Answer
- Odd functions: a, b, c, i
- Even functions: d, e, f, g, h
Work Step by Step
1) Recall that:
- If $f$ is an even function, then $f(-x)=f(x)$
- If $g$ is an odd function, then $g(-x)=-g(x)$
$f$ and $g$ are defined on $(-\infty,\infty)$
2) To know which is odd and which is even, we need to test $f(-x)$ and $g(-x)$. In detail,
(a) $fg$
- We check $f(-x)g(-x)$
- Remember again that $f(-x)=f(x)$ and $g(-x)=-g(x)$
- Therefore, $f(-x)g(-x)=f(x)[-g(x)]=-f(x)g(x)=-fg$
So $fg$ is an odd function (defined on $(-\infty,\infty)$)
(b) $f/g$
$f(-x)/g(-x)=f(x)/-g(x)=-[f(x)/g(x)]=-f/g$
$f/g$ is an odd function (where $g\ne0$)
(c) $g/f$
$g(-x)/f(-x)=-g(x)/f(x)=-[g(x)/f(x)]=-g/f$
$g/f$ is an odd function (where $f\ne0$)
(d) $f^2=ff$
$f(-x)f(-x)=f(x)f(x)=ff=f^2$
$f^2$ is an even function.
(e) $g^2=gg$
$g(-x)g(-x)=-g(x)[-g(x)]=g(x)g(x)=gg=g^2$
$g^2$ is an even function.
(f) $f\circ g$
$f(-x)\circ g(-x)=f(-g(-x))=f(-(-g(x)))=f(g(x))=(f\circ g)(x)=f\circ g$
$f\circ g$ is an even function.
(g) $g\circ f=(g\circ f)(x)=g(f(x))$
$(g\circ f)(-x)=g(f(-x))=g(f(x)))$
$g\circ f$ is an even function.
(h) $f\circ f=(f\circ f)(x)=f(f(x))$
$(f\circ f)(-x)=f(f(-x))=f(f(x)))$
$f\circ f$ is an even function.
(i) $g\circ g=(g\circ g)(x)=g(g(x))$
$(g\circ g)(-x)=g(g(-x))=g(-g(x)))=-g(g(x))$ (because $g$ is an odd function)
$g\circ g$ is an odd function.
