For the exponential function f, the value of \(\mathrm{f(1)}\) is k, where k is a constant. Which of the following...

1. TRANSLATE the problem information

• Given information:
  • f is an exponential function
  • \(\mathrm{f(1) = k}\) (where k is a constant)
  • Need to find which form shows k as either the coefficient or the base

• What this means: We need to calculate \(\mathrm{f(1)}\) for each choice, then check if that result equals either the coefficient (the number in front) or the base (the number being raised to a power).

2. INFER the approach

• Strategy: Substitute \(\mathrm{x = 1}\) into each function form
• Look for the choice where \(\mathrm{f(1)}\) equals either the coefficient or the base in that expression
• The coefficient is the number multiplying the exponential term; the base is inside the parentheses

3. SIMPLIFY each choice by evaluating f(1)

Choice A: \(\mathrm{f(x) = 50(2)^{(x+1)}}\)

• \(\mathrm{f(1) = 50(2)^{(1+1)}}\)
  \(\mathrm{= 50(2)^2}\)
  \(\mathrm{= 50(4)}\)
  \(\mathrm{= 200}\)
• \(\mathrm{k = 200}\), but coefficient = 50 and base = 2. No match.

Choice B: \(\mathrm{f(x) = 80(2)^x}\)

• \(\mathrm{f(1) = 80(2)^1}\)
  \(\mathrm{= 80(2)}\)
  \(\mathrm{= 160}\)
• \(\mathrm{k = 160}\), but coefficient = 80 and base = 2. No match.

Choice C: \(\mathrm{f(x) = 128(2)^{(x-1)}}\)

• \(\mathrm{f(1) = 128(2)^{(1-1)}}\)
  \(\mathrm{= 128(2)^0}\)
  \(\mathrm{= 128(1)}\)
  \(\mathrm{= 128}\)
• \(\mathrm{k = 128}\), and coefficient = 128. Perfect match!

Choice D: \(\mathrm{f(x) = 205(2)^{(x-2)}}\)

• \(\mathrm{f(1) = 205(2)^{(1-2)}}\)
  \(\mathrm{= 205(2)^{(-1)}}\)
  \(\mathrm{= 205(1/2)}\)
  \(\mathrm{= 102.5}\)
• \(\mathrm{k = 102.5}\), but coefficient = 205 and base = 2. No match.

4. INFER the final answer

• Only Choice C produces a k-value that matches one of the visible numbers in the function form
• \(\mathrm{k = 128}\) appears as the coefficient in \(\mathrm{f(x) = 128(2)^{(x-1)}}\)

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE reasoning: Students misunderstand what "shows the value of k as the coefficient or the base" means. They might think they need to find k first, then look for it in the choices, rather than recognizing they need to evaluate \(\mathrm{f(1)}\) for each choice and see where that result appears.

This leads to confusion and random guessing rather than systematic evaluation.

Second Most Common Error:

Poor SIMPLIFY execution: Students make calculation errors, especially with exponent rules. They might incorrectly calculate \(\mathrm{2^0}\) or struggle with negative exponents like \(\mathrm{2^{(-1)}}\).

For example, they might think \(\mathrm{2^0 = 0}\) instead of \(\mathrm{2^0 = 1}\), leading them to calculate \(\mathrm{f(1) = 128(0) = 0}\) for Choice C, which would make them reject the correct answer.

The Bottom Line:

This problem tests whether students can systematically evaluate exponential expressions and understand the relationship between function values and the structural elements (coefficient, base) of exponential function forms. The key insight is recognizing that when the exponent becomes zero, the exponential term equals 1, leaving just the coefficient.