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Mobile App Chapter 8: Problem 14
The function \(f(x)=2 \ln (x-2)-x^{2}+4 x+1\) increases in the interval (1) \((-\infty, 1) \cup(2,3)\) (2) \((1,2)\) (3) \((2,3)\) (4) \((3, \infty)\)
Short Answer
Expert verified
Option (1) \((-\text{∞}, 1) \cup (2, 3)\).
Step by step solution
01
Find the derivative
To determine where the function is increasing, find its derivative. The function given is \( f(x) = 2 \ln (x-2) - x^2 + 4x + 1 \). Using differentiation rules: \[ f'(x) = \frac{d}{dx}(2 \ln (x-2)) + \frac{d}{dx}(-x^2) + \frac{d}{dx}(4x) + \frac{d}{dx}(1) \]. This simplifies to \[ f'(x) = \frac{2}{x-2} - 2x + 4 \].
02
Set the derivative to zero
To find the critical points, set the derivative equal to zero and solve for \(x\): \[ \frac{2}{x-2} - 2x + 4 = 0 \]. Multiply through by \(x-2\) to clear the fraction: \[ 2 - 2x(x-2) + 4(x-2) = 0 \]. This simplifies to: \[ 2 - 2x^2 + 4x +4x - 8 = 0 \]. Combine like terms and organize: \[ -2x^2 + 8x - 6 = 0 \].
03
Solve the quadratic equation
Solve the quadratic equation \( -2x^2 + 8x - 6 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = -2 \), \( b = 8 \), and \( c = -6 \). Therefore: \[ x = \frac{-8 \pm \sqrt{(8)^2 - 4(-2)(-6)}}{2(-2)} = \frac{-8 \pm \sqrt{64 - 48}}{-4} = \frac{-8 \pm \sqrt{16}}{-4} = \frac{-8 \pm 4}{-4} \]. This gives solutions \( x = 1 \) and \( x = 3 \).
04
Test intervals around the critical points
Now test intervals around the critical points 1 and 3 to determine where \(f(x)\) is increasing: - For \(x < 1\), choose \(x = 0\): \( f'(0) = \frac{2}{0-2} - 2(0) + 4 = -1 + 4 = 3 > 0\). - For \(1 < x < 3\), choose \(x = 2.5\): \(f'(2.5) = \frac{2}{2.5-2} - 2(2.5) + 4 = 4 - 5 = -1 < 0\). - For \(x > 3\), choose \(x = 4\): \( f'(4) = \frac{2}{4-2} - 2(4) + 4 = 1 - 8 + 4 = -3 < 0\).
05
Determine the intervals
Based on the testing, the function increases on the intervals where the derivative is positive. Therefore, the intervals are \((-\text{∞}, 1)\) and \((2, 3)\).
06
Conclusion
The function \( f(x) \) increases in the intervals \((-\text{∞}, 1)\) and \((2, 3)\). Hence, the correct option is: (1) \((-\text{∞}, 1) \cup (2, 3)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives help us understand how a function changes at any given point. In this problem, we dealt with the function: \[f(x) = 2 \, \ln(x-2) - x^2 + 4x + 1\]To find where this function increases, we first need its derivative. The derivative, noted as \(f'(x)\), tells us the slope of the function at any point \(x\). Using derivative rules, we get: \[f'(x) = \frac{2}{x-2} - 2x + 4\]This equation shows us how the function's rate of change behaves. Remember, where the derivative is positive, the function is increasing. Conversely, where it’s negative, the function is decreasing. This derivative is crucial for locating where the function increases or decreases.
Critical Points
Critical points of a function occur where its derivative is zero or undefined. These points can indicate local maximums, minimums, or points of inflection. To find them, we set our derivative \(f'(x)\) to zero:\[\frac{2}{x-2} - 2x + 4 = 0\]By solving this equation, we locate our potential critical points. In our problem, solving it gives us:\[x = 1 \quad \text{and} \quad x = 3\]These values of \(x\) are crucial as they split the domain into intervals we need to test to see where the function is increasing or decreasing. The points help us determine the behavior of the function around these values.
Intervals of Increase
After finding the critical points, we need to determine the intervals where the derivative is positive. These intervals indicate where the function is increasing. We examine the function's derivative over intervals around our critical points:
- For \( x < 1 \), we chose \( x = 0 \). Calculating \( f'(0) \) yields: \[ f'(0) = \frac{2}{0-2} - 2(0) + 4 = -1 + 4 = 3 > 0\]Here, the derivative is positive, so the function is increasing.
- For \(1 < x < 3\), we chose \( x = 2.5 \). Calculating \( f'(2.5) \) yields: \[ f'(2.5) = \frac{2}{2.5-2} - 2(2.5) + 4 = 4 - 5 = -1 < 0\]Here, the derivative is negative, so the function is decreasing.
- For \( x > 3 \), we chose \( x = 4 \). Calculating \( f'(4) \) yields: \[ f'(4) = \frac{2}{4-2} - 2(4) + 4 = 1 - 8 + 4 = -3 < 0\]Here, the derivative is negative, so the function is decreasing.
