Math Lesson 16.7.2 - How to Prove the Evenness of a Function Analytically

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Welcome to our Math lesson on How to Prove the Evenness of a Function Analytically, this is the second lesson of our suite of math lessons covering the topic of Even and Odd Functions, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

How to Prove the Evenness of a Function Analytically

The procedure described above for evaluating whether a function is even or not is only good for illustration purposes but we can't rely on it too much. This is because it is feasible that two opposite x-values give the same y-value though this may simply be a coincidence. For example, if we take x = -1 and x = 1 in the function f(x) = x³ - x + 1, we obtain the same y-values, i.e.

and

However, this is simply a coincidence, as for other opposite x-values we don't obtain the same y-value. Thus, for x = -2 and x = 2 we have

and

Hence, we cannot rely on numerical substitutions to confirm whether a function is even or not but it is necessary to use a general method for this. Indeed, it is clear that if we are able to obtain two identical expressions for f(-x) and f(x), then the function f(x) is even. In this way, no numerical substitutions are necessary to prove the evenness of a function. This reasoning provides us with the general rule of even functions:

If f(x) = f(-x) for any x-value, then the function f(x) is even.

For example, if we want to know whether the function f(x) = x⁴ + x² + 3 is even or not, we write

and

Thus, since f(x) = f(-x) for any x, the function f(x) = x4 + x² + 3 is even. If you have doubts about the findings above, subtract the expression of f(-x) from f(x). If the difference is zero, then it is confirmed that f(x) is even. Indeed,

In fact, we have not simply obtained a '0' as a result but an equation of the type 0x = 0, which is an identity, as discussed in tutorial 9.3. This means it is always true for any x. Therefore, the expressions belonging to f(x) and f(-x) are identical, so the function f(x) is even. In other words, a function is even if and only if f(x) - f(-x) = 0 for any x.

Example 2

Check whether the following functions are even or not without making any numerical substitutions.

1. f(x) = x² + 4x + 4
2. g(x) = x6 - x² + 1
3. h(x) = 1 - x² + 2x

Solution 2

1. We have
   f(x) = x² + 4x + 4
   and
   f(-x) = (-x)² + 4 ∙ (-x) + 4
   = x² - 4x + 4
   Let's see whether f(x) - f(-x) gives zero as a result or not. Thus,
   f(x) - f(-x) = (x² + 4x + 4) - (x² - 4x + 4)
   = x² + 4x + 4 - x² + 4x - 4
   = 8x
   Therefore, this function is not even, because f(x) - f(-x) ≠ 0.
2. We have
   g(x) = x⁶ - x² + 1
   and
   g(-x) = (-x)⁶ - (-x)² + 1
   = x⁶ - x² + 1
   Hence,
   g(x) - g(-x) = (x⁶ - x² + 1) - (x⁶ - x² + 1)
   = x⁶ - x² + 1 - x² + x² - 1
   = 0
   Therefore, since g(x) - g(-x) = 0, then g(x) = g(-x). Therefore, the function g(x) is even.
3. We have
   h(x) = 1 - x² - 2^x
   and
   h(-x) = 1 - (-x)² - 2^-x
   = 1 - x² - 2^-x
   Subtracting f(-x) from f(x) yields
   h(x) - h(-x) = (1 - x² - 2^x) - (1 - x² - 2^-x)
   = 1 - x² - 2^x - 1 + x² + 2^-x
   = -2^x + 2^-x
   = 2^-x - 2^x
   = 1/2^x - 2^x
   = 1 - (2^x)²)/2^x
   = 1 - 2^2x/2^x
   Thus, since h(x) - h(-x) ≠ 0, the function h(x) is not even.

You have reached the end of Math lesson 16.7.2 How to Prove the Evenness of a Function Analytically. There are 10 lessons in this physics tutorial covering Even and Odd Functions, you can access all the lessons from this tutorial below.

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