Some values of the linear function f are shown in the table above. What is the value of \(\mathrm{f(3)}\)?x\(\mathrm{f(x)}\)0-224616

1. INFER the key property of linear functions

• Given information:
  • \(\mathrm{f(0) = -2}\), \(\mathrm{f(2) = 4}\), \(\mathrm{f(6) = 16}\)
  • \(\mathrm{f}\) is a linear function
• Key insight: Linear functions have a constant rate of change (slope)

2. INFER the solution strategy

• Since the rate of change is constant, I can use any two points to find it
• Then apply this rate to move from a known value to \(\mathrm{f(3)}\)

3. SIMPLIFY to find the rate of change

• Using points (0, -2) and (2, 4):
• Rate = \(\mathrm{\frac{4 - (-2)}{2 - 0}}\)
  \(\mathrm{= \frac{6}{2}}\)
  \(\mathrm{= 3}\)
• This means \(\mathrm{f(x)}\) increases by 3 for every 1-unit increase in x

4. SIMPLIFY to find \(\mathrm{f(3)}\)

• Since \(\mathrm{f(2) = 4}\) and the rate is 3:
• \(\mathrm{f(3) = f(2) + 3}\)
  \(\mathrm{= 4 + 3}\)
  \(\mathrm{= 7}\)

Answer: B. 7

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students may not recognize that they can use the constant rate of change property efficiently. Instead, they might try to set up a full linear equation \(\mathrm{f(x) = mx + b}\), making unnecessary calculations and increasing chance for arithmetic errors.

While this approach can work, it's more complex and time-consuming, potentially leading to computational mistakes that result in selecting Choice A (6), Choice C (8), or Choice D (9).

Second Most Common Error:

Poor SIMPLIFY execution: Students understand the concept but make arithmetic errors when calculating the rate of change. For example, calculating \(\mathrm{\frac{4 - (-2)}{2 - 0}}\) incorrectly as 2/2 = 1 instead of 6/2 = 3.

This leads to \(\mathrm{f(3) = f(2) + 1 = 4 + 1 = 5}\), which isn't among the choices, causing confusion and guessing.

The Bottom Line:

This problem tests whether students can efficiently use the defining property of linear functions rather than getting bogged down in unnecessary algebraic setup. The key insight is recognizing that constant rate of change gives you a direct path to the answer.