Respuesta :
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
<=> 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!)