Answer
a) $f(0)=-\frac{1}{4}$
(b). $f(1)=0$
(c). $f(-1)=0$
(d) $f(-x)=\frac{x^2-1}{4-x}$
(e) $-f(x)=-\frac{x^2-1}{x+4}$ or $-f(x)=\frac{1-x^2}{x+4}$
(f) $f(x+1)=\frac{x^2+2x}{x+5}$
(g) $f(2x)=\frac{4x^2-1}{2x+4}$
(h) $f(x+h)=\frac{x^2+2xh+h^2-1}{x+h+4}$
Work Step by Step
Given $f(x)=\frac{x^2-1}{x+4}$
a) To Evaluate $f(0)$ in the given function, substitute $0$ in the place of $x$.
$f(0)=\frac{0^2-1}{0+4}=\frac{-1}{4}=-\frac{1}{4}$
(b). To Evaluate $f(1)$ in the given function, substitute $1$ in the place of $x$.
$f(1)=\frac{1^2-1}{1+4}=\frac{1-1}{1+4}=-\frac{0}{5}=0$
(c). To Evaluate $f(−1)$ in the given function, substitute $−1$ in the place of $x$.
$f(-1)=\frac{(-1)^2-1}{-1+4}=\frac{1-1}{-1+4}=\frac{0}{3}=0$
(d) To Evaluate $f(−x)$ in the given function, substitute $−x$ in the place of $x$.
$f(-x)= \frac{(-x)^2-1}{-x+4}=\frac{x^2-1}{4-x}$
(e) To Evaluate $−f(x)$ in the given function,
$-f(x)=-\frac{x^2-1}{x+4}$
This can also be written as $-f(x)=\frac{1-x^2}{x+4}$
(f) To Evaluate $f(x+1)$ in the given function, substitute $x+1$ in the place of $x$.
$f(x+1)=\frac{(x+1)^2-1}{(x+1)+4}=\frac{x^2+2x+1-1}{x+1+4}=\frac{x^2+2x}{x+5}$
(g) To Evaluate $f(2x)$ in the given function, substitute $2x$ in the place of $x$.
$f(2x)=\frac{(2x)^2-1}{2x+4}=\frac{4x^2-1}{2x+4}$
(h) To Evaluate $f(x+h)$ in the given function, substitute $x+h$ in the place of $x$.
$f(x+h)=\frac{(x+h)^2-1}{(x+h)+4}=\frac{x^2+2xh+h^2-1}{x+h+4}$
