Question

given that f(x) is even and g(x) is odd, determine whether each function is even, odd, or neither. (f·g)(x) = (g·g)(x) = even odd neither

Explanation:

For \((f \cdot g)(x)\):

Step1: Recall definitions

An even function satisfies \(f(-x) = f(x)\), an odd function satisfies \(g(-x) = -g(x)\). For \((f \cdot g)(x)=f(x)g(x)\), find \((f \cdot g)(-x)\).
\((f \cdot g)(-x)=f(-x)g(-x)\)

Step2: Substitute even/odd properties

Since \(f\) is even, \(f(-x)=f(x)\); \(g\) is odd, \(g(-x)=-g(x)\). So \((f \cdot g)(-x)=f(x)(-g(x))=-f(x)g(x)=-(f \cdot g)(x)\).
Thus, \((f \cdot g)(x)\) is odd.

For \((g \cdot g)(x)\):

Step1: Recall definitions

\((g \cdot g)(x)=g(x)g(x)=[g(x)]^2\). Find \((g \cdot g)(-x)\).
\((g \cdot g)(-x)=g(-x)g(-x)\)

Step2: Substitute odd property

\(g\) is odd, so \(g(-x)=-g(x)\). Then \((g \cdot g)(-x)=(-g(x))(-g(x))=g(x)g(x)=[g(x)]^2=(g \cdot g)(x)\).
Thus, \((g \cdot g)(x)\) is even.

Answer:

\((f \cdot g)(x)\): odd
\((g \cdot g)(x)\): even