Let g be a function defined for all real numbers x by \(\mathrm{g(x) = |x - 2| + |x +...

1. TRANSLATE the problem information

• Given: \(\mathrm{g(x) = |x - 2| + |x + 2|}\) for all real numbers x
• Given: \(\mathrm{g(7) - g(a) = -8}\) and a is positive
• Find: the value of a

2. Calculate g(7)

• \(\mathrm{g(7) = |7 - 2| + |7 + 2|}\)
• \(\mathrm{= |5| + |9|}\)
• \(\mathrm{= 5 + 9}\)
• \(\mathrm{= 14}\)

3. TRANSLATE the constraint equation

• From \(\mathrm{g(7) - g(a) = -8}\):
• \(\mathrm{14 - g(a) = -8}\)
• SIMPLIFY: \(\mathrm{g(a) = 14 + 8 = 22}\)

4. CONSIDER ALL CASES for the absolute value function

Since we need \(\mathrm{|a - 2| + |a + 2| = 22}\), we must analyze different cases based on where a falls relative to the critical points -2 and 2.

Case 1: \(\mathrm{a \lt -2}\)

• Both \(\mathrm{(a - 2)}\) and \(\mathrm{(a + 2)}\) are negative
• \(\mathrm{|a - 2| = -(a - 2) = -a + 2}\)
• \(\mathrm{|a + 2| = -(a + 2) = -a - 2}\)
• \(\mathrm{g(a) = (-a + 2) + (-a - 2)}\)
• \(\mathrm{= -2a}\)
• Setting \(\mathrm{-2a = 22}\) gives \(\mathrm{a = -11}\)

Case 2: \(\mathrm{-2 \leq a \lt 2}\)

• \(\mathrm{(a - 2)}\) is negative, \(\mathrm{(a + 2)}\) is non-negative
• \(\mathrm{|a - 2| = -(a - 2) = -a + 2}\)
• \(\mathrm{|a + 2| = a + 2}\)
• \(\mathrm{g(a) = (-a + 2) + (a + 2)}\)
• \(\mathrm{= 4}\)
• Since \(\mathrm{4 \neq 22}\), no solution exists in this interval

Case 3: \(\mathrm{a \geq 2}\)

• Both \(\mathrm{(a - 2)}\) and \(\mathrm{(a + 2)}\) are non-negative
• \(\mathrm{|a - 2| = a - 2}\)
• \(\mathrm{|a + 2| = a + 2}\)
• \(\mathrm{g(a) = (a - 2) + (a + 2)}\)
• \(\mathrm{= 2a}\)
• Setting \(\mathrm{2a = 22}\) gives \(\mathrm{a = 11}\)

5. APPLY CONSTRAINTS to select final answer

• We found two potential solutions: \(\mathrm{a = -11}\) and \(\mathrm{a = 11}\)
• Since the problem states that a is positive: \(\mathrm{a = 11}\)

Answer: D (11)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES skill: Students often try to solve \(\mathrm{|a - 2| + |a + 2| = 22}\) without systematically considering the different cases based on the critical points.

They might attempt to "remove" the absolute value signs without properly analyzing when the expressions inside are positive or negative. This leads to incorrect equations like \(\mathrm{(a - 2) + (a + 2) = 22}\), giving \(\mathrm{2a = 22}\) and \(\mathrm{a = 11}\), which happens to be correct by luck, but misses the full mathematical reasoning.

Alternatively, they might get confused by the multiple cases and abandon systematic solution, leading to guessing.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students correctly find both \(\mathrm{a = -11}\) and \(\mathrm{a = 11}\) as mathematical solutions, but fail to apply the constraint that a is positive.

This may lead them to select Choice A (-11) or causes confusion about which answer to choose.

The Bottom Line:

This problem challenges students to work systematically with absolute value functions by cases while keeping track of multiple constraints. Success requires methodical case analysis combined with careful attention to the given conditions.