91Ó°ÊÓ

In mathematics, an even function is a function that satisfies the equation \(f(-x) = f(x)\) for all real numbers \(x\). This is the definition we will use for our proof.

We are asked to prove that the product of two even functions is even. Assuming that \(g(x)\) and \(h(x)\) are even functions, we can say that by the definition of an even function, \(g(-x) = g(x)\) and \(h(-x) = h(x)\).\n\nNow, consider the product of \(g(x)\) and \(h(x)\), denoted by \(p(x) = g(x)h(x)\). We want to prove that \(p(x)\) is also an even function. Therefore we have to show that \(p(-x) = p(x)\).\n\nWe replace \(g(-x)\) with \(g(x)\) and \(h(-x)\) with \(h(x)\) (since they are both even functions) in the equation for \(p(-x)\) giving us: \n\n\(p(-x) = g(-x)h(-x) = g(x)h(x) = p(x)\)\n\nThis demonstrates that the product of two even functions is also an even function.

Similarly, in order to prove that the product of two odd functions is even, we first need to understand what an odd function is. By definition, a function \(f(x)\) is said to be odd if for all \(x\) in the domain, \(f(-x) = -f(x)\). Let's assume two functions, \(m(x)\) and \(n(x)\), are both odd functions. So, \(m(-x) = -m(x)\) and \(n(-x) = -n(x)\).\n\nWe now consider their product \(q(x) = m(x)n(x)\), and we want to prove that \(q(x)\) is also an even function, i.e., \(q(-x) = q(x)\).\n\nWe swap \(m(-x)\) with \(-m(x)\) and \(n(-x)\) with \(-n(x)\) in the equation for \(q(-x)\), which gives us:\n\n\(q(-x) = m(-x)n(-x) = -m(x)(-n(x)) = m(x)n(x) = q(x)\).\n\nThus, this satisfies the condition of an even function, making our statement true: the product of two odd functions is also an even function.