Answer
$\begin{array}{llll}
a. & \dfrac{3}{4} & e. & \dfrac{1}{(x+2)^{2}}-1\\\\
b. & \dfrac{8}{9} & f. & 1-\dfrac{1}{(x+3)^{2}}\\\\
c. & 0 & g. & 1-\dfrac{1}{4(x+1)^{2}}\\\\
d. & 1-\dfrac{1}{(x-2)^{2}} & h. & 1-\dfrac{1}{(x+h+2)^{2}}
\end{array}$
Work Step by Step
$f(x)=1-\displaystyle \frac{1}{(x+2)^{2}}$
$\begin{array}{lll}
(a)\ f(0) & (b)\ f(1) & (c)\ f(-1)\quad \\
=1-\frac{1}{(0+2)^{2}} & =1-\frac{1}{(1+2)^{2}} & =1-\frac{1}{(-1+2)^{2}}\\
=1-\frac{1}{4} & =1-\frac{1}{9} & =1-\frac{1}{1}\\
=\frac{3}{4} & =\frac{8}{9} & =0
\end{array}$
$\begin{array}{ll}
(d)\ f(-x) & (e)\ -f(x)\\
=1-\frac{1}{(-x+2)^{2}} & =-(1-\frac{1}{(x+2)^{2}})\\
=1-\frac{1}{(x-2)^{2}}\quad & =\frac{1}{(x+2)^{2}}-1\\
&
\end{array}$
$\begin{array}{ll}
(f)\ f(x+1) & (g)\ f(2x)\\
=1-\frac{1}{(x+1+2)^{2}} & =1-\frac{1}{(2x+2)^{2}}\\
=1-\frac{1}{(x+3)^{2}} & =1-\frac{1}{(2(x+1))^{2}}\\
& =1-\frac{1}{4(x+1)^{2}}\\
& \\
&
\end{array}$
$(h)\displaystyle \ f(x+h)=1-\frac{1}{(x+h+2)^{2}}$
