How do you know if # f(x)=x^3+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 + 1# is #3#, which is odd, making this an odd function.

Dec 16, 2015

Neither.

Explanation:

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

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

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

So, find #f(-x)#.

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

#f(-x)=-x^3+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 #y#-axis and odd functions are reflections of themselves over the #x#-axis.

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

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