If 0 , what is the median of the values (x, x^(-1), x^2, sqrt(x) and x^3?

1. Translate the problem requirements: We need to find the median (middle value when arranged in order) of five expressions involving x: x, \(\mathrm{x}^{-1}\), \(\mathrm{x}^2\), \(\sqrt{\mathrm{x}}\), and \(\mathrm{x}^3\), given that x is between 0 and 1.
2. Analyze the behavior of each expression: Determine how each power of x behaves when \(0 < \mathrm{x} < 1\) by understanding what happens to numbers less than 1 when raised to different powers.
3. Order the five values from smallest to largest: Use the behavioral patterns to arrange x, \(\mathrm{x}^{-1}\), \(\mathrm{x}^2\), \(\sqrt{\mathrm{x}}\), and \(\mathrm{x}^3\) in ascending order.
4. Identify the median: Select the middle value from the ordered list of five numbers.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We have five different expressions all involving the same number x: the original number x, its reciprocal \(\mathrm{x}^{-1}\), its square \(\mathrm{x}^2\), its square root \(\sqrt{\mathrm{x}}\), and its cube \(\mathrm{x}^3\). We need to find which one sits right in the middle when we arrange all five from smallest to largest.

The key constraint is that x is between 0 and 1, meaning x could be something like 0.5, 0.8, or 0.1 - any positive decimal less than 1.

Process Skill: TRANSLATE - Converting the median requirement into ordering five expressions

2. Analyze the behavior of each expression

Now let's think about what happens to numbers between 0 and 1 when we apply different operations. Let's use \(\mathrm{x} = 0.5\) as our concrete example to build intuition:

When x = 0.5:

• x = 0.5 (this stays the same)
• \(\mathrm{x}^{-1} = 1/0.5 = 2\) (reciprocals of small numbers become large)
• \(\mathrm{x}^2 = 0.5 \times 0.5 = 0.25\) (squaring makes small numbers even smaller)
• \(\sqrt{\mathrm{x}} = \sqrt{0.5} \approx 0.71\) (square roots make small numbers bigger, but not as big as 1)
• \(\mathrm{x}^3 = 0.5 \times 0.5 \times 0.5 = 0.125\) (cubing makes small numbers very small)

Let's verify this pattern with another example, \(\mathrm{x} = 0.1\):

• x = 0.1
• \(\mathrm{x}^{-1} = 1/0.1 = 10\) (again, reciprocal is large)
• \(\mathrm{x}^2 = 0.01\) (smaller than original)
• \(\sqrt{\mathrm{x}} = \sqrt{0.1} \approx 0.32\) (bigger than original, but less than 1)
• \(\mathrm{x}^3 = 0.001\) (much smaller than original)

The pattern is clear: For any number between 0 and 1:

• Higher powers (\(\mathrm{x}^2\), \(\mathrm{x}^3\)) make the number smaller
• Fractional powers (\(\sqrt{\mathrm{x}} = \mathrm{x}^{1/2}\)) make the number bigger but keep it less than 1
• Negative powers (\(\mathrm{x}^{-1}\)) make the number greater than 1

Process Skill: INFER - Recognizing the general behavior pattern from specific examples

3. Order the five values from smallest to largest

Based on our analysis, we can now arrange the five expressions in ascending order:

Smallest to Largest:

1. \(\mathrm{x}^3\) (smallest - cubing makes the number very small)
2. \(\mathrm{x}^2\) (small - squaring makes the number smaller than original)
3. x (middle - the original number)
4. \(\sqrt{\mathrm{x}}\) (larger - square root makes it bigger but still less than 1)
5. \(\mathrm{x}^{-1}\) (largest - reciprocal makes it greater than 1)

So our ordered list is: \(\mathrm{x}^3 < \mathrm{x}^2 < \mathrm{x} < \sqrt{\mathrm{x}} < \mathrm{x}^{-1}\)

Let's verify with \(\mathrm{x} = 0.5\): \(0.125 < 0.25 < 0.5 < 0.71 < 2\) ✓

Let's verify with \(\mathrm{x} = 0.1\): \(0.001 < 0.01 < 0.1 < 0.32 < 10\) ✓

Process Skill: APPLY CONSTRAINTS - Using the condition \(0 < \mathrm{x} < 1\) to determine the relative ordering

4. Identify the median

With five numbers arranged in order, the median is the middle value - the 3rd number in our ordered list.

