Question: The function g is defined by \(\mathrm{g(x) = \frac{4}{2x-1}}\). If \(\mathrm{g(a) = 2}\), what is the value of a?1/23/225/23

1. TRANSLATE the problem information

• Given information:
  • Function definition: \(\mathrm{g(x) = \frac{4}{2x-1}}\)
  • Condition: \(\mathrm{g(a) = 2}\)
  • Need to find: the value of a

• What this tells us: We need to substitute a into the function and set it equal to 2

2. TRANSLATE into a solvable equation

• Since \(\mathrm{g(a) = 2}\) and \(\mathrm{g(x) = \frac{4}{2x-1}}\), we substitute:

\(\mathrm{\frac{4}{2a-1} = 2}\)

3. INFER the solution strategy

• To solve for a, we need to eliminate the fraction
• Strategy: Multiply both sides by (2a-1) to clear the denominator

4. SIMPLIFY through algebraic steps

• Multiply both sides by (2a-1):

\(\mathrm{4 = 2(2a-1)}\)

• Distribute the 2:

\(\mathrm{4 = 4a - 2}\)

• Add 2 to both sides:

\(\mathrm{6 = 4a}\)

• Divide by 4:

\(\mathrm{a = \frac{6}{4} = \frac{3}{2}}\)

5. Verify the answer

• Check: \(\mathrm{g(\frac{3}{2}) = \frac{4}{2(\frac{3}{2})-1} = \frac{4}{3-1} = \frac{4}{2} = 2}\) ✓

Answer: B) 3/2

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: When distributing \(\mathrm{2(2a-1)}\), students incorrectly get \(\mathrm{4 = 4a + 2}\) instead of \(\mathrm{4 = 4a - 2}\)

Students think: "2 times (2a-1) equals 2 times 2a plus 2 times negative 1, which is \(\mathrm{4a + 2}\)"

This leads to: \(\mathrm{4 = 4a + 2}\), so \(\mathrm{2 = 4a}\), giving \(\mathrm{a = \frac{1}{2}}\)

This may lead them to select Choice A (1/2)

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly get to \(\mathrm{6 = 4a}\) but make an arithmetic error when simplifying the fraction

They might think: "\(\mathrm{\frac{6}{4}}\) equals 2 because 6 divided by 3 is 2"

This may lead them to select Choice C (2)

The Bottom Line:

This problem requires careful algebraic manipulation, especially when distributing and working with fractions. The key insight is recognizing that function notation creates an equation that can be solved using standard algebraic techniques.