How do you determine if #1/(x^3+1)# is an even or odd function?

Question

Theodore Amidon · Accepted Answer

Evaluate #f(2)# and #f(-2)# to show that it is neither.

A function is even if #f(-x) = f(x)# for any #x# A function is odd if #f(-x) = -f(x)# for any #x# In our example, let #f(x) = 1/(x^3+1)# We find: #f(2) = 1/(8+1) = 1/9# #f(-2) = 1/(-8+1) = 1/-7 = -1/7# So neither #f(-x) = f(x)# for any #x#, nor #f(-x) = -f(x)# for any #x#. So this function is neither even nor odd.

Ariana Alvarez · Answer

undefined To determine if $ \frac{1}{x^3+1} $ is an even or odd function, we can use the properties of even and odd functions:

1. An even function satisfies $ f(-x) = f(x) $ for all $ x $ in the domain.
2. An odd function satisfies $ f(-x) = -f(x) $ for all $ x $ in the domain.

Now, let's analyze $ \frac{1}{x^3+1} $ for even or odd behavior:

1. Substitute $ -x $ into the function:
$$ f(-x) = \frac{1}{(-x)^3+1} = \frac{1}{-x^3+1} $$

2. Compare $ f(-x) $ with $ f(x) $:
$$ f(-x) = \frac{1}{-x^3+1} $$
$$ f(x) = \frac{1}{x^3+1} $$

Since $ f(-x) = \frac{1}{-x^3+1} $ is not equal to $ f(x) = \frac{1}{x^3+1} $, the function $ \frac{1}{x^3+1} $ does not satisfy the condition for being an even function.

Now, let's check if it's an odd function:
$$ f(-x) = \frac{1}{-x^3+1} $$
$$ -f(x) = -\frac{1}{x^3+1} $$

Since $ f(-x) = \frac{1}{-x^3+1} $ is equal to $ -f(x) = -\frac{1}{x^3+1} $, the function $ \frac{1}{x^3+1} $ satisfies the condition for being an odd function.