For which of the following functions f is \(\mathrm{f(x)} = \mathrm{f(1-x)}\) for all x?

1. Translate the problem requirements: We need to find which function satisfies \(\mathrm{f(x) = f(1-x)}\) for ALL values of x. This means the function has a specific symmetry property - whatever output we get for input x, we should get the same output for input \(\mathrm{(1-x)}\).
2. Develop a systematic testing approach: Since we need to verify this property holds for ALL x, we'll substitute \(\mathrm{(1-x)}\) into each function and check if the result equals the original function \(\mathrm{f(x)}\).
3. Apply algebraic verification to each choice: For each answer choice, calculate \(\mathrm{f(1-x)}\) by replacing every x with \(\mathrm{(1-x)}\), then simplify and compare with the original \(\mathrm{f(x)}\).
4. Identify the function with perfect symmetry: The correct answer will be the function where \(\mathrm{f(1-x)}\) simplifies to exactly the same expression as \(\mathrm{f(x)}\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what the condition \(\mathrm{f(x) = f(1-x)}\) means in simple terms. Think of it this way: if we have a function and we put in some number x, we get an output. The condition tells us that if we put in \(\mathrm{(1-x)}\) instead, we should get exactly the same output.

For example, if \(\mathrm{x = 0.3}\), then \(\mathrm{(1-x) = 0.7}\). The condition means \(\mathrm{f(0.3)}\) must equal \(\mathrm{f(0.7)}\). Similarly, if \(\mathrm{x = 0.8}\), then \(\mathrm{(1-x) = 0.2}\), so \(\mathrm{f(0.8)}\) must equal \(\mathrm{f(0.2)}\).

This creates a symmetry around the point \(\mathrm{x = 0.5}\) (since when \(\mathrm{x = 0.5}\), we have \(\mathrm{1-x = 0.5}\) as well).

Process Skill: TRANSLATE - Converting the mathematical condition into concrete, understandable terms

2. Develop a systematic testing approach

To check if a function satisfies \(\mathrm{f(x) = f(1-x)}\) for ALL values of x, we need a reliable method. Here's our plan:

• Take each answer choice function \(\mathrm{f(x)}\)
• Replace every x in the function with \(\mathrm{(1-x)}\) to get \(\mathrm{f(1-x)}\)
• Simplify \(\mathrm{f(1-x)}\) completely
• Compare the simplified result with the original \(\mathrm{f(x)}\)
• If they're identical, we found our answer

This systematic approach ensures we don't miss anything and can confidently eliminate wrong choices.

3. Apply algebraic verification to each choice

Now let's test each function systematically:

Choice A: \(\mathrm{f(x) = 1 - x}\)
\(\mathrm{f(1-x) = 1 - (1-x) = 1 - 1 + x = x}\)
Since \(\mathrm{f(x) = 1 - x}\) and \(\mathrm{f(1-x) = x}\), these are not equal. ❌

Choice B: \(\mathrm{f(x) = 1 - x^2}\)
\(\mathrm{f(1-x) = 1 - (1-x)^2 = 1 - (1 - 2x + x^2) = 1 - 1 + 2x - x^2 = 2x - x^2}\)
Since \(\mathrm{f(x) = 1 - x^2}\) and \(\mathrm{f(1-x) = 2x - x^2}\), these are not equal. ❌

Choice C: \(\mathrm{f(x) = x^2 - (1-x)^2}\)
First, let's simplify the original: \(\mathrm{f(x) = x^2 - (1-2x+x^2) = x^2 - 1 + 2x - x^2 = 2x - 1}\)
Now \(\mathrm{f(1-x) = 2(1-x) - 1 = 2 - 2x - 1 = 1 - 2x}\)
Since \(\mathrm{f(x) = 2x - 1}\) and \(\mathrm{f(1-x) = 1 - 2x}\), these are not equal. ❌

Choice D: \(\mathrm{f(x) = x^2(1-x)^2}\)
\(\mathrm{f(1-x) = (1-x)^2(1-(1-x))^2 = (1-x)^2(1-1+x)^2 = (1-x)^2x^2}\)
Since multiplication is commutative: \(\mathrm{(1-x)^2x^2 = x^2(1-x)^2}\)
Therefore \(\mathrm{f(x) = f(1-x)}\) ✓

Choice E: \(\mathrm{f(x) = \frac{x}{1-x}}\)
\(\mathrm{f(1-x) = \frac{1-x}{1-(1-x)} = \frac{1-x}{1-1+x} = \frac{1-x}{x}}\)
Since \(\mathrm{f(x) = \frac{x}{1-x}}\) and \(\mathrm{f(1-x) = \frac{1-x}{x}}\), these are not equal. ❌

4. Identify the function with perfect symmetry

Only Choice D satisfies our condition. Let's verify this makes intuitive sense:

The function \(\mathrm{f(x) = x^2(1-x)^2}\) involves both x and \(\mathrm{(1-x)}\) in a perfectly balanced way - both terms are squared and multiplied together. When we swap x with \(\mathrm{(1-x)}\), we get \(\mathrm{(1-x)^2 \times x^2}\), which is exactly the same due to the commutative property of multiplication.

This symmetry makes \(\mathrm{f(x) = x^2(1-x)^2}\) the only function that satisfies \(\mathrm{f(x) = f(1-x)}\) for all values of x.

