MAT137Y1 Midterm: 2005 Test 1 solution

• School: UTSG
• Department: Mathematics
• Course: MAT137Y1
• Professor: N/ A

Document Summary

1. (10%) (a) solve for all x which satis es |2x| +|x + 1| = 3. By de nition of absolute value, x 1, x < 1. x 0, x < 0, If x < 1, then |2x| +|x + 1| = ( 2x) + ( x 1) = 3x 1 = 3 valid solution since we are only looking at values x < 1. If 1 x < 0, then |2x|+|x +1| = ( 2x)+(x +1) = x +1 = 3 solution (since x [ 1,0)). Hence, the equation is satis ed for x = 1. 6, which is also a valid solution. (10%) (b) solve the inequality and express your answer as a union of intervals. Note that the inequality does not make sense for x = 1, 5. 0 = x(x + 5) (x 2)(x + 1)