Question

is the function w(x) = -2x^5 + 7x^2 even, odd, or neither?
even
odd
neither

Explanation:

Step1: Recall definitions

A function \( w(x) \) is even if \( w(-x) = w(x) \) for all \( x \), and odd if \( w(-x) = -w(x) \) for all \( x \).

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

Given \( w(x) = -2x^5 + 7x^2 \), substitute \( -x \) for \( x \):
\( w(-x) = -2(-x)^5 + 7(-x)^2 \).

Step3: Simplify \( w(-x) \)

Simplify each term:
\( (-x)^5 = -x^5 \), so \( -2(-x)^5 = -2(-x^5) = 2x^5 \).
\( (-x)^2 = x^2 \), so \( 7(-x)^2 = 7x^2 \).
Thus, \( w(-x) = 2x^5 + 7x^2 \).

Step4: Compare \( w(-x) \) with \( w(x) \) and \( -w(x) \)

• \( w(x) = -2x^5 + 7x^2 \), so \( -w(x) = 2x^5 - 7x^2 \).
• \( w(-x) = 2x^5 + 7x^2 \) does not equal \( w(x) \) (since \( 2x^5

eq -2x^5 \) for \( x
eq 0 \)) and does not equal \( -w(x) \) (since \( 7x^2
eq -7x^2 \) for \( x
eq 0 \)).

Answer:

neither