If f and g are odd functions, which of the following must also be odd?I. f(g(x))II. f(x) + g(x)III. f(x)g(x)

To determine which of the given functions must also be odd, we need to recall the definition of an odd function. A function $h(x)$ is odd if $h(-x) = -h(x)$ for all $x$.

Let's analyze each option step by step:

I. $f(g(x))$

Given that $f$ and $g$ are odd functions:

• $g(-x) = -g(x)$
• $f(-x) = -f(x)$

We need to check if $f(g(x))$ is odd: $f(g(-x)) = f(-g(x))$ Since $g$ is odd: $f(-g(x)) = f(-g(x))$ Since $f$ is odd: $f(-g(x)) = -f(g(x))$

Thus, $f(g(-x)) = -f(g(x))$, which means $f(g(x))$ is odd.

II. $f(x) + g(x)$

Given that $f$ and $g$ are odd functions:

• $f(-x) = -f(x)$
• $g(-x) = -g(x)$

We need to check if $f(x) + g(x)$ is odd: $(f + g)(-x) = f(-x) + g(-x)$ Since $f$ and $g$ are odd: $f(-x) + g(-x) = -f(x) + -g(x) = -(f(x) + g(x))$

Thus, $f(x) + g(x)$ is odd.

III. $f(x)g(x)$

Given that $f$ and $g$ are odd functions:

• $f(-x) = -f(x)$
• $g(-x) = -g(x)$

We need to check if $f(x)g(x)$ is odd: $(f \cdot g)(-x) = f(-x) \cdot g(-x)$ Since $f$ and $g$ are odd: $f(-x) \cdot g(-x) = (-f(x)) \cdot (-g(x)) = f(x) \cdot g(x)$

Thus, $f(x)g(x)$ is even, not odd.

Conclusion:

The functions that must also be odd are:

• I. $f(g(x))$
• II. $f(x) + g(x)$

So, the correct answer is I and II.