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Chapter 1: Problem 70

Determine whether f is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually. \(f(x)=x|x|\)

Short Answer

Expert verified
The function is neither even nor odd.

Step by step solution

01

Understand the Function

The function given is \(f(x) = x|x|\). The term \(|x|\) represents the absolute value of \(x\), which is \(x\) if \(x \geq 0\) and \(-x\) if \(x < 0\). Combine these regions with \(x\) as given in the function.
02

Verify Symmetry for Even Function

A function is even if \(f(-x) = f(x)\) for all \(x\) in the domain. Replace \(x\) by \(-x\) in the function: \(f(-x) = (-x)|-x| = x|x|\) because \(|-x| = x\). Hence, \(f(-x) = x(x) = x^2\). This is not equal to \(f(x) = x|x|\) for all \(x\). Thus, the function is not even.
03

Verify Symmetry for Odd Function

A function is odd if \(f(-x) = -f(x)\) for all \(x\). We already calculated \(f(-x) = x^2\). Now, \(-f(x) = -(x|x|)\). For positive \(x\), this becomes \(-x^2\), and for negative \(x\), this becomes \(x^2\), which is \(-f(x) = x|x|\). These do not satisfy \(f(-x) = -f(x)\) for all \(x\). Thus, the function is not odd.
04

Conclusion

Since the function is neither even nor odd based on the checks in Steps 2 and 3, conclude \(f(x) = x|x|\) is neither even nor odd.

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Key Concepts

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

Symmetry
When exploring mathematical functions, symmetry is a key concept in determining whether a function is even or odd. A function is said to be symmetric if it maintains certain properties when its input is altered in a specific way.
  • An even function exhibits symmetry about the y-axis. This means that for every point \(x, y\) on the graph of the function, there is a corresponding point \(-x, y\) with the same y-value. This symmetry can be mathematically expressed as \(f(-x) = f(x)\).
  • An odd function shows rotational symmetry about the origin, implying that for every point \(x, y\) on the graph, there is a corresponding point \(-x, -y\) that creates a mirror image through the origin. The condition for a function to be odd is given by \(f(-x) = -f(x)\).
To determine if a function is even or odd, you would substitute \(-x\) into the function and compare the result with \(f(x)\) and \(-f(x)\). Understanding this concept is crucial for analyzing functions and predicting their graphical behavior.
Absolute Value Function
The absolute value function, denoted as \(|x|\), is a fundamental concept in mathematics. It represents the distance of a number from zero on the number line, thereby always producing a non-negative result.
  • If \(x \geq 0\), then we have \(|x| = x\), thus the value remains unchanged.
  • If \(x < 0\), then \(|x| = -x\), effectively reflecting the negative value to its positive counterpart.
In the function \(f(x) = x|x|\), the absolute value significantly impacts the behavior and form of the function. It alters how \(f(x)\) is computed for negative and positive values of \(x\). This function's output is quadratically dominant for positive inputs and exhibits different properties on the positive and negative sides of the x-axis, making it interesting to study in terms of symmetry and value distribution.
Function Analysis
Function analysis involves examining the properties and behavior of a given function. It includes determining attributes such as evenness, oddness, domain, range, and intercepts.
  • For \(f(x) = x|x|\), testing for evenness involves checking if \(f(x)\) equals \(f(-x)\). Substituting \(-x\) yields \(f(-x) = x|x|\), which is not the same as \(f(x) = x|x|\) for all \(x\), proving the function is not even.
  • For oddness, you'd verify \(f(-x) = -f(x)\). From our solutions, calculating \(-f(x)\) does not equate with \(f(-x)\) for every \(x\), discounting it as an odd function.
Since it fails the conditions for being even or odd, \(f(x) = x|x|\) is neither. Thorough function analysis allows students to not only solve specific queries but also gain insights into how a function behaves overall, contributing to a deeper understanding of calculus and algebraic principles.

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