Problem 40 - Precalculus : Building Concepts and Connections

Question

Decide if each function is odd, even, or neither by using the definitions. $$f(x)=|x-1|$$

Step-by-Step Solution

Answer

The function \(f(x)=|x-1|\) is neither even nor odd.

Step 1: Determine f(-x)

The first step is to substitute \(-x\) into the function (for \(f(-x)\)). This yields: \(f(-x) = |-x - 1|\)

Step 2: Simplify f(-x)

Next, the absolute function is simplified using the definition. Two cases arise: \(x ≥ 1\) and \(x < 1\). For \(x ≥ 1\), the expression |-x - 1| becomes -x -1. For \(x < 1\), the expression becomes x + 1.

Step 3: Comparison with f(x)

Now compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). It is noted that neither does \(f(-x)\) equals \(f(x)\), nor does \(f(-x)\) equals \(-f(x)\) for all \(x\) in its domain.

Key Concepts

Absolute Value Function

The absolute value function is a fascinating tool in mathematics. To understand it, think of it as a machine that removes any negative sign from a number. The absolute value of a number is how far it is from zero on the number line, without considering directions. For example, the absolute value of both 5 and -5 is 5. This is represented as \(|5| = 5\) and \(|-5| = 5\).

In the function given, \(f(x) = |x-1|\), the expression \(x-1\) is input into the absolute value. This function measures the distance between \(x\) and 1 on the number line. The brief interruption of direction means this function never outputs a negative value, making it unique in structure and analysis.

Absolute value functions form 'V' shaped graphs. They open upwards, have a vertex, and are always non-negative. The vertex in \(f(x) = |x-1|\) occurs at \(x = 1\), where the function value is zero. From this point, the graph climbs as \(|x-1|\) increases with both increasing and decreasing \(x\).

Odd Function

An odd function is one that showcases symmetry about the origin. Mathematically, it holds that \(f(-x) = -f(x)\) for all \(x\) in the function's domain. This means when you reflect the graph of an odd function across both the x-axis and y-axis, it appears unchanged. Common examples include \(f(x) = x^3\) or \(f(x) = \sin(x)\).

When analyzing the function \(f(x) = |x-1|\), step 3 asserts that neither \(f(-x) = f(x)\) nor \(f(-x) = -f(x)\). Specifically, \(f(-x)\) simplifies differently in regions divided by the critical point \(x=1\). Due to this, \(f(x) = |x-1|\) does not maintain the required symmetry to be classified as an odd function.

Even Function

Even functions exhibit symmetry along the y-axis, meaning \(f(x)\) is equal to \(f(-x)\) for every \(x\) in the domain. This results in a graphical representation where the left side mirrors the right side, just as in functions like \(f(x) = x^2\) or \(f(x) = \cos(x)\).

For \(f(x) = |x-1|\), when finding \(f(-x)\), different values surface based on whether \(x\) is greater or less than 1. The function does not consistently yield \(f(x) = f(-x)\) across all domain values due to the directional influence of subtraction within the absolute value. Thus, the depreciation of consistent symmetry indicates \(f(x) = |x-1|\) is not an even function either.

Function Analysis

Function analysis delves into evaluating a function’s behavior and its adherence to specific characteristics, such as whether it's odd, even, or neither. In the step-by-step solution provided, the aim was to determine the nature of the function \(f(x) = |x-1|\).

Through substitution and simplification, we derived that this function does not satisfy conditions of either evenness or oddness.

• When \(f(-x)\) does not match \(f(x)\), the even function criteria get negated.
• When \(f(-x)\) neither mirrors \(-f(x)\), any odd function status is dismissed.

Thus, the careful examination determines that \(f(x) = |x-1|\) is classified as neither even nor odd. A vital insight derived from function analysis is understanding how altering positional elements like subtraction impact symmetry-related classifications.