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