For which of the following functions is \(\mathrm{f(a + b) = f(a) + f(b)}\) for all positive numbers a and...

1. Translate the problem requirements: We need to find which function satisfies the property \(\mathrm{f(a + b) = f(a) + f(b)}\) for all positive numbers a and b. This means that applying the function to a sum should give the same result as applying the function to each number separately and then adding those results.
2. Test each function systematically with concrete values: Use simple positive numbers like a = 1 and b = 2 to check if \(\mathrm{f(a + b) = f(a) + f(b)}\) holds for each given function.
3. Eliminate functions that fail the test: For each function that doesn't satisfy the equality, we can rule it out immediately since the property must hold for ALL positive numbers.
4. Verify the remaining function: Once we identify a function that works with our test values, confirm it satisfies the property algebraically to ensure it works for all positive numbers.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in plain English. We need to find a function where "the function of a sum equals the sum of the functions."

More specifically, if we have two positive numbers a and b, then applying our function to their sum (a + b) should give us the same result as applying the function to each number separately and then adding those results together.

In mathematical terms: \(\mathrm{f(a + b) = f(a) + f(b)}\)

This is a special property that only certain types of functions have. Let's call this the "additivity property."

Process Skill: TRANSLATE - Converting the mathematical property into understandable language

2. Test each function systematically with concrete values

To make this concrete and manageable, let's use simple test values: a = 1 and b = 2. This means we need to check if \(\mathrm{f(3) = f(1) + f(2)}\) for each function.

Let's test each option:

Option A: \(\mathrm{f(x) = x^2}\)

• \(\mathrm{f(3) = 3^2 = 9}\)
• \(\mathrm{f(1) + f(2) = 1^2 + 2^2 = 1 + 4 = 5}\)
• Since \(\mathrm{9 ≠ 5}\), this function fails the test

Option B: \(\mathrm{f(x) = x + 1}\)

• \(\mathrm{f(3) = 3 + 1 = 4}\)
• \(\mathrm{f(1) + f(2) = (1 + 1) + (2 + 1) = 2 + 3 = 5}\)
• Since \(\mathrm{4 ≠ 5}\), this function fails the test

Option C: \(\mathrm{f(x) = \sqrt{x}}\)

• \(\mathrm{f(3) = \sqrt{3} ≈ 1.73}\)
• \(\mathrm{f(1) + f(2) = \sqrt{1} + \sqrt{2} = 1 + \sqrt{2} ≈ 1 + 1.41 = 2.41}\)
• Since \(\mathrm{\sqrt{3} ≠ 1 + \sqrt{2}}\), this function fails the test

Option D: \(\mathrm{f(x) = \frac{2}{x}}\)

• \(\mathrm{f(3) = \frac{2}{3}}\)
• \(\mathrm{f(1) + f(2) = \frac{2}{1} + \frac{2}{2} = 2 + 1 = 3}\)
• Since \(\mathrm{\frac{2}{3} ≠ 3}\), this function fails the test

Option E: \(\mathrm{f(x) = -3x}\)

• \(\mathrm{f(3) = -3(3) = -9}\)
• \(\mathrm{f(1) + f(2) = -3(1) + (-3)(2) = -3 + (-6) = -9}\)
• Since \(\mathrm{-9 = -9}\), this function passes our test!

3. Eliminate functions that fail the test

Based on our testing with a = 1 and b = 2, we can eliminate options A, B, C, and D because they don't satisfy the required property \(\mathrm{f(a + b) = f(a) + f(b)}\).

Only option E passed our concrete test. However, we need to be thorough since the property must work for ALL positive numbers, not just our test case.

Process Skill: APPLY CONSTRAINTS - Ensuring the property holds universally, not just for specific values

4. Verify the remaining function

Let's verify that \(\mathrm{f(x) = -3x}\) works for any positive numbers a and b, not just our test case.

For \(\mathrm{f(x) = -3x}\):

• \(\mathrm{f(a + b) = -3(a + b) = -3a - 3b}\)
• \(\mathrm{f(a) + f(b) = -3a + (-3b) = -3a - 3b}\)

Since \(\mathrm{-3a - 3b = -3a - 3b}\), the equality \(\mathrm{f(a + b) = f(a) + f(b)}\) holds for ALL positive numbers a and b.

This confirms that \(\mathrm{f(x) = -3x}\) has the additivity property we're looking for. This type of function is called a "linear function" (specifically, a linear function that passes through the origin), and all such functions satisfy this property.

Final Answer

The answer is E. \(\mathrm{f(x) = -3x}\)

This is the only function among the given choices that satisfies \(\mathrm{f(a + b) = f(a) + f(b)}\) for all positive numbers a and b. We verified this both through concrete testing and algebraic verification.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the functional equation property
Students often confuse \(\mathrm{f(a + b) = f(a) + f(b)}\) with other functional properties like \(\mathrm{f(ab) = f(a) + f(b)}\) or \(\mathrm{f(a + b) = f(a) × f(b)}\). This fundamental misunderstanding leads them to test the wrong condition entirely, making all subsequent work incorrect.

2. Overlooking the constraint "for all positive numbers"
Many students miss that the property must hold for ALL positive numbers a and b, not just specific values. They might think testing one pair of numbers is sufficient proof, or they might ignore this universal requirement when analyzing the functions.

3. Attempting to solve algebraically without concrete testing first
Some students jump directly into complex algebraic manipulation for each function without first using simple test values. This approach is more time-consuming and error-prone, especially under test conditions where efficiency matters.

Errors while executing the approach

1. Arithmetic calculation errors during testing
When testing concrete values like \(\mathrm{f(3)}\) vs \(\mathrm{f(1) + f(2)}\), students frequently make basic arithmetic mistakes. For example, with \(\mathrm{f(x) = x^2}\), they might calculate \(\mathrm{1^2 + 2^2 = 1 + 4 = 6}\) instead of 5, or miscalculate \(\mathrm{(-3)(1) + (-3)(2) = -3 + (-6) = -8}\) instead of -9.

2. Incorrect algebraic manipulation during verification
When verifying that \(\mathrm{f(x) = -3x}\) works generally, students might incorrectly expand \(\mathrm{-3(a + b)}\) as \(\mathrm{-3a + 3b}\) instead of \(\mathrm{-3a - 3b}\), or make sign errors when working with the negative coefficient.

3. Using inappropriate test values
Students might choose test values that accidentally work for multiple functions (like a = 0, b = 0, which isn't allowed since we need positive numbers) or values that make calculations unnecessarily complex, increasing the chance of computational errors.

Errors while selecting the answer

1. Stopping after finding the first function that works for test values
Students might test the functions in order and select the first one that works for their specific test case without checking if it works universally or completing the verification step for all remaining options.

2. Selecting a function that works for specific values but not generally
Even after testing, students might convince themselves that a function works based on limited testing. For instance, they might find special values where \(\mathrm{f(x) = x + 1}\) appears to satisfy the condition and select option B without proper verification.