Answer
$\begin{array}{llll}
a. & 0 & e. & \frac{-x}{x^{2}+1}\\
b. & \frac{1}{2} & f. & \frac{x+1}{x^{2}+2x+2}\\
c. & -\frac{1}{2} & g. & \frac{2x}{4x^{2}+1}\\
d. & \frac{-x}{x^{2}+1} & h. & \frac{x+h}{x^{2}+2xh+h^{2}+1}
\end{array}$
Work Step by Step
$f(x)=\displaystyle \frac{x}{x^{2}+1}$
$\begin{array}{lll}
(a)\ f(0) & (b)\ f(1) & (c)\ f(-1)\quad \\
=\frac{0}{0^{2}+1} & =\frac{1}{1^{2}+1} & =\frac{-1}{(-1)^{2}+1}\\
=0 & =\frac{1}{2} & =-\frac{1}{2}\\
& &
\end{array}$
$\begin{array}{ll}
(d)\ f(-x) & (e)\ -f(x)\\
=\frac{-x}{(-x)^{2}+1} & =-(\frac{x}{x^{2}+1})\\
=\frac{-x}{x^{2}+1}\quad & =\frac{-x}{x^{2}+1}\\
&
\end{array}$
$\begin{array}{lll}
(f)\ f(x+1) & (g)\ f(2x) & \\
=\frac{x+1}{(x+1)^{2}+1} & =\frac{2x}{(2x)^{2}+1} & \\
=\frac{x+1}{x^{2}+2x+1+1} & =\frac{2x}{4x^{2}+1} & \\
=\frac{x+1}{x^{2}+2x+2} & & \\
& & \\
& &
\end{array}$
$(h)\displaystyle \ f(x+h)=\frac{x+h}{(x+h)^{2}+1}$
$=\displaystyle \frac{x+h}{x^{2}+2xh+h^{2}+1}$
