Find all the solutions of the equation |x + 1| - |x| + 3|x - 1| - 2|x - 2| = x +

Question

Find all the solutions of the equation
|x + 1| - |x| + 3|x - 1| - 2|x - 2| = x + 2 .

Explanation / Answer

Let f (x) = |x + 1| - |x| + 3|x - 1| - 2|x - 2| - (x + 2). We have to solve f (x) = 0 in the ?ve regions of the x-axis determined by the modulus functions, namely x 6 -1 ; -1 6 x 6 0 ; 0 6 x 6 1 ; 1 6 x 6 2 ; 2 6 x . In the separate regions, we have f (x) = ? ? ? (x + 1) - x + 3(x - 1) - 2(x - 2) - (x + 2) = 0 for 2 6 x < 8 (x + 1) - x + 3(x - 1) + 2(x - 2) - (x + 2) = 4x - 8 for 1 6 x 6 2 (x + 1) - x - 3(x - 1) + 2(x - 2) - (x + 2) = -2x - 2 for 0 6 x 6 1 (x + 1) + x - 3(x - 1) + 2(x - 2) - (x + 2) = -2 for -1 6 x 6 0 -(x + 1) + x - 3(x - 1) + 2(x - 2) - (x + 2) = -2x - 4 for -8 2 Here, f (x) = 0 so the equation is satis?ed for all values of x. f (-2) = 0 and f (x) = 0 for any x > 2. (ii) 1 6 x 6 2 Here, f (x) = 0 only if x = 2. (iii) 0 6 x 6 1 Here, f (x) = 0 only if x = -1. This is not a solution since the point x = -1 does not lie in the region 0 6 x 6 1. (iv) -1 6 x 6 0 Here, f (x) = -2 so there is no solution. (v) x 6 -1 Here, f (x) = 0 only if x = -2. This is a solution since the point x = -2 does lie in the region x 6 -1. The equation f (x) = 0 is therefore satis?ed by x = -2 and by any x greater than, or equal to, 2.

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