How do you know if f(x)=x^3+1f(x)=x3+1 is an even or odd function?

2 Answers
Dec 15, 2015

Odd.

Explanation:

Quite simply, whether a function is even or odd is whether the degree (the largest exponent) is even or odd.
The degree of x^3 + 1x3+1 is 33, which is odd, making this an odd function.

Dec 16, 2015

Neither.

Explanation:

An function is even if: f(-x)=f(x)f(x)=f(x)

A function is odd if: f(-x)=-f(x)f(x)=f(x)

If f(-x)=x^3+1f(x)=x3+1, the function is even.
If f(-x)=-x^3-1f(x)=x31, the function is odd.

So, find f(-x)f(x).

f(-x)=(-x)^3+1f(x)=(x)3+1

f(-x)=-x^3+1f(x)=x3+1

This function is neither odd nor even.

A good way to check is by recognizing that even functions are reflections of themselves over the yy-axis and odd functions are reflections of themselves over the xx-axis.

This is the graph of x^3+1x3+1: graph{x^3+1 [-10, 10, -5, 5]}

Notice that if the graph were shifted down one unit (the function x^3x3), the graph would be a reflection of itself over the xx-axis and would indeed be odd.