If \(\mathrm{f(x) = g(x - 1), where g(x) = |x| + 1}\), which of the following must be true?

1. Translate the problem requirements: We need to understand what \(\mathrm{f(x)}\) equals by substituting the given functions, and determine which statement about \(\mathrm{f(x)}\) must always be true regardless of the input value.
2. Express f(x) in terms of x: Use the given relationship \(\mathrm{f(x) = g(x-1)}\) and the definition of \(\mathrm{g(x)}\) to write \(\mathrm{f(x)}\) as a single expression.
3. Analyze the behavior of the absolute value expression: Examine what happens to \(\mathrm{|x-1| + 1}\) for any real number x to determine the range of possible values.
4. Verify the conclusion with test values: Check our reasoning by testing specific values of x to confirm our general conclusion.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked. We have two functions that are related to each other, and we need to figure out what must always be true about one of them.

We're given:

- A function \(\mathrm{g(x) = |x| + 1}\)
- A function \(\mathrm{f(x) = g(x - 1)}\)
- We need to determine which statement about \(\mathrm{f(x)}\) must be true for ANY value of x

The key word here is "must" - this means we're looking for something that's always true, no matter what number we plug in for x.

Process Skill: TRANSLATE

2. Express f(x) in terms of x

Now let's figure out what \(\mathrm{f(x)}\) actually looks like. We know that \(\mathrm{f(x) = g(x - 1)}\), and we know that \(\mathrm{g(x) = |x| + 1}\).

Think of it this way: if g takes any number and gives us the absolute value of that number plus 1, then f takes any number x, subtracts 1 from it first, and then applies the same rule.

So \(\mathrm{f(x) = g(x - 1) = |x - 1| + 1}\)

This means \(\mathrm{f(x) = |x - 1| + 1}\)

Let's check this with a simple example: if \(\mathrm{x = 3}\), then \(\mathrm{f(3) = |3 - 1| + 1 = |2| + 1 = 2 + 1 = 3}\).

3. Analyze the behavior of the absolute value expression

Now we need to think about what \(\mathrm{|x - 1| + 1}\) does for any possible value of x.

Remember that absolute value always gives us a non-negative result. No matter what number we put inside the absolute value bars, we either get zero or a positive number. Never negative.

So \(\mathrm{|x - 1| \geq 0}\) for any real number x.

This means \(\mathrm{|x - 1| + 1 \geq 0 + 1 = 1}\)

In plain English: \(\mathrm{f(x)}\) is always at least 1, which means \(\mathrm{f(x)}\) is always positive.

Process Skill: INFER

4. Verify the conclusion with test values

Let's test this with several different values to make sure our reasoning is correct:

- When \(\mathrm{x = 1}\): \(\mathrm{f(1) = |1 - 1| + 1 = |0| + 1 = 0 + 1 = 1}\) (positive ✓)
- When \(\mathrm{x = 0}\): \(\mathrm{f(0) = |0 - 1| + 1 = |-1| + 1 = 1 + 1 = 2}\) (positive ✓)
- When \(\mathrm{x = -5}\): \(\mathrm{f(-5) = |-5 - 1| + 1 = |-6| + 1 = 6 + 1 = 7}\) (positive ✓)
- When \(\mathrm{x = 10}\): \(\mathrm{f(10) = |10 - 1| + 1 = |9| + 1 = 9 + 1 = 10}\) (positive ✓)

Every single test confirms that \(\mathrm{f(x) > 0}\).

Looking at our answer choices:
- (A) \(\mathrm{f(x) < 0}\): This is never true
- (B) \(\mathrm{f(x) = 0}\): This is never true
- (C) \(\mathrm{f(x) > 0}\): This is always true
- (D) \(\mathrm{f(x) = 2}\): This is sometimes true, but not always
- (E) \(\mathrm{f(x) = g(x)}\): This would mean \(\mathrm{|x - 1| + 1 = |x| + 1}\), which is not always true

Final Answer

The answer is (C) \(\mathrm{f(x) > 0}\).

Since \(\mathrm{f(x) = |x - 1| + 1}\), and absolute values are always non-negative, we have \(\mathrm{|x - 1| \geq 0}\), which means \(\mathrm{f(x) = |x - 1| + 1 \geq 1}\). Therefore, \(\mathrm{f(x)}\) is always positive, making choice (C) the statement that must be true.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding function composition: Students often struggle with the concept that \(\mathrm{f(x) = g(x - 1)}\) means we substitute (x - 1) into the function g. They might incorrectly think this means \(\mathrm{f(x) = g(x) - 1}\) or confuse the order of operations.

2. Missing the "must be true" requirement: Students may not pay sufficient attention to the word "must" in the question, leading them to look for what "could be true" rather than what is "always true" for any value of x.

3. Overlooking the domain consideration: Students might not recognize that since we're dealing with absolute value functions, we need to consider all real numbers as possible inputs, not just specific cases.

Errors while executing the approach

1. Incorrect substitution in function composition: When finding \(\mathrm{f(x) = g(x - 1)}\), students may incorrectly substitute and write \(\mathrm{f(x) = |x| - 1 + 1 = |x|}\) instead of the correct \(\mathrm{f(x) = |x - 1| + 1}\).

2. Misunderstanding absolute value properties: Students might forget that absolute value expressions are always non-negative \(\mathrm{(\geq 0)}\), which is crucial for determining that \(\mathrm{|x - 1| \geq 0}\) and therefore \(\mathrm{f(x) \geq 1}\).

3. Insufficient testing of edge cases: Students may only test positive values of x and miss testing negative values or \(\mathrm{x = 1}\) (where \(\mathrm{|x - 1| = 0}\)), failing to verify that \(\mathrm{f(x) > 0}\) holds for all possible inputs.

Errors while selecting the answer

1. Confusing "greater than" vs "greater than or equal to": Since \(\mathrm{f(x) = |x - 1| + 1 \geq 1}\), students might think \(\mathrm{f(x) \geq 1}\) means \(\mathrm{f(x)}\) could equal 1, but they need to recognize that \(\mathrm{f(x)}\) is always strictly greater than 0, making choice (C) correct.

2. Selecting a sometimes-true answer: Students might choose option (D) \(\mathrm{f(x) = 2}\) because they verified this works for specific values (like \(\mathrm{x = 0}\)), not recognizing that the question asks for what "must" always be true.

3. Misinterpreting the minimum value: Students correctly find that the minimum value of \(\mathrm{f(x)}\) is 1 (when \(\mathrm{x = 1}\)), but then incorrectly conclude that \(\mathrm{f(x) \geq 1}\) means \(\mathrm{f(x)}\) could equal 0, missing that \(\mathrm{f(x) > 0}\) is the appropriate choice.