Prove that, if f and g are functions, then f g is a function by showing that f g

Question

Prove that, if f and g are functions, then f g is a function by showing that f g = where A = implies (x, y) that is, . Now suppose Since A = Dom(f), there is w B such that (x, w) f. Hence from and we have . This shows Therefore, Claim. If f and functions on A, and then Proof. Suppose and Since This given Since , by cancellation we have Claim. If f, g, and are functions on A, then Proof. Using associatively and Theorems 4.2.3 and 4.2.4, Claim. If on an open interval (a, b), then f is increasing on (a, b). Proof. Assume that k on the interval (a, b) . Suppose x1 and x2 are in (a, b) and . We must show that We know from calculus that since f is differentiable on (x1, x2). By the Mean Value Theorem, there exists c in (x1, x2) such that Therefore, . By hypothesis and since .Therefore, We conclude that

Explanation / Answer

Let f and g be odd functions and let h=f*g.
Since f and g are odd functions f(-x)=-f(x) and g(-x)=-g(x) for all x.

h(-x) = f(-x)*g(-x)
= -f(x) * -g(x)
= f(x)*g(x)
= h(x)

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