From our ordering: \(\mathrm{x}^3 < \mathrm{x}^2 < \mathrm{x} < \sqrt{\mathrm{x}} < \mathrm{x}^{-1}\)

The middle value is x.

Final Answer

The median of the five values x, \(\mathrm{x}^{-1}\), \(\mathrm{x}^2\), \(\sqrt{\mathrm{x}}\), and \(\mathrm{x}^3\) when \(0 < \mathrm{x} < 1\) is x.

The answer is A.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint \(0 < \mathrm{x} < 1\)
Students may not fully grasp what this constraint means for the behavior of different powers and roots. They might think that since x is positive, all expressions will behave the same way as they do for numbers greater than 1, leading to incorrect ordering assumptions.

2. Attempting to solve algebraically without testing values
Some students may try to find a general algebraic solution or set up inequalities without first building intuition through concrete examples. This approach is much more complex and error-prone for this type of problem.

3. Forgetting that median requires ordering all five values
Students might focus on comparing just two expressions at a time rather than understanding they need to establish a complete ordering of all five expressions to find the middle value.

Errors while executing the approach

1. Incorrectly calculating powers and roots for test values
When using concrete examples like \(\mathrm{x} = 0.5\), students may make arithmetic errors such as calculating \(\sqrt{0.5}\) incorrectly or confusing \(\mathrm{x}^{-1} = 1/\mathrm{x}\) with negative values instead of reciprocals.

2. Mixing up the ordering of expressions
Students may correctly calculate individual values but then incorrectly order them. For example, they might place \(\mathrm{x}^2 > \mathrm{x}\) or \(\sqrt{\mathrm{x}} < \mathrm{x}\), forgetting how these operations affect numbers between 0 and 1.

3. Using only one test value and assuming it applies universally
Students might test with just one value (like \(\mathrm{x} = 0.5\)) and not verify their ordering holds for other values in the range \(0 < \mathrm{x} < 1\), missing potential exceptions or building false confidence.

Errors while selecting the answer

1. Confusing median with mean or mode
After correctly ordering the five expressions, students might accidentally select the largest, smallest, or most frequently appearing value instead of the middle value.

2. Selecting the wrong position in the ordered list
Even with correct ordering \(\mathrm{x}^3 < \mathrm{x}^2 < \mathrm{x} < \sqrt{\mathrm{x}} < \mathrm{x}^{-1}\), students might count incorrectly and select the 2nd or 4th value instead of recognizing that the median of 5 values is the 3rd value.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a strategic value for x

Since \(0 < \mathrm{x} < 1\), let's choose \(\mathrm{x} = 1/4 = 0.25\). This fraction is ideal because:

• It's clearly between 0 and 1
• Its powers and roots yield clean, comparable decimal values
• It makes the relative magnitudes of different expressions easy to see

Step 2: Calculate each expression with \(\mathrm{x} = 1/4\)

• \(\mathrm{x} = 1/4 = 0.25\)
• \(\mathrm{x}^{-1} = 1/(1/4) = 4\)
• \(\mathrm{x}^2 = (1/4)^2 = 1/16 = 0.0625\)
• \(\sqrt{\mathrm{x}} = \sqrt{1/4} = 1/2 = 0.5\)
• \(\mathrm{x}^3 = (1/4)^3 = 1/64 \approx 0.0156\)

Step 3: Arrange the calculated values in ascending order

From smallest to largest:

• \(\mathrm{x}^3 = 0.0156\) (smallest)
• \(\mathrm{x}^2 = 0.0625\)
• \(\mathrm{x} = 0.25\) (middle value)
• \(\sqrt{\mathrm{x}} = 0.5\)
• \(\mathrm{x}^{-1} = 4\) (largest)

Step 4: Identify the median

With 5 values arranged in order, the median is the 3rd value (middle position).

The median is \(\mathrm{x} = 0.25\).

Step 5: Verify with another smart number

Let's confirm with \(\mathrm{x} = 1/9 \approx 0.111\):

• \(\mathrm{x}^3 \approx 0.00137\) (smallest)
• \(\mathrm{x}^2 \approx 0.0123\)
• \(\mathrm{x} \approx 0.111\) (middle)
• \(\sqrt{\mathrm{x}} \approx 0.333\)
• \(\mathrm{x}^{-1} = 9\) (largest)

Again, x is the median value.

Answer: A