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If \(-1 < \mathrm{x} < 1\), which of the following must be true?
Let's start by understanding what we're looking for. We have the condition that x is between -1 and 1, but not equal to -1 or 1. This means x could be any number like -0.9, -0.5, 0, 0.3, 0.99, etc.
We need to find which of the five given expressions is ALWAYS true for EVERY possible value of x in this range. If even one value of x makes an expression false, then that expression is not our answer.
Process Skill: TRANSLATE - Converting the interval notation \(-1 < \mathrm{x} < 1\) into concrete understanding of what values we're testing
The interval from -1 to 1 has two distinct regions that behave very differently:
- Negative values: from -1 to 0 (like -0.5, -0.1)
- Positive values: from 0 to 1 (like 0.2, 0.8)
- The special case: \(\mathrm{x} = 0\)
This is important because when we square negative numbers, they become positive, and when we cube them, they stay negative. These sign changes will affect our expressions differently in each region.
Let's test each option with simple values. I'll use \(\mathrm{x} = -0.5\), \(\mathrm{x} = 0.5\), and \(\mathrm{x} = 0\) as our test cases.
Option A: \(\mathrm{x} > \mathrm{x}^3\)
For \(\mathrm{x} = -0.5\): \(-0.5 > (-0.5)^3 = -0.5 > -0.125\). This gives us \(-0.5 > -0.125\), which is FALSE (since -0.5 is more negative).
Since we found a counterexample, Option A is eliminated.
Option B: \(\mathrm{x} + \mathrm{x}^2 > 0\)
For \(\mathrm{x} = -0.5\): \(-0.5 + (-0.5)^2 = -0.5 + 0.25 = -0.25 < 0\). This is FALSE.
Option B is eliminated.
Option C: \(\mathrm{x}^2 > \mathrm{x}^3\)
For \(\mathrm{x} = -0.5\): \((-0.5)^2 > (-0.5)^3\) gives us \(0.25 > -0.125\), which is TRUE.
For \(\mathrm{x} = 0.5\): \((0.5)^2 > (0.5)^3\) gives us \(0.25 > 0.125\), which is TRUE.
This looks promising, but let's check \(\mathrm{x} = 0\): \(0^2 > 0^3\) gives us \(0 > 0\), which is FALSE.
Option C is eliminated.
Option D: \((\mathrm{x}+1/2)(\mathrm{x}-1/2) < 0\)
For \(\mathrm{x} = 0.6\): \((0.6 + 0.5)(0.6 - 0.5) = (1.1)(0.1) = 0.11 > 0\). This is FALSE since we need \(< 0\).
Option D is eliminated.
Option E: \((\mathrm{x}^2+1)(\mathrm{x}-1) < 0\)
Let's check this systematically for all our test values:
For \(\mathrm{x} = -0.5\): \((0.25 + 1)(-0.5 - 1) = (1.25)(-1.5) = -1.875 < 0\). TRUE.
For \(\mathrm{x} = 0.5\): \((0.25 + 1)(0.5 - 1) = (1.25)(-0.5) = -0.625 < 0\). TRUE.
For \(\mathrm{x} = 0\): \((0 + 1)(0 - 1) = (1)(-1) = -1 < 0\). TRUE.
Option E survived our testing, so let's verify it works for ALL values in the interval using sign analysis.
For expression \((\mathrm{x}^2+1)(\mathrm{x}-1) < 0\):
First factor: \((\mathrm{x}^2+1)\)
Since \(\mathrm{x}^2\) is always non-negative for any real number, and we're adding 1, we get \(\mathrm{x}^2 + 1 \geq 1\) for all x.
Therefore, \((\mathrm{x}^2+1)\) is ALWAYS POSITIVE.
Second factor: \((\mathrm{x}-1)\)
Since our constraint is \(-1 < \mathrm{x} < 1\), we know that \(\mathrm{x} < 1\) for all values in our interval.
