Suppose g is an odd function and let h = f.g. Is h always an odd function?

Solution:

Given: g is an odd function and  h = f.g

A function f is odd if f(-x) = -f(x) ∀ x∈ D(f) and

even if f(-x) = f(x) ∀ x ∈ D(f).

There are two cases to discuss this.

Case(1): When f is even function:

Let g(x) = -x and f(x) = x²

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

= f(-(-x))

= f(x) = x²

h(x) = f(g(x)) = f(g(x))

= f(-x)

= (-x)²

= x²

h(-x) = h(x)

Hence, we can say that h is an even function

Case(2): when f is an odd function:

Let g(x) = x and f(x) = x + 1

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

= f(-x)

= -x + 1

h(x) = f(g(x))

= f(x)

= x + 1

h(-x) = h(x)

Hence, we can say that h is an odd function

Therefore, h is not always odd function.

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Suppose g is an odd function and let h = f.g. Is h always an odd function?

Summary:

If suppose g is an odd function and h = f.g then h is not always an odd function.