Final Answer

The answer is D. \(\mathrm{f(x) = x^2(1-x)^2}\)

This is the only function where \(\mathrm{f(x) = f(1-x)}\) for all x, demonstrated by our systematic verification that \(\mathrm{f(1-x) = (1-x)^2x^2 = x^2(1-x)^2 = f(x)}\).

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the symmetry condition: Students often struggle to grasp what \(\mathrm{f(x) = f(1-x)}\) actually means. They might think it's asking for \(\mathrm{f(x) = f(-x)}\) (even symmetry) or get confused about the "1-x" transformation. The key insight that this creates symmetry around \(\mathrm{x = 0.5}\) is frequently missed.

2. Choosing inefficient verification methods: Instead of the systematic algebraic approach shown in the solution, students might try plugging in specific numerical values (like \(\mathrm{x = 0.2, x = 0.8}\)) to test the condition. While this can eliminate some choices, it's unreliable for proving a function works for ALL x values, and students might incorrectly conclude a function is correct based on limited testing.

3. Overlooking the "for all x" requirement: Students might miss that the condition must hold for every possible value of x, not just specific values. This leads to incomplete verification strategies that don't guarantee the function works universally.

Errors while executing the approach

1. Algebraic expansion errors: When computing \(\mathrm{f(1-x)}\), students frequently make mistakes expanding expressions like \(\mathrm{(1-x)^2}\). For example, incorrectly expanding \(\mathrm{(1-x)^2}\) as \(\mathrm{(1-x^2)}\) instead of \(\mathrm{(1-2x+x^2)}\), or making sign errors during the expansion process.

2. Incomplete simplification: Students often stop simplifying too early or make errors during simplification. For instance, in Choice C, they might not fully simplify \(\mathrm{f(x) = x^2 - (1-x)^2}\) to get \(\mathrm{2x-1}\), leading to incorrect comparisons between \(\mathrm{f(x)}\) and \(\mathrm{f(1-x)}\).

3. Order confusion in verification: When checking if \(\mathrm{f(x)}\) equals \(\mathrm{f(1-x)}\), students might get confused about which expression they're comparing to which, especially when both expressions look similar after partial simplification.

Errors while selecting the answer

1. Stopping at the first "working" choice: Students might incorrectly verify an earlier choice (like A or B) due to algebraic errors and select it without checking the remaining options, missing the correct answer D.

2. Misidentifying equivalent expressions: In Choice D, students might not recognize that \(\mathrm{(1-x)^2x^2}\) and \(\mathrm{x^2(1-x)^2}\) are identical due to the commutative property of multiplication, leading them to incorrectly reject the correct answer.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose strategic test values
Since we need \(\mathrm{f(x) = f(1-x)}\) for ALL x, we can test specific values that will quickly reveal which functions satisfy this property. Let's choose \(\mathrm{x = 0}\) and \(\mathrm{x = 1/2}\) as our smart numbers:

• \(\mathrm{x = 0}\): We need \(\mathrm{f(0) = f(1-0) = f(1)}\)
• \(\mathrm{x = 1/2}\): We need \(\mathrm{f(1/2) = f(1-1/2) = f(1/2)}\) (automatically satisfied)

Step 2: Test each function with \(\mathrm{x = 0}\)
For the condition \(\mathrm{f(0) = f(1)}\):

Choice A: \(\mathrm{f(x) = 1 - x}\)
\(\mathrm{f(0) = 1 - 0 = 1}\)
\(\mathrm{f(1) = 1 - 1 = 0}\)
Since \(\mathrm{1 \neq 0}\), Choice A fails.

Choice B: \(\mathrm{f(x) = 1 - x^2}\)
\(\mathrm{f(0) = 1 - 0^2 = 1}\)
\(\mathrm{f(1) = 1 - 1^2 = 0}\)
Since \(\mathrm{1 \neq 0}\), Choice B fails.

Choice C: \(\mathrm{f(x) = x^2 - (1-x)^2}\)
\(\mathrm{f(0) = 0^2 - (1-0)^2 = 0 - 1 = -1}\)
\(\mathrm{f(1) = 1^2 - (1-1)^2 = 1 - 0 = 1}\)
Since \(\mathrm{-1 \neq 1}\), Choice C fails.

Choice D: \(\mathrm{f(x) = x^2(1-x)^2}\)
\(\mathrm{f(0) = 0^2(1-0)^2 = 0 \times 1 = 0}\)
\(\mathrm{f(1) = 1^2(1-1)^2 = 1 \times 0 = 0}\)
Since \(\mathrm{0 = 0}\), Choice D passes this test.

Choice E: \(\mathrm{f(x) = \frac{x}{1-x}}\)
\(\mathrm{f(0) = \frac{0}{1-0} = \frac{0}{1} = 0}\)
\(\mathrm{f(1) = \frac{1}{1-1} = \frac{1}{0}}\) = undefined
Choice E fails (undefined at \(\mathrm{x = 1}\)).

Step 3: Verify Choice D with another value
Let's test \(\mathrm{x = 1/4}\) for Choice D:
\(\mathrm{f(1/4) = (1/4)^2(1-1/4)^2 = (1/16)(3/4)^2 = (1/16)(9/16) = 9/256}\)
\(\mathrm{f(1-1/4) = f(3/4) = (3/4)^2(1-3/4)^2 = (9/16)(1/4)^2 = (9/16)(1/16) = 9/256}\)
Since \(\mathrm{f(1/4) = f(3/4)}\), Choice D passes this test as well.

Conclusion: Only Choice D satisfies \(\mathrm{f(x) = f(1-x)}\) for our test values, confirming it's the correct answer.