Problem 48
Question
Suppose \(f:[2,2] \rightarrow R\) is defined by, $$ f(x)= \begin{cases}-1 & \text { for }-2 \leq x \leq 0 \\ x-1 & \text { for } 0 \leq x \leq 2\end{cases} $$ then \(\\{x \in(-2,2): x \leq 0\) and \(f(x \mid)=x\\}=\) (A) \(\\{-1\\}\) (B) \(\\{0\\}\) (C) \(\\{-1 / 2\\}\) (D) \(\phi\)
Step-by-Step Solution
Verified Answer
(A) \{-1\}.
1Step 1: Understand the Conditions and Function Definitions
The function \(f\) is defined on the interval \([-2, 2]\) with piecewise conditions. For \(-2 \leq x \leq 0\), the function is given by \(f(x) = -1\). For \(0 \leq x \leq 2\), it is given by \(f(x) = x - 1\). Our task is to find \(x\) in \((-2, 2)\) such that \(x \leq 0\) and \(f(x) = x\).
2Step 2: Evaluate Conditions for \(-2 \leq x \leq 0\)
In this case, \(f(x) = -1\). We need to check for any \(x\) in \([-2, 0]\) for which \(-1 = x\). Solving this, we find that \(-1 = x\) means \(x = -1\). This value satisfies \(-2 \leq x \leq 0\).
3Step 3: Evaluate Conditions for \(0 < x < 2\)
For the interval \(0 \leq x \leq 2\), \(f(x) = x - 1\). We don't need to evaluate this part for the function since the conditions require \(x \leq 0\). Thus, only the interval \(-2 \leq x \leq 0\) and its related condition is relevant.
4Step 4: Conclude with Correct Solution
Given \(x = -1\), which satisfies both \(f(x) = x\) and \(x \leq 0\), \(x = -1\) is a solution. Therefore, the set of \(x\) is \{-1\}.
Key Concepts
Interval AnalysisFunction EvaluationMathematical Problem Solving
Interval Analysis
When dealing with piecewise functions, interval analysis is crucial for understanding how different function rules apply over distinct regions of the input variable. For the given function \(f\), our domain is the interval \([-2, 2]\). Here, we're working with two sub-intervals: \([-2, 0]\) and \([0, 2]\). During interval analysis, each sub-interval is approached as a separate entity, making it easier to apply the given rules for each segment.
To analyze the interval \([-2, 0]\), we look at the constraint \(-2 \leq x \leq 0\), where the function simplifies to \(f(x) = -1\). In contrast, for the interval \([0, 2]\), the function changes to \(f(x) = x - 1\). To solve problems involving piecewise functions, always check which interval your input variable \(x\) falls into, and apply the corresponding expression.
This approach helps identify where certain conditions are met, making problem-solving more straightforward and systematic.
To analyze the interval \([-2, 0]\), we look at the constraint \(-2 \leq x \leq 0\), where the function simplifies to \(f(x) = -1\). In contrast, for the interval \([0, 2]\), the function changes to \(f(x) = x - 1\). To solve problems involving piecewise functions, always check which interval your input variable \(x\) falls into, and apply the corresponding expression.
This approach helps identify where certain conditions are met, making problem-solving more straightforward and systematic.
Function Evaluation
Function evaluation involves determining the output of a function based on its input. For the piecewise function \(f\) used in this exercise, each segment has different rules for evaluation. When \(x\) is in \([-2, 0]\), evaluate the function using \(f(x) = -1\). This segment assigns a constant output for any input \(x\) within this range.
For inputs that fall in the interval \([0, 2]\), evaluate using the condition \(f(x) = x - 1\). This part of the function yields a linear output that directly depends on the input. Understanding how each section of a piecewise function behaves can demystify function evaluation, allowing you to correctly compute \(f(x)\) for any given \(x\).
Being meticulous about the conditions and intervals helps avoid miscalculations and ensures that each piece of the function is applied accurately.
For inputs that fall in the interval \([0, 2]\), evaluate using the condition \(f(x) = x - 1\). This part of the function yields a linear output that directly depends on the input. Understanding how each section of a piecewise function behaves can demystify function evaluation, allowing you to correctly compute \(f(x)\) for any given \(x\).
Being meticulous about the conditions and intervals helps avoid miscalculations and ensures that each piece of the function is applied accurately.
Mathematical Problem Solving
Mathematical problem-solving often requires synthesizing information from different parts of the problem. With the exercise involving the piecewise function \(f\), we begin by identifying relevant intervals and applying appropriate expressions for each sub-section.
When solving problems, especially those involving conditions like \(f(x) = x\), it's essential to methodically check each interval that applies. Here, the task is to find \(x\) such that \(x \leq 0\) and \(f(x) = x\). Start by setting \(f(x)\) equal to \(x\) in each interval expression separately. For \([-2, 0]\), solve \(-1 = x\) to find \(x = -1\), which is indeed within the interval and satisfies the condition.
By breaking down the problem into smaller, manageable parts — understanding intervals, evaluating pieces of the function, and logically working through conditions — a seemingly complex problem becomes much more approachable. This organized approach increases efficiency and reduces errors, aiding in effective mathematical problem-solving.
When solving problems, especially those involving conditions like \(f(x) = x\), it's essential to methodically check each interval that applies. Here, the task is to find \(x\) such that \(x \leq 0\) and \(f(x) = x\). Start by setting \(f(x)\) equal to \(x\) in each interval expression separately. For \([-2, 0]\), solve \(-1 = x\) to find \(x = -1\), which is indeed within the interval and satisfies the condition.
By breaking down the problem into smaller, manageable parts — understanding intervals, evaluating pieces of the function, and logically working through conditions — a seemingly complex problem becomes much more approachable. This organized approach increases efficiency and reduces errors, aiding in effective mathematical problem-solving.
Other exercises in this chapter
Problem 43
The function $$ f(x)=\sin ^{-1}\left(x-x^{2}\right)+\sqrt{1-\frac{1}{|x|}}+\frac{1}{\left[x^{2}-1\right]} $$ is defined in the interval (where [-] is the greate
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View solution Problem 49
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View solution Problem 50
If \(q^{2}-4 p r=0, p>0\), then the domain of the function \(f(x)=\log \left[p x^{3}+(p+q) x^{2}+(q+r) x+r\right]\) is (A) \(R-\left\\{-\frac{q}{2 p}\right\\}\)
View solution