Problem 50 - Thomas Calculus

Question

In Exercises \(47-58,\) say whether the function is even, odd, or neither. Give reasons for your answer. $$ f(x)=x^{2}+x $$

Step-by-Step Solution

Answer

The function \( f(x) = x^2 + x \) is neither even nor odd.

Step 1: Determine Even Function Criteria

A function is considered even if it satisfies the condition \( f(-x) = f(x) \) for all \( x \) in its domain. Let's first find \( f(-x) \).

Step 2: Calculate f(-x)

Substitute \( -x \) into the function: \[ f(-x) = (-x)^2 + (-x) = x^2 - x \].

Step 3: Compare f(x) and f(-x) for Evenness

We found \( f(x) = x^2 + x \) and \( f(-x) = x^2 - x \). Since \( f(-x) \) is not equal to \( f(x) \), the function is not even.

Step 4: Determine Odd Function Criteria

A function is considered odd if it satisfies the condition \( f(-x) = -f(x) \) for all \( x \) in its domain. Let's see if this holds.

Step 5: Check f(-x) = -f(x) for Oddness

Calculate \(-f(x)\): \[-f(x) = -(x^2 + x) = -x^2 - x \]. Now compare it with \( f(-x) = x^2 - x \). These are not equal, therefore the function is not odd.

Step 6: Conclude Based on Even and Odd Criteria

Since the function is neither even nor odd based on our criteria, \( f(x) = x^2 + x \) is neither even nor odd.

Key Concepts

Function Behavior

When analyzing a function, understanding its behavior helps us determine various characteristics about its graph. This includes identifying if the function is increasing or decreasing, any points of inflection, and recognizing if it's an even or odd function.
To start, note how functions behave as their input values change. For example, in our exercise, the function \( f(x) = x^2 + x \) has different behaviors as \( x \) changes. As \( x \) increases, both \( x^2 \) and \( x \) terms will increasingly add to the function's value. The upward-trending output indicates that the function is not entirely decreasing or increasing but shows a mix of both behaviors depending on the values of \( x \).
Observing function behavior provides a foundation for determining other properties such as symmetry, which we'll explore next.

Symmetry in Functions

Symmetry in functions helps us understand how a function behaves around the origin or the y-axis. A function can be symmetric in different ways, often described as either even or odd.

• **Even Functions**: A function is classified as even if it satisfies the condition \( f(-x) = f(x) \). This results in a graph that is symmetric about the y-axis. It means every point on the function when flipped over the y-axis remains unchanged.
• **Odd Functions**: A function is classified as odd if it satisfies the condition \( f(-x) = -f(x) \). This results in a graph that has rotational symmetry about the origin. Thus, when rotated 180 degrees around the origin, the graph looks the same.

For the function \( f(x) = x^2 + x \), through evaluation, we see that neither condition is satisfied. Calculating \( f(-x) = x^2 - x \), it's clear that it does not match \( f(x) = x^2 + x \) (so it's not even), nor does it result in the negative equivalent \( -f(x) = -x^2 - x \), hence, it's not odd.
This analysis indicates that \( f(x) = x^2 + x \) lacks reflective or rotational symmetry about the axes typically seen in purely even or odd functions.

Polynomial Functions

Polynomial functions have distinct characteristics and include terms with non-negative integer exponents, such as \( a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0 \). Each term in the polynomial has a coefficient and a power of \( x \).
The function \( f(x) = x^2 + x \) is a polynomial function of the form \( ax^2 + bx \), where \( a = 1 \) and \( b = 1 \). Recognizing its structure is crucial for analyzing and even graphing the function. Each part influences the overall graph:

• **Quadratic term \( x^2 \)** adds a parabolic shape. As a result, for large positive or negative \( x \), the function shoots upwards, showing a characteristic U-shape.
• **Linear term \( x \)** shifts the graph, adding a tilt that influences the symmetry and other characteristics.

Through dissecting polynomial functions, we gain insights into their behavior, aiding in the determination of their symmetry, as seen in the step-by-step solution provided.