Absolute value Question

Question

how to solve the following absolute values.   E=element

A.     l-x^2+4x-5l     the answer is -(-x^2+4x-5) if x E (-infinity,+infinity)

B.     lxl + lx+3l        the answer is 2x+3 if x is E (0, +infinity), 3 if X E (-3,0) , -2x-3 if x E (-infinity, -3)

 

 

how do I do these. Thankyou for your help

 

Answer 1

l A l = A if A ≥ 0 and l A l = -A if A < 0.

 

-x²+4x-5 = -[x²-4x+5] = -[x²-4x+4+1] = -[(x-2)²+1]

                                                       < 0 for all x

 

So, l-x²+4x-5l = -(x²+4x-5) for x ε (-∞,∞).

 

------------------------------------------------------------------------------------

If x < -3, then x < 0 and x + 3 < 0, so lxl + lx+3l = -x + -(x+3)

                                                                      = -2x - 3

 

If -3 ≤ x < 0, then x+3 > 0 and x < 0, so lxl + lx+3l = -x + x + 3

                                                                          = 3

 

If x ≥ 0 , then x + 3 > 0, so lxl + lx+3l = x + x + 3 = 2x + 3