Respuesta :

PSDHE
f(x) = 15 <=> 4|x-5| + 3 = 15
<=> 4|x-5| = 12
<=> |x-5| = 3
<=> x-5 = 3 (1) <=> x-5 = -3 (2)
<=> x = 8 (1) <=> x = 2 (2)
x = 8; x = 2

Start with the definition of absolute value |x|:

|x| = x  for x>=0

|x| = -x  for x<0

We are given a function f(x) and are asking when is that function = 15.

This means we are asking:

4|x-5| + 3 = 15, or simplified:

|x-5| = 3

but at this point further simplification seems stuck on the absolute value. Time to use the above definition, and we split this into two cases:

|x-5| = 3 -->

Case 1: (x-5) = 3   for (x-5)>=0

Case 2: -(x-5) = 3   for (x-5)<0

and now we can proceed and solve each case separately:

Case 1: x = 8  for x >= 5    

Case 2: x = 2  for x < 5

Give it sharp look and realize that both cases are valid because each case has a solution of an x that is "allowed" by the inequality condition (meaning it is not contradicting it), therefore both solutions

x1 = 2 and x2 = 8

are valid solutions to the equation 4|x-5|+3=15. For both the function f(x) will have the value 15 (and please do verify for yourself!)