问题 1.
For $f(x)=3 x^{3}-x$ we have
$$
\begin{aligned}
f(-x) &=3(-x)^{3}-(-x)=-3 x^{3}+x \
&=-\left(3 x^{3}-x\right)=-f(x)
\end{aligned}
$$
So $3 x^{3}-x$ is an odd function.
For $f(x)=\frac{x^{2}}{1+x^{2}}$
$$
f(-x)=\frac{(-x)^{2}}{1+(-x)^{2}}=\frac{x^{2}}{1+x^{2}}=f(x)
$$
so this is even.
证明 .
If $f(x)=\frac{2 x}{x^{2}-1}$ then
$$
f(-x)=\frac{2(-x)}{(-x)^{2}-1}=-\frac{2 x}{x^{2}-1}=-f(x)
$$
so $f(x)$ is odd.
If $f(x)=\frac{x^{2}}{x+1}$ we have
$$
f(-x)=\frac{(-x)^{2}}{-x+1}=\frac{x^{2}}{1-x}
$$
This is not equal to $f(x)$ or $-f(x)$ and so this function is neither odd nor even.
ENG1005 COURSE NOTES ：
$$
y=f(x)=\frac{1}{x-2} \quad(x \neq 2)
$$
If $y=4$, find $x$.
We have
$$
4=\frac{1}{x-2}
$$
so
$$
4(x-2)=1=4 x-8
$$
and therefore
$$
4 x=9
$$
and
$$
x=\frac{9}{4}
$$