Question

I don't know what to do. This is about odd and even functions.

Answer 1

Answer:

C.

Step-by-step explanation:

We are given $f,g$ are odd which means:

$f(-x)=-f(x)$

$g(-x)=-g(x)$

We can tell if a function,$h$, is even if $h(-x)=h(x)$.

We can tell if a function,$h$, is odd if $h(-x)=-h(x)$.

So let's test your I,II,III.

We will be replacing x with -x to find out.

I.

$p(x)=f(g(x))$

$p(-x)=f(g(-x))$

$p(-x)=f(-g(x))$

$p(-x)=-f(g(x))$

$p(-x)=-p(x)$

So $p$ is odd.

II

$r(x)=f(x)+g(x)$

$r(-x)=f(-x)+g(-x)$

$r(-x)=-f(x)+-g(x)$

$r(-x)=-(f(x)+g(x))$

$r(-x)=-(r(x))$

$r(-x)=-r(x)$

So $r$ is odd.

III

$s(x)=f(x)\cdot g(x)$

$s(-x)=f(-x) \cdot g(-x)$

$s(-x)=-f(x) \cdot -g(x)$

$s(-x)=f(x) \cdot g(x)$

$s(-x)=s(x)$

So $s$ is even.

So I and II are odd and III is even.

C. is the answer.

Answer 2

The answer will be C