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