1. TRANSLATE the problem information
- Given: Function \(\mathrm{h(x) = \frac{45}{3^x}}\)
- Need to find: \(\mathrm{h(-1)}\), \(\mathrm{h(0)}\), and \(\mathrm{h(1)}\)
- What this means: Substitute each x-value and calculate the result
2. INFER the approach
- We need to substitute each x-value into the function
- The key challenge will be handling the negative exponent correctly
- Work through each value systematically
3. SIMPLIFY for x = -1
- \(\mathrm{h(-1) = \frac{45}{3^{-1}}}\)
- Apply negative exponent rule: \(\mathrm{3^{-1} = \frac{1}{3}}\)
- So: \(\mathrm{h(-1) = \frac{45}{\frac{1}{3}} = 45 \times 3 = 135}\)
4. SIMPLIFY for x = 0
- \(\mathrm{h(0) = \frac{45}{3^0}}\)
- Apply zero exponent rule: \(\mathrm{3^0 = 1}\)
- So: \(\mathrm{h(0) = \frac{45}{1} = 45}\)
5. SIMPLIFY for x = 1
- \(\mathrm{h(1) = \frac{45}{3^1} = \frac{45}{3} = 15}\)
6. TRANSLATE results to table format
- \(\mathrm{x = -1}\) gives \(\mathrm{h(x) = 135}\)
- \(\mathrm{x = 0}\) gives \(\mathrm{h(x) = 45}\)
- \(\mathrm{x = 1}\) gives \(\mathrm{h(x) = 15}\)
- This matches table D: 135, 45, 15
Answer: D
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak conceptual knowledge of negative exponents: Students often think \(\mathrm{3^{-1} = -3}\) instead of \(\mathrm{\frac{1}{3}}\). They apply the negative sign to the base rather than understanding it creates a reciprocal.
Using this incorrect reasoning: \(\mathrm{h(-1) = \frac{45}{-3} = -15}\), which would lead them to select Choice B (-15, 45, 135).
Second Most Common Error:
Poor SIMPLIFY execution with order of operations: Students might incorrectly think they should calculate \(\mathrm{\frac{45}{3}}\) first to get 15, then apply exponent rules to get \(\mathrm{15^{-1}}\), \(\mathrm{15^0}\), \(\mathrm{15^1}\).
This flawed approach gives them \(\mathrm{\frac{1}{15}}\), 15, 15 (or similar confused results), potentially leading them to select Choice C (15, 45, 15) by partially matching some values.
The Bottom Line:
This problem tests whether students truly understand negative exponents as reciprocals rather than just "making numbers negative." The exponential decay pattern \(\mathrm{135 \rightarrow 45 \rightarrow 15}\) should make sense once students correctly apply exponent rules.
