Identifying Even Functions A Step By Step Guide

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Determining whether a function is even is a fundamental concept in mathematics, particularly in the study of functions and their properties. An even function exhibits a specific type of symmetry, which can be identified both graphically and algebraically. In this article, we will delve into the characteristics of even functions, explore how to identify them, and apply this knowledge to determine which of the given functionsβ€”g(x)=(xβˆ’1)2+1g(x)=(x-1)^2+1, g(x)=2x2+1g(x)=2x^2+1, g(x)=4x+2g(x)=4x+2, and g(x)=2xg(x)=2xβ€”is indeed even. Understanding even functions is not only crucial for solving mathematical problems but also provides a deeper insight into the behavior and nature of various mathematical expressions.

Understanding Even Functions

To accurately identify an even function, it's essential to first grasp the formal definition and underlying principles. A function is classified as even if it satisfies a specific symmetry condition: for every value of x in the function's domain, the function value at x is the same as the function value at -x. Mathematically, this condition is expressed as g(x) = g(-x). This definition implies that the graph of an even function is symmetric with respect to the y-axis. This means if you were to fold the graph along the y-axis, the two halves would perfectly overlap. Visually, this symmetry makes even functions easily recognizable on a graph. However, to rigorously confirm whether a function is even, we must rely on the algebraic definition.

The Algebraic Test for Even Functions

The algebraic test is the most reliable method for verifying if a function is even. This test involves substituting -x into the function and simplifying the expression. If the simplified expression is identical to the original function, then the function is even. Let’s break down the process step-by-step:

  1. Substitute -x for x: Replace every instance of x in the function's equation with -x. This is the first crucial step in determining whether the function meets the even function criteria.
  2. Simplify the Expression: After the substitution, carefully simplify the resulting expression. This often involves dealing with negative signs, exponents, and other algebraic operations. Accuracy in this step is vital, as any error can lead to an incorrect conclusion.
  3. Compare with the Original Function: Once the expression is simplified, compare it with the original function g(x). If the simplified expression is exactly the same as g(x), then the function is even. If it is different, the function is not even.

Understanding this algebraic test is paramount because it provides a definitive way to classify functions. Visual inspection of graphs can sometimes be misleading, especially with complex functions. The algebraic test provides a concrete, mathematical foundation for determining evenness.

Examples of Even Functions

Before we dive into the given options, let's look at some classic examples of even functions to solidify our understanding:

  • g(x) = x^2: This is one of the most basic and commonly cited examples. Substituting -x gives us g(-x) = (-x)^2 = x^2, which is the same as the original function. The graph of g(x) = x^2 is a parabola that opens upwards, symmetric about the y-axis.
  • g(x) = cos(x): The cosine function is another standard example. We know that cos(-x) = cos(x), which satisfies the condition for even functions. The cosine function's graph exhibits symmetry about the y-axis, oscillating evenly on both sides.
  • g(x) = |x|: The absolute value function is also even. Since the absolute value of a number is its distance from zero, |-x| = |x|. The graph of g(x) = |x| is V-shaped and symmetric about the y-axis.

These examples illustrate the common characteristics of even functions: symmetry about the y-axis and the algebraic property that g(x) = g(-x). Keeping these examples in mind will help us analyze the functions given in the problem.

Analyzing the Given Functions

Now, let's apply our understanding of even functions to the given options and determine which one meets the criteria. We will go through each function systematically, applying the algebraic test to confirm whether g(x) = g(-x).

1. g(x)=(xβˆ’1)2+1g(x) = (x-1)^2 + 1

To determine if g(x)=(xβˆ’1)2+1g(x) = (x-1)^2 + 1 is even, we need to substitute -x for x and simplify:

g(βˆ’x)=((βˆ’x)βˆ’1)2+1g(-x) = ((-x)-1)^2 + 1

Now, let's simplify the expression:

g(βˆ’x)=(βˆ’xβˆ’1)2+1g(-x) = (-x-1)^2 + 1 g(βˆ’x)=(βˆ’(x+1))2+1g(-x) = (-(x+1))^2 + 1 g(βˆ’x)=(x+1)2+1g(-x) = (x+1)^2 + 1 g(βˆ’x)=(x2+2x+1)+1g(-x) = (x^2 + 2x + 1) + 1 g(βˆ’x)=x2+2x+2g(-x) = x^2 + 2x + 2

Comparing this to the original function, g(x)=(xβˆ’1)2+1=x2βˆ’2x+2g(x) = (x-1)^2 + 1 = x^2 - 2x + 2, we see that g(βˆ’x)g(-x) is not equal to g(x)g(x). Therefore, this function is not even.

2. g(x)=2x2+1g(x) = 2x^2 + 1

Next, we examine g(x)=2x2+1g(x) = 2x^2 + 1. Substitute -x for x:

g(βˆ’x)=2(βˆ’x)2+1g(-x) = 2(-x)^2 + 1

Simplify the expression:

g(βˆ’x)=2(x2)+1g(-x) = 2(x^2) + 1 g(βˆ’x)=2x2+1g(-x) = 2x^2 + 1

In this case, g(βˆ’x)=2x2+1g(-x) = 2x^2 + 1 is exactly the same as the original function g(x)=2x2+1g(x) = 2x^2 + 1. Thus, this function is even. The term x^2 ensures that the sign of x does not affect the function value, and adding a constant (1 in this case) maintains the symmetry about the y-axis.

3. g(x)=4x+2g(x) = 4x + 2

Now, let's analyze g(x)=4x+2g(x) = 4x + 2. Substitute -x for x:

g(βˆ’x)=4(βˆ’x)+2g(-x) = 4(-x) + 2

Simplify the expression:

g(βˆ’x)=βˆ’4x+2g(-x) = -4x + 2

Comparing g(βˆ’x)=βˆ’4x+2g(-x) = -4x + 2 with the original function g(x)=4x+2g(x) = 4x + 2, we see that they are not the same. Therefore, this function is not even. The presence of the term 4x (a linear term) breaks the symmetry required for an even function.

4. g(x)=2xg(x) = 2x

Finally, let's consider g(x)=2xg(x) = 2x. Substitute -x for x:

g(βˆ’x)=2(βˆ’x)g(-x) = 2(-x)

Simplify the expression:

g(βˆ’x)=βˆ’2xg(-x) = -2x

Comparing g(βˆ’x)=βˆ’2xg(-x) = -2x with the original function g(x)=2xg(x) = 2x, we see that they are not the same. In fact, g(βˆ’x)g(-x) is the negative of g(x)g(x), which indicates that this function is odd, not even. Odd functions satisfy the condition g(-x) = -g(x) and are symmetric about the origin.

Conclusion: Identifying the Even Function

After analyzing all the given functions, we can confidently conclude that the function g(x)=2x2+1g(x) = 2x^2 + 1 is the only even function among the options. This conclusion is based on the algebraic test, where substituting -x for x resulted in the original function, thus satisfying the condition g(x) = g(-x). Understanding the characteristics and tests for even functions is essential for various mathematical analyses and problem-solving scenarios. Recognizing even functions not only simplifies calculations but also provides insights into the symmetrical properties of functions and their graphs. In summary, the correct answer is g(x)=2x2+1g(x) = 2x^2 + 1 because it meets the criteria for an even function by maintaining its form when x is replaced with -x. This systematic approach to function analysis reinforces the importance of both algebraic manipulation and conceptual understanding in mathematics.