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Chapter 1: Problem 13

Find the number \((\mathrm{s}) x,\) if any, where \(f\) takes on the value 1. $$f(x)=|2-x|$$

Short Answer

Expert verified
The numbers \(x\), where \(f(x) = 1\) for \(f(x) = |2-x|\) are \(x = 1\) and \(x = 3\).

Step by step solution

01

Solve for (2-x)>0

Here, we start with the equation \(|2-x| = 1\). Since we are considering the case where (2-x) is positive, this equation simplifies to \(2-x = 1\). Simplify this equation for x and find the corresponding value.
02

Solve for (2-x)

Here, we start with the equation \(|2-x| = 1\). Since we are considering the case where (2-x) is negative, this equation simplifies to \(- (2-x) = 1\). Simplify this equation for x and find the corresponding value.
03

Put the Results Together

After solving for x in both cases, it is important to put the results together in a unique set because the absolute value function can have more than one solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Absolute Value Equations
When solving absolute value equations, like \( |2-x| = 1 \), it is crucial to consider all potential cases. The absolute value of a number represents its distance from zero, which implies it can be positive or negative. Hence, when solving \(|A| = B\) where \(B\) is positive, you need to address two scenarios:
  • Case 1: \(A = B\)
  • Case 2: \(A = -B\)
These cases ensure that you cover both possible values of \(A\). In the original problem, you solve for \((2-x) = 1\) and \(-(2-x) = 1\) to find the values of \(x\). This results in two separate solutions, which might both be valid depending on the context of the function.
Understanding Piecewise Functions
Piecewise functions operate in separate sections depending on specific conditions. For functions like \(f(x) = |2-x|\), the usage of an absolute value naturally divides the function into multiple pieces. In essence, the entire function behaves differently based on the value of the expression inside the absolute value:
  • If \(2-x > 0\), the function simplifies to \(f(x) = 2-x\).
  • If \(2-x < 0\), it changes to \(f(x) = -(2-x) = x-2\).
This means the function has distinct expressions depending on whether \(x\) is less than, greater than, or equal to 2. Recognizing these segments helps in correctly analyzing the function's behavior and solving any related equations.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and simplifying equations to find desired variables. Here, once you decide on the scenario for the absolute value equation, you simplify it to solve for \(x\):
  • Starting with \(2-x = 1\), add \(x\) to both sides and subtract 1 to get: \(x = 1\).
  • For the alternative scenario \(-(2-x) = 1\): start by distributing the negative, turning it into \(-2 + x = 1\), which simplifies to \(x = 3\) after solving.
Algebraic manipulation allows you to isolate \(x\) by carefully performing operations on both sides of the equation. This process ensures that each potential solution fits within the boundaries of the defined conditions for the function.

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Most popular questions from this chapter

Find \(g\) such that \(f \circ g=F\) given that $$f(x)=x^{2}+1 , F(x)=\left(2 x^{3}-1\right)^{2}+1$$

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