Question

is the function $r(x) = 3x^6 + x^2$ even, odd, or neither?
even
odd
neither

Explanation:

Step1: Recall the definitions

A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \) in the domain. A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \) in the domain.

Step2: Compute \( r(-x) \)

Given \( r(x) = 3x^6 + x^2 \), substitute \( -x \) for \( x \):

$$ r(-x)=3(-x)^6+(-x)^2 $$

Since \( (-x)^n = x^n \) when \( n \) is even, we have \( (-x)^6 = x^6 \) and \( (-x)^2 = x^2 \). So:

$$ r(-x)=3x^6 + x^2 $$

Step3: Compare \( r(-x) \) and \( r(x) \)

We see that \( r(-x)=3x^6 + x^2 = r(x) \).

Answer:

A. even