Is the function #f(x)=1/(x^3+1)# even, odd or neither?

Question

Is the function #f(x)=1/(x^3+1)#  even, odd or neither?

Lincoln Anderson · Accepted Answer

It is neither .

A function f(x) is even if f(-x)=f(x) and odd if f(-x)=-f(x) Putting x=-x we get f(x)=#1/(-x^3+1)# which is not equal to either f(x) or f(-x). So its neither of the two. Hope it helps!!

Serenity Jones · Answer

undefined To determine if the function $f(x) = \frac{1}{x^3 + 1}$ is even, odd, or neither, we need to check its symmetry properties.

1. **Even function**: A function $f(x)$ is even if $f(x) = f(-x)$ for all $x$ in the function's domain.

2. **Odd function**: A function $f(x)$ is odd if $f(x) = -f(-x)$ for all $x$ in the function's domain.

Let's check the given function:

$$f(x) = \frac{1}{x^3 + 1}$$

Now, let's evaluate $f(-x)$:

$$f(-x) = \frac{1}{(-x)^3 + 1} = \frac{1}{-x^3 + 1}$$

For this function to be even, we need $f(x) = f(-x)$:

$$\frac{1}{x^3 + 1} = \frac{1}{-x^3 + 1}$$

However, this doesn't hold true, as the denominator changes signs when $x$ changes sign, resulting in different function values.

Now, let's check if it's odd:

For the function to be odd, we need $f(x) = -f(-x)$:

$$\frac{1}{x^3 + 1} = -\frac{1}{-x^3 + 1}$$

But this also doesn't hold true, as the function values are not negatives of each other.

Thus, the function $f(x) = \frac{1}{x^3 + 1}$ is neither even nor odd.