The function f is defined by \(\mathrm{f(x) = 6 + \sqrt{x}}\). What is the value of \(\mathrm{f(36)}\)?

1. TRANSLATE the question into mathematical operations

• Given information:
  • Function definition: \(\mathrm{f(x) = 6 + \sqrt{x}}\)
  • Need to find: \(\mathrm{f(36)}\)
• What this means: Substitute \(\mathrm{x = 36}\) into the function expression

2. SIMPLIFY by substituting and evaluating

• Replace x with 36 in the function:
  \(\mathrm{f(36) = 6 + \sqrt{36}}\)
• Evaluate the square root:
  \(\mathrm{\sqrt{36} = 6}\)
• Complete the addition:
  \(\mathrm{f(36) = 6 + 6 = 12}\)

Answer: 12

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make calculation errors with the square root, perhaps forgetting that \(\mathrm{\sqrt{36} = 6}\) or confusing it with other operations.

For example, some students might think \(\mathrm{\sqrt{36} = 18}\) (confusing square root with doubling) or get confused about which perfect squares they know. This leads to incorrect final calculations like \(\mathrm{f(36) = 6 + 18 = 24}\) or other wrong values.

Second Most Common Error:

Poor TRANSLATE reasoning: Students might not fully understand function notation and what it means to evaluate \(\mathrm{f(36)}\).

They might treat \(\mathrm{f(36)}\) as multiplication (\(\mathrm{f \times 36}\)) or get confused about how to use the function definition. This leads to completely wrong approaches and typically results in guessing.

The Bottom Line:

This problem tests whether students can perform the fundamental operation of function evaluation - substituting a value and simplifying the result. Success requires both understanding function notation and accurately computing with square roots.