Therefore, \((\mathrm{x}-1)\) is ALWAYS NEGATIVE.
Product analysis:
(positive number) × (negative number) = negative number
Since the product is always negative, \((\mathrm{x}^2+1)(\mathrm{x}-1) < 0\) is always true for \(-1 < \mathrm{x} < 1\).
Process Skill: APPLY CONSTRAINTS - Using the given interval condition \(\mathrm{x} < 1\) to determine the sign of \((\mathrm{x}-1)\)
The answer is E: \((\mathrm{x}^2+1)(\mathrm{x}-1) < 0\)
This expression is always true because for any x in the interval \(-1 < \mathrm{x} < 1\):
- The factor \((\mathrm{x}^2+1)\) is always positive (at least 1)
- The factor \((\mathrm{x}-1)\) is always negative (since \(\mathrm{x} < 1\))
- Therefore, their product is always negative, making the inequality true.
Faltering Point 1: Misunderstanding "must be true" vs "could be true"
Students often confuse questions asking for what "must be true" (always true) with what "could be true" (sometimes true). They may select an option that works for some values of x but not all values in the given interval. For example, they might choose option C \((\mathrm{x}^2 > \mathrm{x}^3)\) because it works for most values they test, without realizing it fails when \(\mathrm{x} = 0\).
Faltering Point 2: Incomplete constraint analysis
Students may focus only on testing specific numerical values without fully utilizing the constraint \(-1 < \mathrm{x} < 1\). They might miss that this constraint directly tells us \(\mathrm{x} < 1\), which immediately makes \((\mathrm{x}-1)\) negative for ALL values in the interval. Instead, they rely solely on plug-in methods without leveraging the algebraic implications of the given bounds.
Faltering Point 3: Inadequate test value selection
Students often choose convenient test values like \(\mathrm{x} = 0.5\) or \(\mathrm{x} = -0.5\), but fail to test critical boundary cases or special values like \(\mathrm{x} = 0\). This can lead them to incorrectly eliminate or select options. For instance, option C appears correct when tested with \(\pm 0.5\), but fails at \(\mathrm{x} = 0\).
Faltering Point 1: Sign errors with powers of negative numbers
When working with negative values of x, students frequently make errors with powers. They might incorrectly calculate \((-0.5)^3 = 0.125\) instead of \(-0.125\), or confuse whether even/odd powers preserve or change the sign. This is particularly problematic when evaluating options A and C.
Faltering Point 2: Arithmetic mistakes in factored expressions
Students may make computational errors when evaluating expressions like \((\mathrm{x} + 1/2)(\mathrm{x} - 1/2)\) or \((\mathrm{x}^2 + 1)(\mathrm{x} - 1)\). For example, when \(\mathrm{x} = 0.6\), they might calculate \((0.6 + 0.5)(0.6 - 0.5)\) incorrectly, leading to wrong conclusions about whether the expression is positive or negative.
Faltering Point 3: Incomplete sign analysis
Even when students attempt sign analysis, they may incorrectly determine the sign of individual factors. For option E, they might wrongly conclude that \((\mathrm{x}^2 + 1)\) could be negative, or fail to properly analyze that \((\mathrm{x} - 1)\) is always negative when \(-1 < \mathrm{x} < 1\).
Faltering Point 1: Stopping after finding a "working" option
Students may select the first option that works for their test values without checking if it works for ALL possible values in the interval. They might choose option C after seeing it works for \(\mathrm{x} = \pm 0.5\), without testing \(\mathrm{x} = 0\) where it fails.
Faltering Point 2: Misreading inequality directions
When checking final answers, students may confuse inequality directions. For instance, they might calculate \((\mathrm{x}^2 + 1)(\mathrm{x} - 1) = -1.875\) correctly but then mistakenly think this contradicts the requirement for \(< 0\), not realizing that negative values do satisfy the "less than zero" condition.