Solve the given polynomial and rational inequalities x3 Solu

Solve the given polynomial and rational inequalities x^3

Solution

a) We have the inequality x³ < 4x or, x³ – 4x < 0.

If the polynomial x³4x is 0 , then x( x+2)( x – 2) = 0 so that either x= 2, or, x =2, or, 2. At x < -2 (say x = 3), the value of the polynomial is < 0; at 0 < x < 2 ( say x = 1) the value of the polynomial is < 0. At x > 2, ( say x = 3), the value of the polynomial is > 0. Also if -2< x < 0 ( say x = -1), then the value of the polynomial is > 0.

From this we see the polynomial is < 0 for x < -2   and 0 < x < 2. Thus, the solution is x ( - , -2) U ( 0, 2)

b) (x -2)/ (2x + 2) 0

If (x-2)/ ( 2x +2 ) = 0 , then (x -2) = 0 * ( 2x +2) = 0 so that x = 2 ( provided that x -1, because then 2 x + 2 = 0). Thus, the critical values are x = -2 and x = - 1. Now, if x -2, say x = -2 or -3 then (x-2)/ ( 2x +2 ) = -4/ -2 = 2 > 0   or, (x -2)/ (2x + 2) = -5/ -4 = 5/4 > 0. Further, if -2 < x < -1 , say x = - 3/2 , then (x-2)/ ( 2x +2 ) = ( -7/2) / ( -1) = 7/2 > 0 . Also, if x > -1 , say x = 0, then (x-2)/ ( 2x +2 ) = -2/2 = -1 < 0.

From this we see that (x -2)/ (2x + 2) 0 if x < -2 and if -2 < x < -1. Thus, x ( - , -2] U ( -2, -1) = ( -, -1)

c) If y varies directly with the square root of x, then y = a x , where a is a constant. When x = 9, x = 3 , and y = 3a = 0.12 so that a = 1/3 ( 0.12) = 0.04. Thus y = (0.04)x so that, when x = 2500, we have y = ( 0.04) 2500 = ( 0.04)* 50 = 2