The table below gives the values of the function g for some values of x. Which of the following equations...

1. INFER the function pattern from the table

• Given data points:
  • When \(\mathrm{x = 0}\), \(\mathrm{g(x) = 8}\)
  • When \(\mathrm{x = 1}\), \(\mathrm{g(x) = 4}\)
  • When \(\mathrm{x = 2}\), \(\mathrm{g(x) = 2}\)
  • When \(\mathrm{x = 3}\), \(\mathrm{g(x) = 1}\)

• Key insight: As x increases by 1, g(x) is divided by 2 each time (8 → 4 → 2 → 1). This constant ratio indicates an exponential function with decay.

2. TRANSLATE the answer choices into testable expressions

• We need to test which equation produces our known points
• Start with the easiest point to check: (0, 8)

3. SIMPLIFY by testing x = 0 for each choice

• (A) \(\mathrm{g(0) = 8(2^{(0 + 1)})}\)
  \(\mathrm{= 8(2^1)}\)
  \(\mathrm{= 16}\) ❌
• (B) \(\mathrm{g(0) = 8(2^0)}\)
  \(\mathrm{= 8(1)}\)
  \(\mathrm{= 8}\) ✓
• (C) \(\mathrm{g(0) = 8(2^{(-0 + 1)})}\)
  \(\mathrm{= 8(2^1)}\)
  \(\mathrm{= 16}\) ❌
• (D) \(\mathrm{g(0) = 8(2^{(-0)})}\)
  \(\mathrm{= 8(2^0)}\)
  \(\mathrm{= 8(1)}\)
  \(\mathrm{= 8}\) ✓

Only choices B and D work for x = 0.

4. SIMPLIFY by testing x = 1 for remaining choices

• (B) \(\mathrm{g(1) = 8(2^1)}\)
  \(\mathrm{= 8(2)}\)
  \(\mathrm{= 16 \neq 4}\) ❌
• (D) \(\mathrm{g(1) = 8(2^{-1})}\)
  \(\mathrm{= 8(\frac{1}{2})}\)
  \(\mathrm{= 4}\) ✓

Only choice D works for both points.

5. INFER verification is complete

• Since D works for the first two points and represents exponential decay (which matches our pattern), it must be correct
• Optional check: \(\mathrm{g(2) = 8(2^{-2})}\)
  \(\mathrm{= 8(\frac{1}{4})}\)
  \(\mathrm{= 2}\) ✓

Answer: D

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the halving pattern as exponential decay. Instead, they might try to find a linear relationship or guess randomly among the choices without systematic testing.

Without recognizing that g(x) is being halved each time x increases by 1, students miss the key insight that this is exponential decay requiring a negative exponent. This leads to confusion and guessing among the choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students make errors when evaluating negative exponents, particularly confusing \(\mathrm{2^{-x}}\) with \(\mathrm{-2^x}\), or incorrectly calculating expressions like \(\mathrm{2^{-1} = \frac{1}{2}}\).

For example, they might incorrectly calculate \(\mathrm{g(1) = 8(2^{-1})}\) as \(\mathrm{8(-2) = -16}\) instead of \(\mathrm{8(\frac{1}{2}) = 4}\). This may lead them to select Choice B (\(\mathrm{g(x) = 8(2^x)}\)) because it seems to work for x = 0 but they miscalculate the other points.

The Bottom Line:

This problem tests whether students can connect table patterns to exponential function forms. The key breakthrough is recognizing that constant division (halving) indicates exponential decay with a negative exponent, not exponential growth.