8. Suppose function \( f \) is odd and function \( g \) is even. Is the function \[ \begin{array}{llll}\text { (a) } f g & \text { (b) } f \circ g & \text { (c) } g \circ f & \text { (d) } f-g\end{array} \]

Let's explore the nature of the combinations of odd and even functions! When you multiply an odd function \( f \) with an even function \( g \) (like in part (a) \( fg \)), the result is an odd function. This is because \( f(-x) = -f(x) \) and \( g(-x) = g(x) \), so \( (fg)(-x) = f(-x)g(-x) = (-f(x))(g(x)) = -fg(x) \). Now, if we apply an even function \( g \) to an odd function \( f \) (as in part (b) \( f \circ g \)), we find that the composition \( f(g(x)) \) does not guarantee a specific even or odd nature. This is because we can't simply determine the parity of the result without knowing more about how \( g \) interacts with \( f \). For part (c), \( g(f(x)) \) also doesn't yield a predictable parity since \( g \) is even and works on \( f(x) \), which is odd. Thus, the outcome can be mixed depending on the specific equations involved! Lastly, in part (d), \( f - g \) would combine odd and even functions and, intriguingly, the result \( (f-g)(-x) = f(-x) - g(-x) = -f(x) - g(x) \) yields \( -(f(x) - g(x)) \), which showcases that \( f - g \) is also odd. To summarize: - (a) \( fg \) is odd. - (b) \( f \circ g \) is mixed and needs more info. - (c) \( g \circ f \) is also mixed. - (d) \( f - g \) is odd! Understanding these properties can help you master the mathematical symphony of odd and even functions! 🎶