Problem 55
Question
Let \(f(x)=x+1\) and \(\phi(x)=x-2\), then the values of \(x\) satisfying| \(f(x)+\phi(x)|=| f(x)|+| \phi(x) \mid\) are (A) \((-\infty, 1)\) (B) \((2, \infty)\) (C) \((-\infty,-2)\) (D) \((1, \infty)\)
Step-by-Step Solution
Verified Answer
(A) \((-
fty, 1)\) and (B) \((2,
fty)\) are correct.
1Step 1: Understand the Functions
The functions provided are \( f(x) = x + 1 \) and \( \phi(x) = x - 2 \). This means that they are simple linear functions. \( f(x) \) increases and translates by 1, while \( \phi(x) \) decreases and shifts to the left by 2.
2Step 2: Set up the Equation
The equation given is \(|f(x) + \phi(x)| = |f(x)| + |\phi(x)|\). Substituting the expressions for \(f(x)\) and \(\phi(x)\) in, we get: \(|(x + 1) + (x - 2)| = |x + 1| + |x - 2|\). Simplifying inside the absolute values gives us \(|2x - 1| = |x + 1| + |x - 2|\).
3Step 3: Analyze Breakpoints
Identify key points where the expressions inside the absolute values change signs: namely where \(x + 1 = 0\), \(x - 2 = 0\), and \(2x - 1 = 0\). These points are \(x = -1\), \(x = 2\), and \(x = 0.5\), respectively.
4Step 4: Solve in Intervals
Use these breakpoints to solve the equation in different intervals: \((-fty, -1)\), \((-1, 0.5)\), \((0.5, 2)\), and \((2, fty)\). 1. In \((-fty, -1)\): \(|2x - 1| = -(2x - 1)\), \(|x + 1| = -(x + 1)\), \(|x - 2| = -(x - 2)\). Then, \(-(2x-1) = -(x+1) - (x-2)\) giving a true equation: \(x < 1\).2. In \((-1, 0.5)\), the functions provide another scenario: calculations show the condition does not hold.3. In \((0.5, 2)\), calculations also negate the condition.4. Finally, in \((2, fty)\), all positive expressions once more, the condition holds.
5Step 5: Conclusion of Valid Intervals
From calculations of each interval, only \((-fty, -1)\) and \((2, fty)\) satisfy the equation throughout their range, meaning these are the solutions where \(|f(x) + \phi(x)| = |f(x)| + |\phi(x)|\) holds true.
Key Concepts
Linear FunctionsInterval AnalysisPiecewise Functions
Linear Functions
Linear functions are among the simplest types of functions to understand. They graph as straight lines, hence the name "linear." In a linear function, every increase in the independent variable, often called \( x \), results in a constant change in the dependent variable, frequently called \( y \). This relationship can be expressed with the formula \( f(x) = mx + b \), where \( m \) is the slope of the line and \( b \) is the y-intercept.
In the exercise, \( f(x) = x + 1 \) and \( \phi(x) = x - 2 \) are both linear functions. The function \( f(x) = x + 1 \) has a slope of 1 and a y-intercept of 1, shifting the line up by 1 unit. \( \phi(x) = x - 2 \) also has a slope of 1 and a y-intercept of -2, meaning it shifts the line down by 2 units.
Essentially, both functions change at the same rate, but they are positioned differently along the y-axis due to their intercepts. Understanding these basics makes it easier to solve complex problems like absolute value equations involving these functions.
In the exercise, \( f(x) = x + 1 \) and \( \phi(x) = x - 2 \) are both linear functions. The function \( f(x) = x + 1 \) has a slope of 1 and a y-intercept of 1, shifting the line up by 1 unit. \( \phi(x) = x - 2 \) also has a slope of 1 and a y-intercept of -2, meaning it shifts the line down by 2 units.
Essentially, both functions change at the same rate, but they are positioned differently along the y-axis due to their intercepts. Understanding these basics makes it easier to solve complex problems like absolute value equations involving these functions.
Interval Analysis
Interval analysis is a method often used with absolute value equations to solve them piece by piece. Absolute value equations can have different conditions based on the sign of the values inside the absolute value brackets.
In the exercise, solving \(|f(x) + \phi(x)| = |f(x)| + |\phi(x)|\), we found breakpoints where the expressions inside the absolute values change signs. These breakpoints, \(x = -1\), \(x = 2\), and \(x = 0.5\), segment the real number line into different intervals.
By assessing each interval like \((-\infty, -1)\), \((-1, 0.5)\), \((0.5, 2)\), and \((2, \infty)\), we can understand how each segment behaves relative to the absolute values. In the exercise, the values satisfying the condition are found by examining these segments individually. This method allows for isolating behaviors of the function in specific ranges.
In the exercise, solving \(|f(x) + \phi(x)| = |f(x)| + |\phi(x)|\), we found breakpoints where the expressions inside the absolute values change signs. These breakpoints, \(x = -1\), \(x = 2\), and \(x = 0.5\), segment the real number line into different intervals.
By assessing each interval like \((-\infty, -1)\), \((-1, 0.5)\), \((0.5, 2)\), and \((2, \infty)\), we can understand how each segment behaves relative to the absolute values. In the exercise, the values satisfying the condition are found by examining these segments individually. This method allows for isolating behaviors of the function in specific ranges.
Piecewise Functions
Piecewise functions are functions defined by multiple sub-functions, each applying to a certain interval. These are vital when dealing with absolute value expressions, as these expressions can cause different results depending on the input values.
In the exercise, when managing the absolute value settings for \(|2x - 1|\), \(|x + 1|\), and \(|x - 2|\), each expression's behavior changes depending on \(x\).
The process of breaking the problem down into sections like \((-\infty, -1)\), \((-1, 0.5)\), \((0.5, 2)\), and \((2, \infty)\) is similar to treating our equation as a piecewise function. Each interval acts like a different "piece" of the function where rules apply differently. Comprehending this lets you apply different mathematical rules to each interval, proving beneficial in simplifying the problem into manageable parts.
In the exercise, when managing the absolute value settings for \(|2x - 1|\), \(|x + 1|\), and \(|x - 2|\), each expression's behavior changes depending on \(x\).
The process of breaking the problem down into sections like \((-\infty, -1)\), \((-1, 0.5)\), \((0.5, 2)\), and \((2, \infty)\) is similar to treating our equation as a piecewise function. Each interval acts like a different "piece" of the function where rules apply differently. Comprehending this lets you apply different mathematical rules to each interval, proving beneficial in simplifying the problem into manageable parts.
Other exercises in this chapter
Problem 53
If \(f(n+1)=\frac{2 f(n)+1}{2}, n=1,2, \ldots\) and \(f(1)=2\), then \(f(101)\) equals (A) 52 (B) 49 (C) 48 (D) 51
View solution Problem 54
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View solution Problem 56
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If \(\alpha \in\left(0, \frac{\pi}{2}\right)\) then \(\sqrt{x^{2}+x}+\frac{\tan ^{2} \alpha}{\sqrt{x^{2}+x}}\) is always greater than or equal to (A) \(2 \tan \
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