For all nonzero real numbers s, a function p satisfies \(\mathrm{s \cdot p(s) = 18}\). What is the value of...

1. TRANSLATE the problem information

• Given information:
  • For all nonzero real numbers s: \(\mathrm{s \cdot p(s) = 18}\)
  • Need to find: \(\mathrm{p(-24)}\)

• What this tells us: We have a functional equation that must work for any nonzero value of s

2. INFER the approach

• To find \(\mathrm{p(-24)}\), we need to know what \(\mathrm{p(s)}\) equals in general
• Since \(\mathrm{s \cdot p(s) = 18}\), we can solve for \(\mathrm{p(s)}\) by dividing both sides by s
• Then substitute \(\mathrm{s = -24}\) into our general formula

3. SIMPLIFY to find the general form

• Starting with: \(\mathrm{s \cdot p(s) = 18}\)
• Divide both sides by s: \(\mathrm{p(s) = \frac{18}{s}}\)
• This gives us the general form of the function

4. SIMPLIFY to evaluate p(−24)

• Substitute \(\mathrm{s = -24}\) into \(\mathrm{p(s) = \frac{18}{s}}\):

\(\mathrm{p(-24) = \frac{18}{-24}}\)

• Perform the division:

\(\mathrm{\frac{18}{-24} = \frac{-18}{24}}\)

• Reduce the fraction:

\(\mathrm{\frac{-18}{24} = \frac{-3}{4}}\)

Answer: \(\mathrm{\frac{-3}{4}}\) (Choice B)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize they need to solve for \(\mathrm{p(s)}\) first before substituting \(\mathrm{s = -24}\).

Instead, they might try to substitute \(\mathrm{s = -24}\) directly into \(\mathrm{s \cdot p(s) = 18}\), getting \(\mathrm{(-24) \cdot p(-24) = 18}\), and then feel confused about how to proceed since they still have \(\mathrm{p(-24)}\) as an unknown. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly find \(\mathrm{p(s) = \frac{18}{s}}\) but make arithmetic errors when computing \(\mathrm{\frac{18}{-24}}\).

Common mistakes include getting \(\mathrm{\frac{18}{24} = \frac{3}{4}}\) (forgetting the negative sign) or incorrectly reducing fractions. This may lead them to select Choice (C) \(\mathrm{\frac{3}{4}}\) or Choice (D) \(\mathrm{\frac{4}{3}}\).

The Bottom Line:

This problem tests whether students can work backwards from a functional equation to find the general form of a function, then apply it to a specific case. The key insight is recognizing that solving for \(\mathrm{p(s)}\) first makes the evaluation straightforward.