Question:\(\mathrm{h(x) = 6(3)^x}\)The function h is defined by the given equation. If \(\mathrm{j(x) = h(x - 1)}\), which of the...
GMAT Advanced Math : (Adv_Math) Questions
\(\mathrm{h(x) = 6(3)^x}\)
The function h is defined by the given equation. If \(\mathrm{j(x) = h(x - 1)}\), which of the following equations defines the function j?
\(\mathrm{j(x) = 2(3)^x}\)
\(\mathrm{j(x) = 18(3)^x}\)
\(\mathrm{j(x) = 6(2)^x}\)
\(\mathrm{j(x) = 6(3)^{(x-1)}}\)
1. TRANSLATE the function relationship
- Given information:
- \(\mathrm{h(x) = 6(3)^x}\)
- \(\mathrm{j(x) = h(x - 1)}\)
- What this means: Wherever we see x in h(x), we replace it with (x - 1)
2. TRANSLATE into mathematical form
- Substitute (x - 1) into h(x):
\(\mathrm{j(x) = h(x - 1) = 6(3)^{(x-1)}}\) - This gives us \(\mathrm{j(x) = 6(3)^{(x-1)}}\)
3. SIMPLIFY using exponent rules
- Apply the rule \(\mathrm{a^{(m-n)} = a^m \times a^{(-n)}}\):
\(\mathrm{6(3)^{(x-1)} = 6 \times 3^x \times 3^{(-1)}}\) - Since \(\mathrm{3^{(-1)} = \frac{1}{3}}\):
\(\mathrm{= 6 \times 3^x \times \left(\frac{1}{3}\right)}\) - Multiply the constants:
\(\mathrm{= \left(\frac{6}{3}\right) \times 3^x = 2 \times 3^x = 2(3)^x}\)
Answer: A
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY execution: Students correctly get to \(\mathrm{j(x) = 6(3)^{(x-1)}}\) but then struggle with the exponent rules. They might try to "distribute" the exponent: \(\mathrm{6(3)^{(x-1)} = 6(3^x - 3^1) = 6(3^x - 3) = 6(3^x) - 18}\), leading to confusion since this doesn't match any answer choice format. This leads to guessing among the choices.
Second Most Common Error:
Incomplete TRANSLATE reasoning: Some students see \(\mathrm{j(x) = h(x-1)}\) and think it means "shift the coefficient" rather than "substitute (x-1) for x." They might incorrectly reason that since \(\mathrm{h(x) = 6(3)^x}\), then j(x) should be \(\mathrm{6(3)^x}\) multiplied by 3 (thinking about horizontal shifts incorrectly), leading them toward Choice B (\(\mathrm{18(3)^x}\)).
The Bottom Line:
This problem tests whether students can properly execute function substitution followed by careful algebraic simplification. The key insight is that \(\mathrm{j(x) = h(x-1)}\) means literal substitution, not a transformation of the function's graph properties.
\(\mathrm{j(x) = 2(3)^x}\)
\(\mathrm{j(x) = 18(3)^x}\)
\(\mathrm{j(x) = 6(2)^x}\)
\(\mathrm{j(x) = 6(3)^{(x-1)}}\)