( x x - 1) < 0. If 1 <x ,then( x x - 1) > 0. Hence, x 2 + x - 2 < 0 on the interval ( - 2 , 1). Example. Solve graphically: x 4 x +1. Solution. From the graph below, x = - . 70 ,y = . 24 and x =1 . 21 =2 . 16 are approximate points where the graphs of x 4 and x +1 intersect. Hence the approximate solution is the interval [ - . 70 , 1 . 21]. –5 0 5 –2 –1 1 2 y = x 4 , y = x +1 Example. Solve f ( x )= x - 1 x - 2 > 0. Graphical solution. The graph below indicates the solution consists of x in the intervals ( -∞ , 1) and (2 , ). –10 –5 0 5 10 1234 y = x - 1 x - 2 Analytical solution. The strategic values of x where f ( x ) may change sign are x , 2. If -∞ 1, then f ( x ) > 0 If 1 2, then f ( x ) < 0 If 2

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ss_3_8 - Section 3.8 Solving Polynomial Inequalities...

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