Which of the following is a value of x for which x^(-9) - x^(-7) > 0?

1. Translate the problem requirements: We need to find which x value makes \(\mathrm{x}^{-9} - \mathrm{x}^{-7}\) positive, where negative exponents mean 1 divided by the positive power
2. Rewrite using fraction form: Convert negative exponents to fractions to make the inequality easier to visualize and manipulate
3. Analyze the sign behavior: Determine when the difference of two fractions with the same sign in the denominator becomes positive
4. Verify with direct substitution: Test the promising answer choice by calculating the actual values to confirm the inequality holds

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this inequality is actually asking us. We have \(\mathrm{x}^{-9} - \mathrm{x}^{-7} > 0\), which means we need the first term to be larger than the second term.

When we see negative exponents, remember that \(\mathrm{x}^{-n}\) simply means "1 divided by \(\mathrm{x}^n\)". So \(\mathrm{x}^{-9}\) means \(\frac{1}{\mathrm{x}^9}\) and \(\mathrm{x}^{-7}\) means \(\frac{1}{\mathrm{x}^7}\).

So our inequality becomes: "1 divided by x to the 9th power" minus "1 divided by x to the 7th power" must be positive.

In mathematical notation: \(\frac{1}{\mathrm{x}^9} - \frac{1}{\mathrm{x}^7} > 0\)

Process Skill: TRANSLATE - Converting the negative exponent notation into fraction form makes the problem much more approachable

2. Rewrite using fraction form

Now let's work with fractions to make this clearer. We have:
\(\frac{1}{\mathrm{x}^9} - \frac{1}{\mathrm{x}^7} > 0\)

To subtract fractions, we need a common denominator. The common denominator for \(\mathrm{x}^9\) and \(\mathrm{x}^7\) is \(\mathrm{x}^9\) (since \(\mathrm{x}^9 = \mathrm{x}^7 \times \mathrm{x}^2\)).

So we can rewrite \(\frac{1}{\mathrm{x}^7}\) as \(\frac{\mathrm{x}^2}{\mathrm{x}^9}\).

Our inequality becomes:
\(\frac{1}{\mathrm{x}^9} - \frac{\mathrm{x}^2}{\mathrm{x}^9} > 0\)

Combining the fractions:
\(\frac{1 - \mathrm{x}^2}{\mathrm{x}^9} > 0\)

This is much clearer! We need the fraction \(\frac{1 - \mathrm{x}^2}{\mathrm{x}^9}\) to be positive.

3. Analyze the sign behavior

For a fraction to be positive, either both the numerator and denominator are positive, or both are negative.

Let's look at each part:

• Numerator: \((1 - \mathrm{x}^2)\)
  • This is positive when \(1 - \mathrm{x}^2 > 0\), which means \(\mathrm{x}^2 < 1\), so \(-1 < \mathrm{x} < 1\)
  • This is negative when \(\mathrm{x}^2 > 1\), so \(\mathrm{x} > 1\) or \(\mathrm{x} < -1\)
• Denominator: \(\mathrm{x}^9\)
  • Since 9 is odd, \(\mathrm{x}^9\) has the same sign as x
  • \(\mathrm{x}^9\) is positive when \(\mathrm{x} > 0\)
  • \(\mathrm{x}^9\) is negative when \(\mathrm{x} < 0\)

For the whole fraction to be positive, we need:
Case 1: Both positive → \((1 - \mathrm{x}^2) > 0\) AND \(\mathrm{x}^9 > 0\) → \(-1 < \mathrm{x} < 1\) AND \(\mathrm{x} > 0\) → \(0 < \mathrm{x} < 1\)
Case 2: Both negative → \((1 - \mathrm{x}^2) < 0\) AND \(\mathrm{x}^9 < 0\) → \((\mathrm{x} > 1 \text{ or } \mathrm{x} < -1)\) AND \(\mathrm{x} < 0\) → \(\mathrm{x} < -1\)

So our solution is: \(\mathrm{x} < -1\) or \(0 < \mathrm{x} < 1\)

Process Skill: CONSIDER ALL CASES - We must check when both parts of the fraction have the same sign

4. Verify with direct substitution

Looking at our answer choices, let's see which ones fall in our solution regions:

• "-2": This is less than -1, so it should work
• "-1": This is exactly -1, not less than -1, so it shouldn't work
• "-1/2": This is between -1 and 0, so it shouldn't work
• "1": This is exactly 1, not between 0 and 1, so it shouldn't work
• "2": This is greater than 1, so it shouldn't work

Let's verify that x = -2 works by substituting:
\(\mathrm{x}^{-9} - \mathrm{x}^{-7} = (-2)^{-9} - (-2)^{-7}\)
\(= \frac{1}{(-2)^9} - \frac{1}{(-2)^7}\)
\(= \frac{1}{-512} - \frac{1}{-128}\)
\(= -\frac{1}{512} - (-\frac{1}{128})\)
\(= -\frac{1}{512} + \frac{1}{128}\)
\(= -\frac{1}{512} + \frac{4}{512}\)
\(= \frac{3}{512}\)

Since \(\frac{3}{512} > 0\), our inequality is satisfied!

5. Final Answer

The answer is A: "-2"

We found that \(\mathrm{x}^{-9} - \mathrm{x}^{-7} > 0\) when \(\mathrm{x} < -1\) or \(0 < \mathrm{x} < 1\). Among the given choices, only x = -2 satisfies \(\mathrm{x} < -1\), and our direct substitution confirmed that \((-2)^{-9} - (-2)^{-7} = \frac{3}{512} > 0\).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting negative exponents

Students often struggle with negative exponents and may incorrectly think that \(\mathrm{x}^{-9}\) means \(-\mathrm{x}^9\) instead of \(\frac{1}{\mathrm{x}^9}\). This fundamental misunderstanding would lead them down a completely wrong path from the start.

2. Not recognizing the need for a common denominator approach

Some students might attempt to work directly with the negative exponents without converting to fractions, making it much harder to visualize when the inequality holds true. They miss that factoring out terms or finding a common denominator will simplify the analysis significantly.

3. Forgetting to consider domain restrictions

Students may not realize that x cannot equal 0 since we have negative exponents (which would make the expressions undefined). This oversight could lead to incorrect analysis of the solution set.

Errors while executing the approach

1. Sign errors when working with odd powers

When analyzing \(\mathrm{x}^9\) in the denominator, students often forget that since 9 is odd, \(\mathrm{x}^9\) has the same sign as x. They might incorrectly assume \(\mathrm{x}^9\) is always positive, leading to wrong conclusions about when the fraction is positive or negative.

2. Incorrect interval analysis for the numerator

When solving \(1 - \mathrm{x}^2 > 0\), students may incorrectly determine the interval. They might think \(\mathrm{x}^2 < 1\) means \(\mathrm{x} < 1\) (forgetting about the negative values) or make errors with the inequality direction when dealing with \(\mathrm{x}^2\).

3. Mistakes in combining the sign analysis cases

Even if students correctly identify when the numerator and denominator are positive or negative separately, they often make errors when combining these conditions to find when both have the same sign (making the overall fraction positive).

Errors while selecting the answer

1. Boundary value confusion

Students might include boundary values like x = -1 or x = 1 in their solution set, not realizing that at these points either the original expression is undefined or the inequality becomes an equality (not strict inequality).

2. Not verifying the answer choice with substitution

Students may correctly find the solution intervals (\(\mathrm{x} < -1\) or \(0 < \mathrm{x} < 1\)) but then incorrectly match this to the given choices, or they may skip the verification step of actually substituting their chosen answer back into the original inequality.