Problem 29
Question
Solve the equation or inequality. $$2(x-2)^{-\frac{1}{3}}-\frac{2}{3} x(x-2)^{-\frac{4}{3}} \leq 0$$
Step-by-Step Solution
Verified Answer
Solution: \(2 < x \leq 3\).
1Step 1: Identifying the common factor
Identify and take out the common factor from both terms. In this case, both terms have \((x-2)^{-\frac{4}{3}}\) as a common factor. Factor the inequality to get:\[(x-2)^{-\frac{4}{3}} \left[ 2(x-2) - \frac{2}{3}x \right] \leq 0.\]
2Step 2: Simplifying the expression
Simplify the expression inside the brackets. The expression is:\[ 2(x-2) - \frac{2}{3}x = 2x -4 - \frac{2}{3}x = \frac{6}{3}x - \frac{2}{3}x - 4 = \frac{4}{3}x - 4.\]So the inequality becomes:\[(x-2)^{-\frac{4}{3}} \left( \frac{4}{3}x - 4 \right) \leq 0.\]
3Step 3: Finding zero points and critical points
Identify the zero points and points where the expression is undefined. The expression \((x-2)^{-\frac{4}{3}}\) is undefined at \(x = 2\). The expression \(\frac{4}{3}x - 4 = 0\) gives \(x = 3\). Thus, the critical points are \(x = 2\) and \(x = 3\).
4Step 4: Testing intervals
Test the intervals determined by critical points: \(( -\infty, 2)\), \((2, 3)\), and \((3, \infty)\). Choose test points such as \(x = 0\), \(x = 2.5\), and \(x = 4\): - For \(x = 0\): \( (0-2)^{-\frac{4}{3}} \left(\frac{4}{3}(0) - 4\right) > 0\).- For \(x = 2.5\): \( (2.5-2)^{-\frac{4}{3}} \left(\frac{4}{3}(2.5) - 4\right) < 0\).- For \(x = 4\): \( (4-2)^{-\frac{4}{3}} \left(\frac{4}{3}(4) - 4\right) > 0\).
5Step 5: Interpreting testing results
For the inequality \((x-2)^{-\frac{4}{3}} \left( \frac{4}{3}x - 4 \right) \leq 0\), the expression is non-positive in the interval \((2, 3]\). Therefore, the solution to the inequality is \(2 < x \leq 3\).
Key Concepts
Common FactorCritical PointsInterval TestingUndefined Expression
Common Factor
The process of finding a common factor in an expression helps simplify complex terms, making them easier to handle. It involves identifying a term that appears in each part of the expression. In the given inequality, both terms contain \((x-2)^{-\frac{4}{3}}\) as a common factor.
This element is factored out, simplifying the inequality altogether.
By factoring, we can express the inequality as \((x-2)^{-\frac{4}{3}}[2(x-2) - \frac{2}{3}x] \leq 0\).
Such simplification is crucial because it prepares the expression for further analysis.
Factoring out the common term reduces potential complexity and allows the identification of critical points more easily.
This element is factored out, simplifying the inequality altogether.
By factoring, we can express the inequality as \((x-2)^{-\frac{4}{3}}[2(x-2) - \frac{2}{3}x] \leq 0\).
Such simplification is crucial because it prepares the expression for further analysis.
Factoring out the common term reduces potential complexity and allows the identification of critical points more easily.
Critical Points
Critical points in the inequality include points where the expression equals zero or becomes undefined.
These points are essential as they dictate the boundaries of the solution interval.
To determine critical points, examine both terms of the simplified inequality:
These points help us focus on specific intervals that determine the inequality's solution.
These points are essential as they dictate the boundaries of the solution interval.
To determine critical points, examine both terms of the simplified inequality:
- The expression \((x-2)^{-\frac{4}{3}}\) becomes undefined at \(x = 2\).
- The expression \(\frac{4}{3}x - 4 = 0\) results in a critical point at \(x = 3\).
These points help us focus on specific intervals that determine the inequality's solution.
Interval Testing
Interval testing is a strategy to determine where an inequality holds true.
After finding critical points, these points define intervals that need testing.
In this problem, critical points \(x = 2\) and \(x = 3\) create three intervals:
For instance:
Thus, interval testing offers a clear picture of truly satisfied regions.
After finding critical points, these points define intervals that need testing.
In this problem, critical points \(x = 2\) and \(x = 3\) create three intervals:
- \((-\infty, 2)\)
- \((2, 3)\)
- \((3, \infty)\)
For instance:
- \(x = 0\) in \((-\infty, 2)\) yields a positive result.
- \(x = 2.5\) within \((2, 3)\) results in a negative value.
- \(x = 4\) in \((3, \infty)\) gives a positive outcome.
Thus, interval testing offers a clear picture of truly satisfied regions.
Undefined Expression
An undefined expression in mathematics occurs when substituting a certain value results in an impossibility, like division by zero.
These undefined points are crucial for analyzing expressions and inequalities.
In our inequality, the term \((x-2)^{-\frac{4}{3}}\) becomes undefined if \(x = 2\) because it results in division by zero.
This makes \(x = 2\) a critical point.
Understanding when an expression is undefined prevents confusion when solving inequalities.
By recognizing these points, you avoid making incorrect assumptions about the expression's value.
Moreover, acknowledging undefined expressions also assists in correctly identifying intervals where the inequality is valid.
These undefined points are crucial for analyzing expressions and inequalities.
In our inequality, the term \((x-2)^{-\frac{4}{3}}\) becomes undefined if \(x = 2\) because it results in division by zero.
This makes \(x = 2\) a critical point.
Understanding when an expression is undefined prevents confusion when solving inequalities.
By recognizing these points, you avoid making incorrect assumptions about the expression's value.
Moreover, acknowledging undefined expressions also assists in correctly identifying intervals where the inequality is valid.
Other exercises in this chapter
Problem 28
Solve the equation or inequality. $$\sqrt[3]{x} \leq x$$
View solution Problem 28
Use \(f(x)=-2 x, g(x)=\sqrt{x}\) and \(h(x)=|x|\) to find and simplify expressions for the following functions and state the domain of each using interval notat
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With the help of your classmates, explain why a function which is either strictly increasing or strictly decreasing on its entire domain would have to be one-to
View solution Problem 29
Use \(f(x)=-2 x, g(x)=\sqrt{x}\) and \(h(x)=|x|\) to find and simplify expressions for the following functions and state the domain of each using interval notat
View solution