\(\mathrm{g(x) = 32(2^x)}\) Which table gives three values of x and their corresponding values of \(\mathrm{g(x)}\) for function g?...

1. TRANSLATE the problem information

• Given function: \(\mathrm{g(x) = 32(2^x)}\)
• Need to: Find g(x) values for x = -1, 0, and 1
• What this means: Substitute each x value into the function and compute the result

2. SIMPLIFY by evaluating each case

For x = -1:

• \(\mathrm{g(-1) = 32(2^{-1})}\)
• Since \(\mathrm{2^{-1} = \frac{1}{2}}\)
• \(\mathrm{g(-1) = 32 \times \frac{1}{2} = 16}\)

For x = 0:

• \(\mathrm{g(0) = 32(2^0)}\)
• Since \(\mathrm{2^0 = 1}\)
• \(\mathrm{g(0) = 32 \times 1 = 32}\)

For x = 1:

• \(\mathrm{g(1) = 32(2^1)}\)
• Since \(\mathrm{2^1 = 2}\)
• \(\mathrm{g(1) = 32 \times 2 = 64}\)

3. APPLY CONSTRAINTS to select the matching table

• Our computed values: \(\mathrm{(-1, 16), (0, 32), (1, 64)}\)
• Only Choice D matches all three values

Answer: D

Why Students Usually Falter on This Problem

Most Common Error Path:

Conceptual gap with exponent rules: Students often forget that \(\mathrm{2^{-1} = \frac{1}{2}}\), thinking instead that negative exponents make the whole result negative, or they might compute \(\mathrm{2^{-1} = -2}\).

This leads to \(\mathrm{g(-1) = 32(-2) = -64}\), making them select Choice A (-64).

Second Most Common Error:

Conceptual confusion about zero exponents: Students sometimes think \(\mathrm{2^0 = 0}\) instead of \(\mathrm{2^0 = 1}\).

This leads to \(\mathrm{g(0) = 32(0) = 0}\), which combined with other errors might make them select Choice A or causes confusion leading to guessing.

The Bottom Line:

This problem tests whether students truly understand exponent rules in a function context. The key insight is recognizing that negative and zero exponents follow specific rules that must be applied before multiplying by the coefficient 32.