The function g is defined by \(\mathrm{g(x) = \frac{|x|}{a} - 14}\), where a lt 0. What is the product of...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{g(x) = \frac{|x|}{a} - 14}\)
  • \(\mathrm{a \lt 0}\)
  • Need to find \(\mathrm{g(15a) \times g(7a)}\)

2. INFER the key insight about signs

• Since \(\mathrm{a \lt 0}\) (given), we can determine:
  • \(\mathrm{15a = (positive) \times (negative) = negative}\)
  • \(\mathrm{7a = (positive) \times (negative) = negative}\)
• This means we'll need the "negative input" case of absolute value

3. TRANSLATE to find g(15a)

• Substitute 15a into the function:

\(\mathrm{g(15a) = \frac{|15a|}{a} - 14}\)

• Since \(\mathrm{15a \lt 0}\), we have \(\mathrm{|15a| = -(15a) = -15a}\)
• So:

\(\mathrm{g(15a) = \frac{-15a}{a} - 14}\)

4. SIMPLIFY the first function value

\(\mathrm{g(15a) = \frac{-15a}{a} - 14}\)

\(\mathrm{= -15 - 14}\)

\(\mathrm{= -29}\)

5. TRANSLATE to find g(7a)

• Substitute 7a into the function:

\(\mathrm{g(7a) = \frac{|7a|}{a} - 14}\)

• Since \(\mathrm{7a \lt 0}\), we have \(\mathrm{|7a| = -(7a) = -7a}\)
• So:

\(\mathrm{g(7a) = \frac{-7a}{a} - 14}\)

6. SIMPLIFY the second function value

\(\mathrm{g(7a) = \frac{-7a}{a} - 14}\)

\(\mathrm{= -7 - 14}\)

\(\mathrm{= -21}\)

7. SIMPLIFY to find the final product

\(\mathrm{g(15a) \times g(7a) = (-29) \times (-21)}\)

\(\mathrm{= 609}\)

Answer: 609

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Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that since \(\mathrm{a \lt 0}\), the expressions 15a and 7a are negative.

They might incorrectly assume \(\mathrm{|15a| = 15a}\) and \(\mathrm{|7a| = 7a}\), leading to:

• \(\mathrm{g(15a) = \frac{15a}{a} - 14 = 15 - 14 = 1}\)
• \(\mathrm{g(7a) = \frac{7a}{a} - 14 = 7 - 14 = -7}\)
• \(\mathrm{Product = 1 \times (-7) = -7}\)

This causes confusion when their answer doesn't match any reasonable expectation, leading to guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify that 15a and 7a are negative but make arithmetic errors in the final steps.

For example, they might get \(\mathrm{g(15a) = -29}\) and \(\mathrm{g(7a) = -21}\) but then compute \(\mathrm{(-29) \times (-21)}\) incorrectly, perhaps getting -609 instead of 609 by forgetting that negative times negative equals positive.

The Bottom Line:

This problem tests whether students can correctly apply the absolute value definition when dealing with expressions involving a negative parameter. The key insight is recognizing the sign implications of the given constraint \(\mathrm{a \lt 0}\).