Answer
$\begin{array}{llll}
a. & -\frac{1}{5} & e. & \frac{-2x-1}{3x-5}\\
b. & -\frac{3}{2} & f. & \frac{2x+3}{3x-2}\\
c. & \frac{1}{8} & g. & \frac{4x+1}{6x-5}\\
d. & \frac{2x-1}{3x+5} & h. & \frac{2x+2h+1}{3x+3h-5}
\end{array}$
Work Step by Step
$f(x)=\displaystyle \frac{2x+1}{3x-5}$
$\begin{array}{lll}
(a)\ f(0) & (b)\ f(1) & (c)\ f(-1)\quad \\
=\frac{2(0)+1}{3(0)-5} & =\frac{2(1)+1}{3(1)-5} & =\frac{2(-1)+1}{3(-1)-5}\\
=\frac{0+1}{0-5} & =\frac{2+1}{3-5} & =\frac{-2+1}{-3-5}\\
=-\frac{1}{5} & =-\frac{3}{2} & =\frac{1}{8}
\end{array}$
$\begin{array}{ll}
(d)\ f(-x) & (e)\ -f(x)\\
=\frac{2(-x)+1}{3(-x)-5} & =-(\frac{2x+1}{3x-5})\\
=\frac{-2x+1}{-3x-5}\quad & =\frac{-2x-1}{3x-5}\\
=\frac{2x-1}{3x+5} &
\end{array}$
$\begin{array}{ll}
(f)\ f(x+1) & (g)\ f(2x)\\
=\frac{2(x+1)+1}{3(x+1)-5} & =\frac{2(2x)+1}{3(2x)-5}\\
=\frac{2x+2+1}{3x+3-5} & =\frac{4x+1}{6x-5}\\
=\frac{2x+3}{3x-2} & \\
& \\
&
\end{array}$
$(h)\displaystyle \ f(x+h)=\frac{2(x+h)+1}{3(x+h)-5}=\frac{2x+2h+1}{3x+3h-5}$
