The function f is exponential and satisfies \(\mathrm{f(x + 1) = \frac{1}{5} f(x)}\) for all real numbers x. It is...

1. TRANSLATE the problem information

• Given information:
  • f is exponential
  • \(\mathrm{f(x + 1) = (1/5) f(x)}\) for all real x
  • \(\mathrm{f(-2) = 625}\)
• What this tells us: We have a recursive relationship and one specific function value

2. INFER the exponential function structure

• Since f is exponential, it has the form \(\mathrm{f(x) = A \cdot b^x}\) for constants A and b
• The key insight: \(\mathrm{f(x + 1) = (1/5) f(x)}\) means the ratio between consecutive values is constant at 1/5
• For \(\mathrm{f(x) = A \cdot b^x}\): \(\mathrm{f(x + 1) = A \cdot b^{(x+1)} = b \cdot (A \cdot b^x) = b \cdot f(x)}\)
• Therefore: \(\mathrm{b = 1/5}\)

3. SIMPLIFY to find the coefficient A

• Now we know \(\mathrm{f(x) = A(1/5)^x}\), but we need to find A
• Use the given point \(\mathrm{f(-2) = 625}\):
  \(\mathrm{f(-2) = A(1/5)^{-2} = 625}\)
• Apply negative exponent rule: \(\mathrm{(1/5)^{-2} = (5/1)^2 = 25}\)
• So: \(\mathrm{A \cdot 25 = 625}\)
• Therefore: \(\mathrm{A = 625/25 = 25}\)

4. INFER the complete function and verify

• Our function is \(\mathrm{f(x) = 25(1/5)^x}\)
• Quick verification:
  • \(\mathrm{f(x+1) = 25(1/5)^{(x+1)} = 25 \cdot (1/5) \cdot (1/5)^x = (1/5) \cdot f(x)}\) ✓
  • \(\mathrm{f(-2) = 25(1/5)^{-2} = 25 \cdot 25 = 625}\) ✓

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that the recursive relationship \(\mathrm{f(x+1) = (1/5)f(x)}\) directly tells them the base of the exponential function.

Instead, they might try to work backwards from the answer choices or get confused about how to use the recursive relationship. Without this key insight, they can't set up the correct general form \(\mathrm{f(x) = A(1/5)^x}\), leading to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution with negative exponents: Students correctly identify that \(\mathrm{f(x) = A(1/5)^x}\) but make computational errors when solving for A.

They might calculate \(\mathrm{(1/5)^{-2} = 1/25}\) instead of 25, leading to \(\mathrm{A = 625 \times 25 = 15,625}\). This massive coefficient doesn't match any answer choice, causing them to doubt their approach and potentially select Choice B [\(\mathrm{(1/5)^x}\)] as the "simplest" exponential form.

The Bottom Line:

This problem requires students to bridge the gap between recursive relationships and exponential function structure - a connection that isn't always obvious to students who typically see these concepts taught separately.